IET CIRCUITS, DEVICES AND SYSTEMS SERIES 23Understandable Electric Circuits Other volumes in this series: Volume 2 Analogue IC design: the current-mode approach C. Toumazou, F.J. Lidgey and D.G. Haigh (Editors) Volume 3 Analogue-digital ASICs: circuit techniques, design tools and applications R.S. Soin, F. Maloberti and J. France (Editors) Volume 4 Algorithmic and knowledge-based CAD for VLSI G.E. Taylor and G. Russell (Editors) Volume 5 Switched currents: an analogue technique for digital technology C. Toumazou, J.B.C. Hughes and N.C. Battersby (Editors) Volume 6 High-frequency circuit engineering F. Nibler et al. Volume 8 Low-power high-frequency microelectronics: a unified approach G. Machado (Editor) Volume 9 VLSI testing: digital and mixed analogue/digital techniques S.L. Hurst Volume 10 Distributed feedback semiconductor lasers J.E. Carroll, J.E.A. Whiteaway and R.G.S. Plumb Volume 11 Selected topics in advanced solid state and fibre optic sensors S.M. Vaezi-Nejad (Editor) Volume 12 Strained silicon heterostructures: materials and devices C.K. Maiti, N.B. Chakrabarti and S.K. Ray Volume 13 RFIC and MMIC design and technology I.D. Robertson and S. Lucyzyn (Editors) Volume 14 Design of high frequency integrated analogue filters Y. Sun (Editor) Volume 15 Foundations of digital signal processing: theory, algorithms and hardware design P. Gaydecki Volume 16 Wireless communications circuits and systems Y. Sun (Editor) Volume 17 The switching function: analysis of power electronic circuits C. Marouchos Volume 18 System on chip: next generation electronics B. Al-Hashimi (Editor) Volume 19 Test and diagnosis of analogue, mixed-signal and RF integrated circuits: the system on chip approach Y. Sun (Editor) Volume 20 Low power and low voltage circuit design with the FGMOS transistor E. Rodriguez-Villegas Volume 21 Technology computer aided design for Si, SiGe and GaAs integrated circuits C.K. Maiti and G.A. Armstrong Volume 22 Nanotechnologies M. Wautelet et al. Understandable Electric Circuits Meizhong Wang The Institution of Engineering and Technology Published by The Institution of Engineering and Technology, London, United Kingdom First edition † 2005 Higher Education Press, China English translation † 2010 The Institution of Engineering and Technology First published 2005 Reprinted 2009 English translation 2010 This publication is copyright under the Berne Convention and the Universal Copyright Convention. All rights reserved. Apart from any fair dealing for the purposes of research or private study, or criticism or review, as permitted under the Copyright, Designs and Patents Act 1988, this publication may be reproduced, stored or transmitted, in any form or by any means, only with the prior permission in writing of the publishers, or in the case of reprographic reproduction in accordance with the terms of licences issued by the Copyright Licensing Agency. Enquiries concerning reproduction outside those terms should be sent to the publisher at the undermentioned address: The Institution of Engineering and Technology Michael Faraday House Six Hills Way, Stevenage Herts, SG1 2AY, United Kingdom www.theiet.org While the author and publisher believe that the information and guidance given in this work are correct, all parties must rely upon their own skill and judgement when making use of them. Neither the author nor publisher assumes any liability to anyone for any loss or damage caused by any error or omission in the work, whether such an error or omission is the result of negligence or any other cause. Any and all such liability is disclaimed. The moral rights of the author to be identified as author of this work have been asserted by him in accordance with the Copyright, Designs and Patents Act 1988. British Library Cataloguing in Publication Data A catalogue record for this product is available from the British Library ISBN 978-0-86341-952-2 (paperback) ISBN 978-1-84919-114-2 (PDF) Typeset in India by MPS Ltd, A Macmillan Company Printed in the UK by CPI Antony Rowe, Chippenham Contents Preface xiii 1 Basic concepts of electric circuits 1 Objectives 1 1.1 Introduction 1 1.1.1 Why study electric circuits? 1 1.1.2 Careers in electrical, electronic and computer engineering 2 1.1.3 Milestones of electric circuit theory 3 1.2 Electric circuits and schematic diagrams 4 1.2.1 Basic electric circuits 4 1.2.2 Circuit schematics (diagrams) and symbols 5 1.3 Electric current 8 1.3.1 Current 8 1.3.2 Ammeter 9 1.3.3 The direction of electric current 9 1.4 Electric voltage 10 1.4.1 Voltage/electromotive force 10 1.4.2 Potential difference/voltage 11 1.4.3 Voltmeter 13 1.5 Resistance and Ohm’s law 14 1.5.1 Resistor 14 1.5.2 Factors affecting resistance 15 1.5.3 Ohmmeter 16 1.5.4 Conductance 17 1.5.5 Ohm’s law 17 1.5.6 Memory aid for Ohm’s law 18 1.5.7 The experimental circuit of Ohm’s law 19 1.5.8 I–V characteristic of Ohm’s law 19 1.5.9 Conductance form of Ohm’s law 20 1.6 Reference direction of voltage and current 20 1.6.1 Reference direction of current 20 1.6.2 Reference polarity of voltage 21 1.6.3 Mutually related reference polarity of current/voltage 22 Summary 23 Experiment 1: Resistor colour code 25 2 Basic laws of electric circuits 31 Objectives 31 2.1 Power and Energy 31 2.1.1 Work 31 2.1.2 Energy 32 2.1.3 Power 32 2.1.4 The reference direction of power 34 2.2 Kirchhoff’s voltage law (KVL) 36 2.2.1 Closed-loop circuit 36 2.2.2 Kirchhoff’s voltage law (KVL) 36 2.2.3 KVL #2 38 2.2.4 Experimental circuit of KVL 38 2.2.5 KVL extension 40 2.2.6 The physical property of KVL 41 2.3 Kirchhoff’s current law (KCL) 41 2.3.1 KCL #1 41 2.3.2 KCL #2 41 2.3.3 Physical property of KCL 44 2.3.4 Procedure to solve a complicated problem 44 2.3.5 Supernode 46 2.3.6 Several important circuit terminologies 47 2.4 Voltage source and current source 47 2.4.1 Voltage source 48 2.4.1.1 Ideal voltage source 48 2.4.1.2 Real voltage source 48 2.4.2 Current source 50 2.4.2.1 Ideal current source 50 2.4.2.2 Real current source 52 2.5 International units for circuit quantities 53 2.5.1 International system of units (SI) 53 2.5.2 Metric prefixes (SI prefixes) 54 Summary 55 Experiment 2: KVL and KCL 57 3 Series–parallel resistive circuits 63 Objectives 63 3.1 Series resistive circuits and voltage-divider rule 63 3.1.1 Series resistive circuits 63 3.1.1.1 Total series voltage 65 3.1.1.2 Total series resistance (or equivalent resistance) 65 3.1.1.3 Series current 66 3.1.1.4 Series power 66 3.1.2 Voltage-divider rule (VDR) 67 3.1.3 Circuit ground 70 vi Understandable electric circuits 3.2 Parallel resistive circuits and the current-divider rule 71 3.2.1 Parallel resistive circuits 71 3.2.1.1 Parallel voltage 73 3.2.1.2 Parallel current 73 3.2.1.3 Equivalent parallel resistance 74 3.2.1.4 Total parallel power 75 3.2.2 Current-divider rule (CDR) 76 3.3 Series–parallel resistive circuits 79 3.3.1 Equivalent resistance 80 3.3.2 Method for analysing series–parallel circuits 81 3.4 Wye (Y) and delta (D) configurations and their equivalent conversions 83 3.4.1 Wye and delta configurations 83 3.4.2 Delta to wye conversion (D!Y) 84 3.4.3 Wye to delta conversion (Y!D) 86 3.4.3.1 R Y and R D 87 3.4.4 Using D!Y conversion to simplify bridge circuits 89 3.4.5 Balanced bridge 90 3.4.6 Measure unknown resistors using the balanced bridge 91 Summary 92 Experiment 3: Series–parallel resistive circuits 95 4 Methods of DC circuit analysis 101 Objectives 101 4.1 Voltage source, current source and their equivalent conversions 101 4.1.1 Source equivalent conversion 101 4.1.2 Sources in series and parallel 104 4.1.2.1 Voltage sources in series 104 4.1.2.2 Voltage sources in parallel 105 4.1.2.3 Current sources in parallel 106 4.1.2.4 Current sources in series 107 4.2 Branch current analysis 108 4.2.1 Procedure for applying the branch circuit analysis 109 4.3 Mesh current analysis 113 4.3.1 Procedure for applying mesh current analysis 114 4.4 Nodal voltage analysis 116 4.4.1 Procedure for applying the node voltage analysis 117 4.5 Node voltage analysis vs. mesh current analysis 121 Summary 122 Experiment 4: Mesh current analysis and nodal voltage analysis 123 5 The network theorems 127 Objectives 127 5.1 Superposition theorem 128 5.1.1 Introduction 128 5.1.2 Steps to apply the superposition theorem 128 Contents vii 5.2 Thevenin’s and Norton’s theorems 133 5.2.1 Introduction 133 5.2.2 Steps to apply Thevenin’s and Norton’s theorems 135 5.2.3 Viewpoints of the theorems 139 5.3 Maximum power transfer 147 5.4 Millman’s and substitution theorems 151 5.4.1 Millman’s theorem 151 5.4.2 Substitution theorem 152 Summary 155 Experiment 5A: Superposition theorem 156 Experiment 5B: Thevenin’s and Norton’s theorems 158 6 Capacitors and inductors 163 Objectives 163 6.1 Capacitor 164 6.1.1 The construction of a capacitor 164 6.1.2 Charging a capacitor 165 6.1.3 Energy storage element 166 6.1.4 Discharging a capacitor 166 6.1.5 Capacitance 167 6.1.6 Factors affecting capacitance 169 6.1.7 Leakage current 170 6.1.8 Breakdown voltage 170 6.1.9 Relationship between the current and voltage of a capacitor 171 6.1.10 Energy stored by a capacitor 173 6.2 Capacitors in series and parallel 174 6.2.1 Capacitors in series 174 6.2.2 Capacitors in parallel 176 6.2.3 Capacitors in series–parallel 178 6.3 Inductor 179 6.3.1 Electromagnetism induction 179 6.3.1.1 Electromagnetic field 179 6.3.1.2 Faraday’s law 180 6.3.1.3 Lenz’s law 181 6.3.2 Inductor 182 6.3.3 Self-inductance 182 6.3.4 Relationship between inductor voltage and current 183 6.3.5 Factors affecting inductance 184 6.3.6 The energy stored by an inductor 185 6.3.7 Winding resistor of an inductor 186 6.4 Inductors in series and parallel 188 6.4.1 Inductors in series 188 6.4.2 Inductors in parallel 188 6.4.3 Inductors in series–parallel 189 viii Understandable electric circuits Summary 190 Experiment 6: Capacitors 191 7 Transient analysis of circuits 195 Objectives 195 7.1 The transient response 195 7.1.1 The first-order circuit and its transient response 195 7.1.2 Circuit responses 196 7.1.3 The initial condition of the dynamic circuit 198 7.2 The step response of an RC circuit 199 7.2.1 The charging process of an RC circuit 199 7.2.2 Quantity analysis for the charging process of the RC circuit 201 7.3 The source-free response of the RC circuit 204 7.3.1 The discharging process of the RC circuit 204 7.3.2 Quantity analysis of the RC discharging process 205 7.3.3 RC time constant t 208 7.3.4 The RC time constant and charging/discharging 209 7.4 The step response of an RL circuit 211 7.4.1 Energy storing process of the RL circuit 212 7.4.2 Quantitative analysis of the energy storing process in an RL circuit 213 7.5 Source-free response of an RL circuit 215 7.5.1 Energy releasing process of an RL circuit 215 7.5.2 Quantity analysis of the energy release process of an RL circuit 216 7.5.3 RL time constant t 218 7.5.4 The RL time constant and the energy storing and releasing 219 Summary 220 Experiment 7: The first-order circuit (RC circuit) 221 8 Fundamentals of AC circuits 227 Objectives 227 8.1 Introduction to alternating current (AC) 227 8.1.1 The difference between DC and AC 227 8.1.2 DC and AC waveforms 228 8.1.3 Period and frequency 229 8.1.4 Three important components of a sine function 230 8.1.5 Phase difference of the sine function 232 8.2 Sinusoidal AC quantity 235 8.2.1 Peak and peak–peak value 235 8.2.2 Instantaneous value 236 8.2.3 Average value 236 8.2.4 Root mean square (RMS) value 237 Contents ix 8.3 Phasors 239 8.3.1 Introduction to phasor notation 239 8.3.2 Complex numbers review 240 8.3.3 Phasor 242 8.3.4 Phasor diagram 243 8.3.5 Rotating factor 244 8.3.6 Differentiation and integration of the phasor 246 8.4 Resistors, inductors and capacitors in sinusoidal AC circuits 248 8.4.1 Resistor’s AC response 248 8.4.2 Inductor’s AC response 250 8.4.3 Capacitor’s AC response 254 Summary 257 Experiment 8: Measuring DC and AC voltages using the oscilloscope 260 9 Methods of AC circuit analysis 265 Objectives 265 9.1 Impedance and admittance 265 9.1.1 Impedance 265 9.1.2 Admittance 266 9.1.3 Characteristics of the impedance 267 9.1.4 Characteristics of the admittance 269 9.2 Impedance in series and parallel 272 9.2.1 Impedance of series and parallel circuits 272 9.2.2 Voltage divider and current divider rules 273 9.2.3 The phasor forms of KVL and KCL 274 9.3 Power in AC circuits 276 9.3.1 Instantaneous power p 276 9.3.2 Active power P (or average power) 279 9.3.3 Reactive power Q 281 9.3.4 Apparent power S 282 9.3.5 Power triangle 284 9.3.6 Power factor (PF) 285 9.3.7 Total power 287 9.4 Methods of analysing AC circuits 290 9.4.1 Mesh current analysis 291 9.4.2 Node voltage analysis 292 9.4.3 Superposition theorem 293 9.4.4 Thevenin’s and Norton’s theorems 296 Summary 299 Experiment 9: Sinusoidal AC circuits 302 x Understandable electric circuits 10 RLC circuits and resonance 307 Objectives 307 10.1 Series resonance 307 10.1.1 Introduction 307 10.1.2 Frequency of series resonance 308 10.1.3 Impedance of series resonance 309 10.1.4 Current of series resonance 309 10.1.5 Phasor diagram of series resonance 310 10.1.6 Response curves of X L , X C and Z versus f 310 10.1.7 Phase response of series resonance 311 10.1.8 Quality factor 312 10.1.9 Voltage of series resonant 313 10.2 Bandwidth and selectivity 315 10.2.1 The bandwidth of series resonance 315 10.2.2 The selectivity of series resonance 316 10.2.3 The quality factor and selectivity 317 10.2.3.1 Series resonance summary 319 10.3 Parallel resonance 319 10.3.1 Introduction 319 10.3.2 Frequency of parallel resonance 320 10.3.3 Admittance of parallel resonance 320 10.3.4 Current of parallel resonance 321 10.3.5 Phasor diagram of parallel resonance 322 10.3.6 Quality factor 322 10.3.7 Current of parallel resonance 323 10.3.8 Bandwidth of parallel resonance 324 10.3.8.1 Parallel resonance summary 324 10.4 The practical parallel resonant circuit 325 10.4.1 Resonant admittance 325 10.4.2 Resonant frequency 326 10.4.3 Applications of the resonance 327 Summary 328 Experiment 10: Series resonant circuit 329 11 Mutual inductance and transformers 333 Objectives 333 11.1 Mutual inductance 333 11.1.1 Mutual inductance and coefficient of coupling 333 11.1.2 Dot convention 335 11.2 Basic transformer 336 11.2.1 Transformer 336 11.2.2 Air-core transformer 337 11.2.3 Iron-core transformer 337 11.2.4 Ideal transformer 338 Contents xi 11.3 Step-up and step-down transformers 340 11.3.1 Step-up transformer 340 11.3.2 Step-down transformer 341 11.3.3 Applications of step-up and step-down transformers 342 11.3.4 Other types of transformers 343 11.4 Impedance matching 344 11.4.1 Maximum power transfer 344 11.4.2 Impedance matching 345 Summary 346 Experiment 11: Transformer 347 12 Circuits with dependent sources 351 Objectives 351 12.1 Dependent sources 352 12.1.1 Dependent (or controlled) sources 352 12.1.2 Equivalent conversion of dependent sources 353 12.2 Analysing circuits with dependent sources 355 Summary 360 Appendix A: Greek alphabet 363 Appendix B: Differentiation of the phasor 364 Bibliography 365 Index 367 xii Understandable electric circuits Preface The book Understandable Electric Circuits is based on my teaching notes for the circuit analysis course that I have taught for many years at Canadian and Chinese institutions. The English version of this book continues in the spirit of its successful Chinese version, which was published by the ‘Higher Education Press’, the largest and the most prominent publisher of educational books in China, in 2005 and reprinted in 2009. This unique and well-structured book provides understandable and effective introduction to the fundamentals of DC/ACcircuits, including current, voltage, power, resistor, capacitor, inductor, impedance, admittance, dependent/independent sources, basic circuit laws/rules (Ohm’s law, KVL/KCL, voltage/current divider rules), series/ parallel and wye/delta circuits, methods of DC/AC analysis (branch current and mesh/node analysis), the network theorems (superposition, Thevenin’s/Norton’s theorems, maximum power transfer, Millman’s and substitution theorems), transient analysis, RLC circuits and resonance, mutual inductance and transformers and more. Key features As an aid to readers, the book provides some noteworthy features: ● Clear and easy-to-understand written style, procedures and examples. ● Outlining (boxing) of all important principles, concepts, laws/rules and for- mulas to emphasize and locate important facts and points. ● Objectives at the beginning of each chapter to highlight to readers the knowl- edge that is expected to be obtained in the chapter. ● Summary at the end of each chapter to emphasize the key points and formulas in the chapter, which is convenient for students reviewing before exams. ● Laboratory experiments at the end of each chapter are convenient for hands-on practice. They also include how to use basic electrical instruments such as the multimeter and oscilloscope. ● Tables organizing and summarizing variables, values and formulas, which clearly present the important information. Suitable readers This book is intended for college and university students, technicians, technologists, engineers or any other professionals who require a solid foundation in the basics of electric circuits. It targets an audience from all sectors in the fields of electrical, electronic and computer engineering such as electrical, electronics, computers, communications, control and automation, embedded systems, signal processing, power electronics, industrial instrumentation, power systems (including renewable energy), electrical apparatus and machines, nanotechnology, biomedical imaging and more. It is also suitable for non-electrical or electronics readers. It provides readers with the necessary foundation for DC/AC circuits in related fields. To make this book more reader friendly, the concepts, new terms, laws/rules and theorems are explained in an easy-to-understand style. Clear step-by-step procedures for applying methods of DC/AC analysis and network theorems make this book easy for readers to learn electric circuits themselves. Acknowledgements Special thanks to Lisa Reading, the commissioning editor for books at the Institu- tion of Engineering and Technology. I really appreciate her belief in my ability to write this book, and her help and support in publishing it. I also appreciate the support from Bianca Campbell, books and journals sales manager, Suzanne Bishop, marketing manager, Felicity Hull, marketing executive, and Jo Hughes, production controller. In addition, I would like to express my sincere gratitude to Ramya Srinivasan (project manager of my production process from MPS Ltd) for her highly efficient work and good guidance/suggestions that have helped to refine the writing of this book. I would also like to express my gratitude to Ying Nan, an electrical and computer engineer, for taking the time to edit some chapters of this book. In addition, it is my good fortune to have help and support from my family members: My husband, Li Wang (an electronics and physics instructor), for translating several chapters of this book from Chinese to English even though he is very weak after having several operations. My daughter Alice Wang (a busy PhD student), who deserves a special acknowledgement for her patience and dedication to editing some chapters and all the experiments in the book, and also proofreading the entire book. And finally, my son Evan Wang has given a hand in editing several chapters. They deserve sincere acknowledgement for their time and energy. My special thanks to all of them. xiv Understandable electric circuits Chapter 1 Basic concepts of electric circuits Objectives After completing this chapter, you will be able to: ● understand the purpose of studying electric circuits ● know the requirements of a basic electric circuit ● become familiar with circuit symbols ● become familiar with the schematics of electric circuits ● understand the concepts of current and voltage ● understand resistance and its characteristics ● become familiar with the ammeter, voltmeter and ohmmeter ● know the difference between the electron flow and the conventional current flow ● know the concept of reference directions of voltage and current ● know how to apply Ohm’s law 1.1 Introduction 1.1.1 Why study electric circuits? Electrical energy is the great driving force and the supporting pillar for modern industry and civilization. Our everyday life would be unthinkable without electricity or the use of electronic products. Any complex electrical and electronic device or control system is founded from the basic theory of electric circuits. Only when you have grasped and understood the basic concepts and principles of electric circuits can you fur- ther study electrical, electronic and computer engineering and other related areas. When you start reading this book, perhaps you have already chosen the electrical or the electronic fields as your professional goal – a wise choice! Electrical, electronic and computer engineering has made and continues to make incredible contributions to most aspects of human society – a truth that cannot be neglected. Moreover, it may have a bigger impact on human civilization in the future. Therefore, experts forecast that demand for profes- sionals in this field will grow continuously. This is good news for people who have chosen these areas of study. Reading this book or other electric circuit book is a first step into the electrical, electronic and computer world that will introduce you to the foun- dation of the professions in these areas. 1.1.2 Careers in electrical, electronic and computer engineering Nowadays, electrical, electronic and computer technology is developing so rapidly that many career options exist for those who have chosen this field. As long as you have gained a solid foundation in electric circuits and electronics, the training that most employers provide in their branches will lead you into a brand new professional career very quickly. There are many types of jobs for electrical and electronic engineering technology. Only a partial list is as follows: ● Electrical engineer ● Electronics engineer ● Electrical design engineer ● Control and automation engineer ● Process and system engineer ● Instrument engineer ● Robotics engineer ● Product engineer ● Field engineer ● Reliability engineer ● Integrated circuits (IC) design engineer ● Computer engineer ● Power electronics engineer ● Electrical and electronics engineering professor/lecturer ● Designer and technologist ● Biomedical engineering technologist ● Electrical and electronics technician ● Hydro technician ● Electrician ● Equipment maintenance technician ● Electronic test technician ● Calibration/lab technician ● Technical writer for electronic products ● Electronic repair Electrical and electronic technicians, technologists, engineers and experts will be in demand in the future, so you definitely do not want to miss this good opportunity. 2 Understandable electric circuits 1.1.3 Milestones of electric circuit theory Many early scientists have made great contributions in developing the theorems of electrical circuits. The laws and physical quantities that they discovered are named after them, and all are important milestones in the field of electric engineering. We list here only the ones that are described in this book. ● Coulomb is the unit of electric charge; it was named in the honour of Charles Augustin de Coulomb (1736–1806), a French physicist. Coulomb developed Coulomb’s law, which is the definition of the electrostatic force of attraction and repulsion, and the principle of charge interactions (attraction or repulsion of positive and negative electric charges). ● Faraday is the unit of capacitance; it was named in the honour of Michael Faraday (1791–1867), an English physicist and chemist. He discovered that relative motion of the magnetic field and conductor can produce electric current, which we know today as the Faraday’s law of electromagnetic induction. Faraday also discovered that the electric current originates from the chemical reaction that occurs between two metallic conductors. ● Ampere is the unit of electric current; it was named in the honour of Andre´ - Marie Ampe` re (1775–1836), a French physicist. He was one of the main discoverers of electromagnetism and is best known for defining a method to measure the flow of current. ● Ohm is the unit of resistance; it was named in the honour of Georg Simon Ohm (1789–1854), a German physicist. He established the relationship between voltage, current and resistance, and formulated the most famous electric circuit law – Ohm’s law. ● Volt is the unit of voltage; it was named in the honour of Alessandro Volta (1745–1827), an Italian physicist. He constructed the first electric battery that could produce a reliable, steady current. ● Watt is the unit of power; it was named in the honour of James Watt (1736–1819), a Scottish engineer and inventor. He made great improve- ments in the steam engine and made important contributions in the area of magnetic fields. ● Lenz’s law was named in the honour of Heinrich Friedrich Emil Lenz (1804– 1865), a Baltic German physicist. He discovered that the polarity of the induced current that is produced in the conductor of the magnetic field always resists the change of its induced voltage; this is known as Lenz’s law. ● Maxwell is the unit of magnetic flux; it was named in the honour of James Clerk Maxwell (1831–1879), a Scottish physicist and mathematician. The German physicist Wilhelm Eduard Weber (1804–1891) shares the honour with Maxwell (1 Wb ¼ 10 8 Mx). Maxwell had established the Maxwell’s equations that represent perfect ways to state the fundamentals of elec- tricity and magnetism. ● Hertz is the unit of frequency; it was named in the honour of Heinrich Rudolf Hertz (1857–1894), a German physicist and mathematician. He Basic concepts of electric circuits 3 was the first person to broadcast and receive radio waves. Through the low-frequency microwave experiment, Hertz confirmed Maxwell’s elec- tromagnetic theory. ● Henry is the unit of inductance; it was named in the honour of Joseph Henry (1797–1878), a Scottish-American scientist. He discovered self- induction and mutual inductance. ● Joule is the unit of energy; it was named in the honour of James Prescott Joule (1818–1889), an English physicist. He made great contributions in discovering the law of the conservation of energy. This law states that energy may transform from one form into another, but is never lost. Joule’s law was named after him and states that heat will be produced in an electrical conductor. The majority of the laws and units of measurement stated above will be used in the later chapters of this book. Being familiar with them will be bene- ficial for further study of electric circuits. 1.2 Electric circuits and schematic diagrams 1.2.1 Basic electric circuits An electric circuit is a closed loop of pathway with electric charges flowing through it. More specifically, an electric circuit can be defined as a sum of all electric components in the closed loop of pathway with flowing electric charges, such as an electric circuit that includes resistors, capacitors, inductors, power sources, switches, wires, etc. (these electric components will be explained later). Electric circuit A closed loop of pathway with electric charges or current flowing through it. A basic electric circuit contains three components: the power supply, the load and the wires (conductors) (Figure 1.1). Wires connect the power Power Supply Load Wire Wire Figure 1.1 Requirements of a basic circuit 4 Understandable electric circuits supply and the load, and carry electric charges through the circuit. A power supply is a device that supplies electrical energy to the load of the circuit; it can convert other forms of energy to electrical energy. The electric battery and generator are examples of power supply. For example: ● the battery converts chemical energy into electrical energy. ● the hydroelectric generator converts hydroenergy (the energy of moving water) into electrical energy. ● the thermo generator converts heat energy into electrical energy. ● the nuclear power generator converts nuclear energy into electrical energy. ● the wind generator converts wind energy into electrical energy. ● the solar generator converts solar energy into electrical energy. Load is a device that is usually connected to the output terminal of an electric circuit. It consumes or absorbs electrical energy from the source. The load may be any device that can receive electrical energy and convert it into other forms of energy. For example: ● electric lamp converts electrical energy into light energy ● electric stove converts electrical energy into heat energy ● electric motor converts electrical energy into mechanical energy ● electric fan converts electrical energy into wind energy ● speaker converts electrical energy into sound energy Therefore, light bulb, electric stove, electric motor, electric fan and speaker are all electric loads. Requirements of a basic circuit ● Power supply (power source) is a device that supplies electrical energy to a load; it can convert the other energy forms into electrical energy. ● Load is a device that is connected to the output terminal of an elec- tric circuit, and consumes electrical energy. ● Wires connect the components in a circuit together, and carry electric charges through the circuit. Figure 1.2 is an example of a simple electric circuit – a flashlight (or electric torch) circuit. In this circuit the battery is the power supply and the small light bulb is the load, and they are connected together by wires. 1.2.2 Circuit schematics (diagrams) and symbols Studying electric circuits usually requires drawing or recognizing circuit diagrams. Circuit diagrams can make electric circuits easier to understand, Basic concepts of electric circuits 5 analyse and calculate. It is not very difficult to draw a realistic pictorial representation of the flashlight circuit as shown in Figure 1.2, but when studying more theories of electric circuits, circuits can be more and more complex and drawing the pictorial representation of the circuits will not be very realistic. The more common electric circuits are usually represented by schematics. A schematic is a simplified circuit diagram that shows the interconnection of circuit components. It uses standard graphic circuit symbols according to the layout of the actual circuit connection. This is a way to draw circuit diagrams far more quickly and easily. The circuit symbols are the idealization and approximation of the actual circuit components. For example, both the battery and the direct current (DC) generator can convert other energy forms into electrical energy and produce DC voltage. Therefore, they are represented by the same circuit symbol – the DC power supply E. The electric lamp, electric stove, electric motor and other loads can be represented by a circuit symbol – the resistor R, since all of them have the same characteristic of converting electrical energy into other forms of energies and consuming electrical energy. The different circuit components are represented by different circuit sym- bols. Table 1.1 lists some commonly used electric circuit symbols in this book. The most commonly used circuit symbols are the resistor, capacitor, inductor, power supply, ground, switch, etc. Schematics are represented by circuit symbols according to the layout of the actual circuit connection. The schematic of the flashlight circuit (Figure 1.2) is shown in Figure 1.3. Further study of this book will help you understand all the circuit elements in Table 1.1. (Switch) Figure 1.2 The flashlight circuit E E 6 Understandable electric circuits Table 1.1 The commonly used circuit schematic symbols Component Circuit symbol DC power supply AC power supply Current source Lamp Connected wires Unconnected wires Fixed resistor Variable resistor or Capacitor Inductor Switch Speaker Ground Fuse Ohmmeter Ω Ammeter A Voltmeter V Transformer Figure 1.3 Schematic of the flashlight circuit Basic concepts of electric circuits 7 1.3 Electric current There are several key circuit quantities in electric circuit theory: electric cur- rent, voltage, power, etc. These circuit quantities are very important to study in electric circuits, and they will be used throughout this book. This section will discuss one of them – the electric current. 1.3.1 Current Although we cannot see electric charges or electric current in the electric cir- cuits, they are analogous to the flow of water in a water hose or pipe. Water current is a flow of water through a water circuit (faucet, pipe or hose, etc.); electric current is a flow of electric charges through an electric circuit (wires, power supply and load). Water is measured in litres or gallons, so you can measure the amount of water that flows out of the tap at certain time intervals, i.e. litres or gallons per minute or hour. Electric current is measured by the amount of electric charges that flows past a given point at a certain time interval in an electric circuit. If Q represents the amount of charges that is moving past a point at time t, then the current I is: Current ¼ Charge Time or I ¼ Q t If you have learned calculus, current also can be expressed by the derivative: i ¼ dq=dt. Electric current I ● Current is the flow of electric charges through an electric circuit. ● Current I is measured by the amount of charges Q that flows past a given point at a certain time t: I ¼ Q/t Note: Italic letters have been used to represent the quantity symbols and non-italic letters to represent unit symbols. Quantity Quantity symbol Unit Unit symbol Charge Q Coulombs C Time t Seconds s Current I Amperes A 8 Understandable electric circuits A current of 1 A means that there is 1 C of electric charge passing through a given cross-sectional area of wire in 1 s: 1 A ¼ 1 C 1 s More precisely, 1 A of current actually means there are about 6.25 610 18 charges passing through a given cross-sectional area of wire in 1 s, since 1 C is approximately equal to 6.25 6 10 18 charges (1 C % 6.25 6 10 18 charges), as shown in Figure 1.4. Example 1.1: If a charge of 100 C passes through a given cross-sectional area of wire in 50 s, what is the current? Solution: Since Q ¼ 100 C and t ¼ 50 s I ¼ Q t ¼ 100 C 50 s ¼ 2 A 1.3.2 Ammeter Ammeter is an instrument that can be used to measure current, and its symbol is A . It must be connected in series with the circuit to measure current, as shown in Figure 1.5. 1.3.3 The direction of electric current When early scientists started to work with electricity, the structure of atoms was not very clear, and they assumed at that time the current was a flow of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.25ϫ10 18 There are 6.25 ؋ 10 18 charges passing through this given cross-sectional area in 1 s Figure 1.4 1 A of current A ϩ Ϫ Figure 1.5 Measuring current with an ammeter Basic concepts of electric circuits 9 positive charges (protons) from the positive terminal of a power supply (such as a battery) to its negative terminal. Which way does electric charge really flow? Later on, scientists discovered that electric current is in fact a flow of negative charges (electrons) from the negative terminal of a power supply to its positive terminal. But by the time the real direction of current flow was discovered, a flow of positive charges (pro- tons) from the positive terminal of a power supply to its negative terminal had already been well established and used commonly in electrical circuitry. Currently, there are two methods to express the direction of electric cur- rent. One is known as the conventional current flow version, in which the cur- rent is defined as a flow of positive charges (protons) from the positive terminal of a power supply to its negative terminal. The other is called electrons flow version, in which the current is defined as a flow of negative charges (electrons) from the negative terminal of a power supply unit to its positive terminal. These two methods are shown in Figure 1.6. Because the charge or current cannot be seen in electric circuits, it will make no difference as to which method is used, and it will not affect the ana- lysis, design, calculation, measurement and applications of the electric circuits as long as one method is used consistently. In this book, the conventional current flow version is used. Conventional and electron current flow version ● Conventional current flow is defined as a flow of positive charges (protons) from the positive terminal of a power supply to its negative terminal. ● Electron flowversion is defined as a flowof negative charges (electrons) from the negative terminal of a power supply to its positive terminal. 1.4 Electric voltage 1.4.1 Voltage/electromotive force We have analysed the flow of water in the water circuit to the flow of electric current in the electric circuit. The concept of a water circuit can help develop an understanding of another important circuit quantity – voltage. I I (a) Conventional current flow (b) Electron flow Figure 1.6 The direction of electric current 10 Understandable electric circuits The concept of voltage works on the principle of a water gun. The trigger of a water gun is attached to a pump that squirts water out of a tiny hole at the muzzle. If there is no pressure from the gun (the trigger is not pressed), there will be no water out of the muzzle. Low-pressure squirting produces thin streams of water over a short distance, while high pressure produces a very powerful stream over a longer distance. Just as water pressure is required for a water gun or water circuit, electric pressure or voltage is required for an electric circuit. Voltage is responsible for the pushing and pulling of electrons or current through an electric circuit. The higher the voltage, the greater the current will be. Let us further analyse the voltage by using the previous flashlight or torch circuit in Figure 1.2. If only a small lamp is connected with wires without a battery in this circuit, the flashlight will not work. Since electric charges in the wire (conductor) randomly drift in different directions, a current cannot form in a specific direction. Once the battery is connected to the load (lamp) by wires, the positive electrode of the battery attracts the negative charges (electrons), and the negative electrode of the battery repels the electrons. This causes the electrons to flow in one direction and produce electric current in the circuit. The battery is one example of a voltage source that produces electromotive force (EMF) between its two terminals. When EMF is exerted on a circuit, it moves electrons around the circuit or causes current to flow through the circuit since EMF is actually ‘the electron-moving force’. It is the electric pressure or force that is supplied by a voltage source, which causes current to flow in a circuit. EMF produced by a voltage source is analogous to water pressure produced by a pump in a water circuit. Voltage is symbolized by V (italic letter), and its unit is volts (non-italic letter V). EMF is symbolized by E, and its unit is also volts (V). Electromotive force (EMF) EMF is an electric pressure or force that is supplied by a voltage source, which causes electric current to flow in a circuit. 1.4.2 Potential difference/voltage Assuming there are two water tanks A and B, water will flow from tank A to B only when tank A has a higher water level than tank B, as shown in Figure 1.7. A B Figure 1.7 Water-level difference Basic concepts of electric circuits 11 Common sense tells us that ‘water flows to the lower end’, so water will only flow when there is a water-level difference. It is the water-level difference that produces the potential energy for tank A, and work is done when water flows from tank A to B. This concept can also be used in the electric circuit. As water will flow between two places in a water circuit only when there is a water-level differ- ence, current will flow between two points in an electric circuit only when there is an electrical potential difference. For instance, if a light bulb is continuously kept on, i.e. to maintain con- tinuous movement of electrons in the circuit, the two terminals of the lamp need to have an electrical potential difference. This potential difference or voltage is produced by the EMF of the voltage source, and it is the amount of energy or work that would be required to move electrons between two points. Work is represented by W and measured in joules (J). The formula may be expressed as: Voltage ¼ Work Charge or V ¼ W Q If you have learned calculus, voltage can also be expressed by the derivative v ¼ dw/dq. Voltage V (or potential difference) V is the amount of energy or work required to move electrons between two points: V ¼ W/Q. For example, if 1 J of energy is used to move a 1 C charge from point a to b, it will have a 1 V potential difference or voltage across two points, as shown in Figure 1.8. W = 1J Q = 1C V = 1V a b Figure 1.8 Potential difference or voltage Quantity Quantity symbol Unit Unit symbol Voltage V Volt V Work (energy) W Joule J Charge Q Coulomb C 12 Understandable electric circuits Although voltage and potential difference are not exactly same, the two are used interchangeably. Current will flow between two points in a circuit only when there is a potential difference. The voltage or the potential difference always exists between two points. There are different names representing voltage or potential difference in electric circuits, such as the source voltage, applied voltage, load voltage, vol- tage drop, voltage rise, etc. What are the differences between them? The EMF can be called source voltage or applied voltage since it is supplied by a voltage source and applied to the load in a circuit. Voltage across the two terminals of the load is called the load voltage. Voltage across a component in a circuit is sometimes called voltage drop when current flows froma higher potential point to a lower potential point in the circuit, or voltage rise when current flows from a lower potential point to a higher potential point in the circuit. Source voltage or applied voltage (E or V S ): EMF can be called source voltage or applied voltage (the EMF is sup- plied by a voltage source and applied to the load in a circuit). Load voltage (V): Voltage across the two terminals of the load. Voltage drop: Voltage across a component when current flows from a higher potential point to a lower potential point. Voltage rise: Voltage across a component when current flows from a lower potential point to a higher potential point. 1.4.3 Voltmeter Voltmeter is an instrument that can be used to measure voltage. Its symbol is V . The voltmeter should be connected in parallel with the circuit component to measure voltage, as shown in Figure 1.9. V Figure 1.9 Measuring voltage with a voltmeter Basic concepts of electric circuits 13 1.5 Resistance and Ohm’s law 1.5.1 Resistor Let us use the water current as an example again to explain the resistor. What will happen when we throw some rocks into a small creek? The speed of the water current will slow down in the creek. This is because the rocks (water resistance) ‘resist’ the flow of water. A similar concept may also be used in an electric circuit. The resistor (current resistance) ‘resists’ the flow of electrical current. The higher the value of resistance, the smaller the current will be. The resistance of a conductor is a measure of how difficult it is to resist the current flow. As mentioned in section 1.2, the lamp, electric stove, motor and other such loads may be represented by resistor R because once this kind of load is con- nected to an electric circuit, it will consume electrical energy, cause resistance and reduce current in the circuit. Sometimes resistor R will need to be adjusted to a different level for dif- ferent applications. For example, the intensity of light of an adjustable lamp can be adjusted by using resistors. A resistor can also be used to maintain a safe current level in a circuit. A resistor is a two-terminal component of a circuit that is designed to resist or limit the flow of current. There are a variety of resistors with different resistance values for different applications. The resistor and resistance of a circuit have different meanings. A resistor is a component of a circuit. The resistance is a measure of a material’s opposition to the flow of current, and its unit is ohms (O). Resistor (R): A two-terminal component of a circuit that limits the flow of current. Resistance (R): The measure of a material’s opposition to the flow of current. Resistors are of many different types, materials, shapes and sizes, but all of them belong to one of the two categories, either fixed or variable. A fixed resistor has a ‘fixed’ resistance value and cannot be changed. A variable resis- tor has a resistance value that can be easily changed or adjusted manually or automatically. Symbols of the resistor ● Fixed resistor ● Variable resistor or 14 Understandable electric circuits 1.5.2 Factors affecting resistance There is no ‘perfect’ electrical conductor; every conductor that makes up the wires has some level of resistance no matter what kind of material it is made from. There are four main factors affecting the resistance in a conductor: the cross-sectional area of the wire (A), length of the conductor (‘), tempera- ture (T) and resistivity of the material (r) (Figure 1.10). ● Cross-sectional area of the wire A: More water will flow through a wider pipe than that through a narrow pipe. Similarly, the larger the diameter of the wire, the greater the cross-sectional area, the less the resistance in the wire and the more the flow of current. ● Length ‘: The longer the wire, the more the resistance and the more the time taken for the current to flow. ● Resistivity r: It is a measure for the opposition to flowing current through a material of wire, or how difficult it is for current to flow through a material. The different materials have different resistivity, i.e. more or less resistance in the materials. ● Temperature T: Resistivity of a material is dependent upon the tempera- ture surrounding the material. Resistivity increases with an increase in temperature for most materials. Table 1.2 lists resistivity of some materials at 20 8C. A ᐉ r Figure 1.10 Factors affecting resistance Table 1.2 Table of resistivities (r) Material Resistivity r (O Á m) Copper 1.68 610 78 Gold 2.44 610 78 Aluminium 2.82 610 78 Silver 1.59 610 78 Iron 1.0 6 10 77 Brass 0.8 6 10 77 Nichrome 1.1 6 10 76 Tin 1.09 610 77 Lead 2.2 6 10 77 Basic concepts of electric circuits 15 Factors affecting resistance can be mathematically expressed with the following formula: R ¼ r ‘ A Factors affecting resistance R ¼ r ‘ A À Á where A is the cross-sectional area, ‘ the length, T the tem- perature and r the resistivity (conducting ability of a material for a wire). Note: r is a Greek letter pronounced ‘rho’ (see Appendix A for a list of Greek letters). Example 1.3: There is a copper wire 50 m in length with a cross-sectional area of 0.13 cm 2 . What is the resistance of the wire? Solution: ‘ ¼ 50 m ¼ 5 000 cm; A ¼ 0.13 cm 2 ; r ¼ 1.68 610 78 O Á m ¼ 1.68 610 76 O Á cm (copper) R ¼ r ‘ A ¼ ð1:68 Â10 À6 O Á cmÞð5 000 cmÞ 0:13 cm 2 % 0:0646 O The resistance of this copper wire is 0.0646 O. Although there is resistance in the copper wire, it is very small. A 50-m-long wire only has 0.0646 O resistance; thus we can say that copper is a good conducting material. Copper and alu- minium are commonly used conducting materials with reasonable price and better conductivity. 1.5.3 Ohmmeter Ohmmeter is an instrument that can be used to measure resistance. Its symbol is Ω . The resistor must be removed from the circuit to measure resistance as shown in Figure 1.11. E R Ω Figure 1.11 Measuring resistance with an ohmmeter 16 Understandable electric circuits Voltmeter V : An instrument that is used to measure voltage; it should be connected in parallel with the component. Ammeter A : An instrument that is used to measure current; it should be connected in series in the circuit. Ohmmeter Ω : An instrument that is used to measure resistance, and the resistor must be removed from the circuit to measure the resistance. 1.5.4 Conductance Conductance (G) is a term that is opposite of the term resistance. It is the ability of a material to pass current rather than resist it, or how easy rather than how difficult it is for current to flow through a circuit. Conductance is the con- ductivity of the material; the less the resistance R of the material, the greater the conductance G, the better the conductivity of the material, and vice versa. The factors that affect resistance are the same for conductance, but in the opposite way. Mathematically, conductance is the reciprocal of resistance, i.e. G ¼ 1 R or G ¼ A r‘ ,R ¼ r ‘ A Increasing the cross-sectional area (A) of the wire or reducing the wire length (‘) can get better conductivity. This can be seen from the equation of conductance. It is often preferable and more convenient to use conductance in parallel cir- cuits. This will be discussed in later chapters. Conductance G G is the reciprocal of resistance: G ¼ 1/R The SI unit of conductance is the siemens (S). Some books use a unit mho ( O ) for conductance, which was derived from spelling ohm backwards and with an upside-down Greek letter omega O . Mho actually is the reciprocal of ohm, just as conductance G is the reciprocal of resistance R. Example 1.4: What is the conductance if the resistance R is 22 O? Solution: G ¼ 1/R ¼ 1/22 O % 0.0455 S or 0.0455 O 1.5.5 Ohm’s law Ohm’s law is a very important and useful equation in electric circuit theory. It precisely expresses the relationship between current, voltage and resistance Basic concepts of electric circuits 17 with a simple mathematical equation. Ohm’s law states that current through a conductor in a circuit is directly proportional to the voltage across it and inversely proportional to the resistance in it, i.e. I ¼ V R or I ¼ E R Any form of energy conversion from one type to another can be expressed as the following equation: Effect ¼ Cause Opposition In an electric circuit, it is the voltage that causes current to flow, so current flow is the result or effect of voltage, and resistance is the opposition to the current flow. Replacing voltage, current and resistance into the above expression will obtain Ohm’s law: Current ¼ Voltage Resistance Ohm’s law ● Ohm’s law expresses the relationship between I, V and R. ● I through a conductor is directly proportional to V, and inversely proportional to R: I ¼ V/R or I ¼ E/R. 1.5.6 Memory aid for Ohm’s law Using mathematics to manipulate Ohm’s law, and solving for V and R respectively, we can write Ohm’s law in several different forms: V ¼ IR; I ¼ V=R; R ¼ V=I These three equations can be illustrated in Figure 1.12 as a memory aid for Ohm’s law. By covering one of the three variables from Ohm’s law in the diagram, we can get the right form of Ohm’s law to calculate the unknown. I R V (a) V I R (b) I R V (c) Figure 1.12 Memory aid for Ohm’s law. (a) V ¼ IR. (b) I ¼ V/R. (c) R ¼ V/I 18 Understandable electric circuits 1.5.7 The experimental circuit of Ohm’s law The experimental circuit with a resistor of 125 O in Figure 1.13 may prove Ohm’s law. If a voltmeter is connected in the circuit and the source voltage is measured, E ¼ 2.5 V. Also, connecting an ammeter and measuring the current in the circuit will result in I ¼ 0.02 A. With Ohm’s law we can confirm that current in the circuit is indeed 0.02 A: I ¼ E=R ¼ 2:5 V=125 O ¼ 0:02 A 1.5.8 I–V characteristic of Ohm’s law Using a Cartesian coordinate system, voltage V (x-axis) is plotted against current I (y-axis); this graph of current versus voltage will be a straight line, as shown in Figure 1.14. When voltage V is 10 V and current is 1 A, R ¼ V/I ¼ 10 V/1 A ¼ 10 O. When voltage V is 5 V and current is 0.5 A, R ¼ V/I ¼ 5 V/0.1 A ¼ 10 O. So the straight line in Figure 1.14 describes the current–voltage relation- ship of a 10-O resistor. The different lines with different slopes on the I–V characteristic can represent the different values of resistors. For example, a 20-O resistor can be illustrated as in Figure 1.15. Since I–V characteristic shows the relationship between current I and vol- tage V for a resistor, it is called the I–V characteristic of Ohm’s law. A R = 125 Ω 2.5 V 0.02 A V E ϩ Ϫ Figure 1.13 The experimental circuit of Ohm’s law V I R 0 1 A 0.1 A 5 V 10 V Figure 1.14 I–V characteristics (R ¼ 10 O) Basic concepts of electric circuits 19 The I–V characteristic of the straight line illustrates the behaviour of a linear resistor, i.e. the resistance does not change with the voltage or current. If the voltage decreases from 10 to 5 V, the resistance still equals 20 O as shown in Figure 1.15. When the relationship of voltage and current is not a straight line, the resultant resistor will be a non-linear resistor. 1.5.9 Conductance form of Ohm’s law Ohm’s law can be written in terms of conductance as follows: I ¼ GV since G ¼ 1=R; and I ¼ V=R ð Þ 1.6 Reference direction of voltage and current 1.6.1 Reference direction of current When performing circuit analysis and calculations in many situations, the actual current direction through a specific component or branch may change sometimes, and it may be difficult to determine the actual current direction for a component or branch. Therefore, it is convenient to assume an arbitrarily chosen current direction (with an arrow), which is the concept of reference direction of current. If the resultant mathematical calculation for current through that component or branch is positive (I 4 0), the actual current direction is consistent with the assumed or reference direction. If the resultant mathematical calculation for the current of that component is negative (I 50), the actual current direction is opposite to the assumed or reference direction. As shown in Figure 1.16, the solid line arrows indicate the reference V I R 0 0.5 A 0.25 A 5 V 10 V Figure 1.15 I–V characteristics (R ¼ 20 O) 5 Ω 10 V 5 Ω 10 V I = 2 A I > 0 I = –2 A I < 0 Figure 1.16 Reference direction of current 20 Understandable electric circuits current directions and the dashed line arrows indicate the actual current directions. Reference direction of current Assuming an arbitrarily chosen direction as the reference direction of current I: ● If I 4 0 the actual current direction is consistent with the reference current direction. ● If I 5 0 the actual current direction is opposite to the reference current direction. Figure 1.17 shows two methods to represent the reference direction of current: ● Expressed with an arrow, the direction of the arrow indicates the reference direction of current. ● Expressed with a double subscription, for instance I ab , indicates the reference direction of current is from point a to b. 1.6.2 Reference polarity of voltage Similar to the current reference direction, the voltage reference polarity is also an assumption of arbitrarily chosen polarity. If the resultant calculation for voltage across a component is positive (V 40), the actual voltage polarity is consistent with the assumed reference polarity. If the resultant calculation is negative (V50), the actual voltage polarity is opposite to the assumed reference polarity. As shown in Figure 1.18, the positive (þ) and negative (7) polarities represent the reference voltage polarities, and arrows represent the actual voltage polarities. I R (a) (b) R b a I ab Figure 1.17 Reference direction of current I. (a) Arrow indicates the reference I direction. (b) Double subscription indicates the reference I direction Basic concepts of electric circuits 21 Reference polarity of voltage Assuming an arbitrarily chosen voltage polarity as the reference polarity of voltage: ● If V 4 0 the actual voltage polarity is consistent with the reference voltage polarity. ● If V 5 0 the actual voltage polarity is opposite to the reference vol- tage polarity. Figure 1.19 shows three methods to indicate the reference polarity of voltage: ● Expressed with an arrow, the direction of the arrow points from positive to negative. ● Expressed with polarities, positive sign (þ) indicates a higher potential position, and negative sign (7) indicates a lower potential position. ● Expressed with a double subscription, for instance V ab , indicates that the potential position a is higher than the potential position b. 1.6.3 Mutually related reference polarity of current/voltage If the reference direction of current is assigned by flow from the positive side to the negative side of voltage across a component (the reference arrow pointing from þ to 7), then the reference current direction and reference voltage polarity is consistent. In other words, along with the current reference direction + – + V – V + – V > 0 V < 0 Figure 1.18 Reference polarity of voltage R R + b – a V V ab V R Figure 1.19 Methods indicating the reference polarity of voltage 22 Understandable electric circuits is the voltage from positive to negative polarity. This is called the mutually related reference direction or polarity of current/voltage. In this case, if we only know one reference direction or polarity, it is also possible to determine the other, and this is shown in Figure 1.20. Mutually related reference polarity of V and I If the reference I direction is assigned by an arrow pointing from þ to 7 of voltage across a component, then the reference I direction and refer- ence V polarity is consistent. Summary Milestones of the electric circuits Name of scientist Nationality Name of unit/law Named for Charles Augustin de Coulomb French Coulomb Unit of charge (C) Alessandro Volta Italian Volt Unit of voltage (V) Andre´ -Marie Ampe` re French Ampere Unit of current (A) Georg Simon Ohm German Ohm Unit of resistance (O) James Watt Scottish Watt Unit of power (W) Friedrich Emil Lenz German Lenz Lenz’s law James Clerk Maxwell Scottish Maxwell Unit of flux (maxwell), Maxwell’s magnetic field equation Wilhelm Eduard Weber German Webber Unit of flux (weber) 1 Wb ¼ 10 8 Mx Heinrich Rudolf Hertz German Hertz Unit of frequency (Hz) Kirchhoff German Kirchhoff Kirchhoff’s current and voltage laws Joseph Henry Scottish-American Henry Unit of inductance (H) James Prescott Joule British Joule Unit of energy (J) Michael Faraday British Faraday Unit of capacitance (F) I I R R + + – – V V (a) (b) Figure 1.20 (a) Mutually related reference polarity of I and V. (b) Non-mutually related reference polarity of I and V Basic concepts of electric circuits 23 Basic concepts ● Electric circuit: A closed loop of pathway with electric current flowing through it. ● Requirements of a basic circuit: ● Power supply (power source): A device that supplies electrical energy to a load. ● Load: A device that is connected to the output terminal of a circuit, and consumes electrical energy. ● Wires: Wires connect the power supply unit and load together, and carry current flowing through the circuit. ● Schematic: A simplified circuit diagram that shows the interconnection of circuit components, and is represented by circuit symbols. ● Circuit symbols: The idealization and approximation of the actual circuit components. ● Electric current (I): A flow of electric charges through an electric circuit: I ¼ Q/t (or I ¼ dq/dt). ● Current direction: ● Conventional current flow version: A flow of positive charge (proton) from the positive terminal of a power supply to its negative terminal. ● Electron flow version: A flow of negative charge (electron) from the negative terminal of a power supply unit to its positive terminal. ● Ammeter: An instrument used for measuring current, represented by the symbol A . It should be connected in series in the circuit. ● Electromotive force (EMF): An electric pressure or force supplied by a voltage source causing current to flow in a circuit. ● Voltage (V) or potential difference: The amount of energy or workthat would be required to move electrons between two points: V ¼ W/Q (or v ¼ dw/dt). ● Source voltage or applied voltage (E or V S ): EMF can be called source voltage or applied voltage. The EMF is supplied by a voltage source and applied to the load in a circuit. ● Load voltage (V): Voltage across two terminals of the load. ● Voltage drop: Voltage across a component when current flows from a higher potential point to a lower potential point in a circuit. ● Voltage rise: Voltage across a component when current flows from a lower point to a higher point in a circuit. ● Voltmeter: An instrument used for measuring voltage. Its symbol is V and it should be connected in parallel with the component. ● Resistor (R): A two-terminal component of a circuit that limits the flow of current. ● Resistance (R): Measure of a material’s opposition to the flow of current. ● Factors affecting resistance: R ¼ rð‘=AÞ, where cross-sectional area (A), length (‘), temperature (T) and resistivity (r). ● Ohmmeter: An instrument used for measuring resistance. Its symbol is Ω and the resistor must be removed from the circuit to measure the resistance. 24 Understandable electric circuits ● Conductance (G): It is the reciprocal of resistance: G ¼ 1/R. ● Ohm’s law: It expresses the relationship between current I, voltage V and resistance R. I ¼ V R or I ¼ E R ● Conductance form of Ohm’s law: I ¼ GV. ● Reference direction of current: Assuming an arbitrarily chosen current direction as the reference direction of current: ● If I 40 actual current direction is consistent with the reference current direction. ● If I 5 0 actual current direction is opposite to the reference current direction. ● Reference polarity of voltage: Assuming an arbitrarily chosen voltage polarity as the reference polarity of voltage: ● If V 40 actual voltage polarity is consistent with the reference voltage polarity. ● If V 5 0 actual voltage polarity is opposite to the reference voltage polarity. ● Mutually related polarity of voltage and current: If the reference current direction is assigned by an arrow pointing from þ to 7 voltage of the component, then the reference current direction and reference voltage polarity is consistent. ● Symbols and units of electrical quantities: Experiment 1: Resistor colour code Objectives ● Become familiar with the breadboard ● Interpret the colour code for resistors ● Measure resistors with a multimeter (ohmmeter function) Quantity Quantity symbol Unit Unit symbol Charge Q Coulomb C EMF E Volt V Work (energy) W Joule J Resistance R Ohm O Resistivity r Ohm Á metres O Á m Conductance G Siemens or mho S or O Current I Ampere A Voltage V or E Volt V Basic concepts of electric circuits 25 Equipment and components ● Breadboard ● Resistors: 12 O (2), 100 O (2), 2.7 kO (2), 3.9 kO, 8.2 kO, 1.1 MO, 15 kO, 470 O, 18 O, 56 kO, 4.7 kO ● Digital multimeter Background information Breadboard guide ● The Universal Solderless Breadboard, or usually known as the bread- board, is one type of circuit board. It offers an easy way to change com- ponents or wire connections on the breadboard without soldering. ● The breadboard is a good training tool and is usually used in the lab to perform experiments on electric or electronic circuits, or for professionals to build temporary electrical or electronic circuits to try out ideas for circuit designs. ● Figure L1.1 is a photograph of a small breadboard, and Figure L1.2 is what the underneath of the breadboard looks like. ● The breadboard contains an array of holes where the leads of components and jumper wires can be inserted. The bottom of the board has many strips of metal, which is laid out as shown in Figure L1.2. These strips connect the holes on top of the board. The top and bottom rows will be used to connect the power supply. Figure L1.3(a) is a simple circuit, and Figure L1.3(b) shows how to build this circuit on the breadboard. Resistor colour code guide (four band) ● Most resistors are very small and it is hard to print the values on them. Usually the small resistors have different colour bands on them, and the Figure L1.1 A breadboard Figure L1.2 The underneath of the breadboard 26 Understandable electric circuits standard resistor colour code can be used to interpret the values of dif- ferent resistors. ● To determine the value of a resistor from the colour band markings, hold the resistor so that the colour bands are closest to the left end as shown in Figure L1.4. ● The first two colour bands on the left side of the resistor represent two digits (0–9), the third band represents the number of zeros to add to the integers (multiplier), and the fourth band represents the tolerance of the resistance. ● The resistor colour code is shown in Table L1.1, and the tolerance of the resistance is shown in Table L1.2. Note: Memory aid: Better Be Right Or Your Great Big Venture Goes West Source: http://www.wikihow.com/Discussion:Remember-Electrical-Resistor- Color-Codes Table L1.1 Resistor colour code Colour Digit Black 0 Brown 1 Red 2 Orange 3 Yellow 4 Green 5 Blue 6 Violet 7 Grey 8 White 9 Figure L1.4 Resistor colour bands + - R E (a) (b) Figure L1.3 Building a simple circuit on the breadboard Basic concepts of electric circuits 27 Example: If a resistor has colour bands of brown (1), black (0), red (add two zeros) and silver (tolerance is 10%) from left to right side respectively, it indi- cates that its resistance lies between 900 and 1100 O: R ¼ 1 000 O Æ 10% ¼ 1 000 Æ100 ¼ 900 to 1100 O Example: If one needs to find a resistor with the value of 470 O, its colour bands should be yellow (4), violet (7) and brown (add one zero). Multimeter guide ● A multimeter or VOM (volt–ohm–millimetre) is an electrical and elec- tronic measuring instrument that combines functions of voltmeter, ammeter, ohmmeter, etc. There are two types of multimeters: digital multimeter (DMM) and analog multimeter. DMM is a very commonly used instrument, since it is easier to use and has a higher level of accuracy. ● Method for measuring resistance with a DMM (ohmmeter function): ● Turn off the power supply if the resistor has been connected in the circuit. ● Insert the multimeter’s leads into the sockets labelled COM and V/O as shown in Figure L1.5, and turn on the multimeter. ● Turn the central selector switch pointing to the ohms range (with O sign), and to where the maximum range of the estimated resistor value is closed. ● Make the measurement by connecting the resistor in parallel with the two leads of the multimeter (connect or touch one lead from the multimeter to one end of the resistor, and connect or touch the other lead of the multimeter to the other end of the resistor). ● If measuring an unknown resistor, adjust the multimeter range from the maximum to the lower range until suitable resistance is read. ● Turn off the multimeter. Note: To get more accurate measurement result, be sure not to grab the resis- tor’s leads with your hands when you are measuring resistors, since you will add your own resistance (in parallel) to the resistor. Better insert the resistors into the holes of the breadboard to measure them. Table L1.2 Tolerance of resistance Colour Tolerance (%) Gold +5 Silver +10 28 Understandable electric circuits Procedure 1. Familiarize with the resistor colour code. Use the resistor colour code chart in Table L1.1 to find the six resistors listed in Table L1.3 from the lab. List the colour band identification for these six resistors, and fill in the ‘Colour code’ and the ‘Resistance range’ columns in Table L1.3. Table L1.3 Resistor Colour code Resistance range Measured value Example: 470 O Yellow, violet, brown, gold 446.5 O7493.5 O 470.5 O 12 O 100 O 2.7 kO 3.9 kO 82 kO 1.1 MO Figure L1.5 Multimeter Basic concepts of electric circuits 29 2. Get the multimeter to function as an ohmmeter, and measure the six resistors in Table L1.3 with multimeter and fill in the ‘Measured value’ column in Table L1.3. 3. Construct the circuits shown in Figure L1.6 on the breadboard with the right resistors. Show your circuits to the instructor to get check-up and signature. Conclusion The conclusion may include the following information: ● lab objectivities accomplished ● results, errors and error analysis ● problems encountered during the experiment and their solutions ● knowledge and skills obtained from the lab 18 Ω 56 kΩ 470 Ω ϩ ϩ Ϫ Ϫ ϩ Ϫ 4.7 kΩ 15 kΩ 12 Ω 100 Ω 2.7 kΩ 12 Ω 100 Ω 2.7 kΩ 3.9 kΩ 1.1 MΩ 8.2 kΩ Figure L1.6 Construct circuits on the breadboard 30 Understandable electric circuits Chapter 2 Basic laws of electric circuits Objectives After completing this chapter, you will be able to: ● define energy and power ● calculate power ● know the reference directions of power ● analyse and calculate circuits with Kirchhoff’s voltage law (KVL) ● analyse and calculate circuits with Kirchhoff’s current law (KCL) ● define the branch, node, network and loop ● understandthe concepts of the ideal voltage source andthe actual voltage source. 2.1 Power and energy 2.1.1 Work You may have learned in physics that work is the result when a force acts on an object and causes it to move a certain distance. Work (W) is the product of the force (F) and the displacement (S) in the direction of the motion. Work W ¼ F6S where W is, if using for example a force of 1 N to lift an object to 1 m, the 1 J of work done in overcoming the downward force of gravity as shown in Figure 2.1. Quantity Quantity Symbol Unit Unit symbol Work W Joule J Force F Newton N Displacement S Meter m Note: When the force (F) and displacement (S) do not point in the same direction, the formula to calculate work will be: W ¼ (F cos y)S ● where the angle y is the angle between force (F) and displacement (S), ● when y is 0 degree, cos08 ¼ 1, W ¼ (Fcosy)S ¼ FS. It is the same in an electric circuit. Work is done after the electrons or charges are moved to a certain distance in a circuit as a result of applying an electric field force from the power supply. 2.1.2 Energy Energy is the ability to do work; it is not work itself, but a transfer of energy. Even though you can’t ever really see it, you use energy to do work every day. For example, after you eat and sleep, your body converts the stored energy to keep you doing daily work, such as walking, running, reading, writing, etc. The law of conservation of energy is one of most important rules in natural science. It states that energy can neither be created nor destroyed, but can only be converted from one form to another. ‘Converted’ means ‘never disappeared’ in physics terms. For example: Electrical generator: mechanical energy ! electrical energy. Lamp: electrical energy ! light energy. Battery: chemical energy ! electrical energy. 2.1.3 Power Power refers to the speed of energy conversion or consumption; it is a measure of how fast energy is transforming or being used. For example in Figure 2.1, 1 N object lifted to 1 m may have different time rates depending on the amount of power applied. If a higher power is applied to the object (an adult is lifting it), it will take a shorter period of time to lift it; and if a lower power is applied to the object (a kid is lifting it), it will take a longer period of time to lift it. So power is defined as the rate of doing work, or the amount of work done per unit of time. Our daily consumption of electricity is electrical energy, and not electrical power. The hydro bill that you receive is for electrical power – the amount of electrical energy consumed in 1 or 2 months. 1 N 1 m F 1 N Figure 2.1 Work 32 Understandable electric circuits Energy and power ● Energy is the ability to do work. ● Power is the speed of energy conversion, or work done per unit of time: Power ¼ Work=time or P ¼ W=t Electrical power is the speed of electrical energy conversion or consumption in an electric circuit, and it is a measure of how fast electrons or charges are moving in a circuit. Since current is the amount of charge (Q) that flows past a given point at the certain time: I ¼ Q=t and voltage is the amount of work that is required to move electrons between two points: V ¼ W=Q or Wc ¼ QV: Substituting work W into the power equation gives P¼W=t ¼QV=t ¼IV: It can also be expressed as the form of a derivative: p ¼ ðdw=dtÞ ¼ ðdw=dqÞðdq=dtÞ ¼ vi Substituting Ohm’s law into the power equation P ¼ IV obtains the other two different power equations: P ¼ VI ¼ ðIRÞI ¼ I 2 R ðOhm’s law : V ¼ IRÞ P ¼ VI ¼ V V R ¼ V 2 R Ohm’s law : I ¼ V R (Electrical) Power P P ¼ IV ¼ I 2 R ¼ V 2 =R ðor P ¼ IE ¼ E 2 =RÞ Quantity Quantity symbol Unit Unit symbol Electrical Power: Work or Energy W Joule J Time t Second s Power P Watt W Or: Kilowatt-hour kWh Hour h Watt W Quantity Quantity symbol Unit Unit symbol Power P Watt W Basic laws of electric circuits 33 The above three power equations can be illustrated in Figure 2.2 as the mem- ory aid for power equations. By covering power in any diagram, the correct equation will be obtained to calculate the unknown power. Example 2.1: In a circuit, voltage V ¼ 10 V, current I ¼ 1 A and resistance R ¼ 10 O, calculate the power in this circuit by using three power equations, respectively. Solution: P ¼ IV ¼ ð1 AÞð10 VÞ ¼ 10 W P ¼ I 2 R ¼ ð1 AÞ 2 ð10 OÞ ¼ 10 W P ¼ V 2 =R ¼ ð10 VÞ 2 =10 O ¼ 10 W Example 2.1 proved that the three power equations are equivalent since each equation leads to the same value of power at 10 W. If power is given in a circuit, using mathematical skill to manipulate the power equations and solving for current I and voltage V, respectively, we can express current I and voltage V as follows: since P ¼ I 2 R or I 2 ¼ P/R, so I ¼ ffiffiffiffiffiffiffiffiffiffi P=R p since P ¼ V 2 /R or V 2 =PR, so V ¼ ffiffiffiffiffiffiffi PR p Example 2.2: If power consumed on a 2.5 O resistor is 10 W in a circuit, cal- culate the current flowing through this resistor. Solution: I ¼ ffiffiffiffiffiffiffiffiffi P=R p ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 10 W=2:5 O p ¼ 2 A 2.1.4 The reference direction of power When a component in a circuit has mutually related reference polarity of cur- rent and voltage (refer to chapter 1, section 1.6.3), power is positive, i.e. P 40, meaning the component absorption (or consumption) of energy. When a component in a circuit has non-mutually related reference polarity of current and voltage, power is negative, i.e. P 5 0, meaning the component releasing (or providing) of energy. The concept of the reference direction of power can be illustrated in Figure 2.3. V I P P I 2 R P R V 2 Figure 2.2 Memory aid for power equations 34 Understandable electric circuits The reference direction of power ● If a circuit has mutually related reference polarity of current and voltage: P 40 (absorption energy). ● If a circuit has non-mutually related reference polarity of current and voltage: P 50 (releasing energy). Example 2.3: Determine the reference direction of power in Figure 2.4(a and b). Solution: (a) P ¼ IV ¼ (2 A)(3 V) ¼ 6 W (P 4 0, the resistor absorbs energy). (b) P ¼ I (7V) ¼ (2 A)(73 V) ¼ 76 W (P 50, the resistor releases energy). Example 2.4: I ¼ 2 A, V 1 ¼ 6 V, V 2 ¼ 14 V and E ¼ 20 V in a circuit as shown in Figure 2.5. Determine the powers dissipated on the resistors R 1 , R 2 , and R 1 and R 2 in series in this figure. Solution: Power for R 1 (a to b): P 1 ¼ V 1 I ¼ (6 V)(2 A) ¼ 12 W (absorption). Power for R 2 (b to c): P 2 ¼ V 2 I ¼ (14 V)(2 A) ¼ 28 W (absorption). Power for R 1 and R 2 (a to d): P 3 ¼ (7E)I ¼ (720 V)(2 A) ¼ 740 W (releasing). P 1 þ P 2 þ P 3 ¼ 12 W þ 28 W 7 40 W ¼ 0 (energy conservation). I ϩ Ϫ V Ϫ ϩ V I (a) P > 0 (b) P < 0 Figure 2.3 The reference direction of power I ϭ 2 A ϩ Ϫ V ϭ 3 V I ϭ 2 A Ϫ ϩ V ϭ 3 V (a) (b) Figure 2.4 Illustrations for Example 2.3 Basic laws of electric circuits 35 2.2 Kirchhoff’s voltage law (KVL) In 1847, a German physicist, physics professor Kirchhoff (Gustav Kirchhoff, 1824–1887) at Berlin University developed the two laws that established the relationship between voltage and current in an electric circuit. Kirchhoff’s laws are the most important fundamental circuit laws for analysing and calculating electric circuits after Ohm’s law. 2.2.1 Closed-loop circuit A closed-loop circuit is a conducting path in a circuit that has the same starting and ending points. If the current flowing through a circuit from any point returns current to the same starting point, it would be a closed-loop circuit. As current flows through a closed-loop circuit, it is same as having a round trip, so the starting and ending points are the same, and they have the same potential positions. For example, Figure 2.6 is a closed-loop circuit, since current I starts at point a, passes through points b, c, d and returns to the starting point a. 2.2.2 Kirchoff’s voltage law #1 KVL #1 states that the algebraic sum of the voltage or potential difference along a closed-loop circuit is always equal to zero at any moment, or the sum E R 1 R 2 I = 2 A ϩ V 1 V 2 Ϫ Ϫ ϩ a a c d b ϩ Ϫ Figure 2.5 Circuit for Example 2.4 I d a b c V 2 = 10 V V 1 = 10 V E 1 = 10 V E 2 = 10 V V 3 = 10 V R 3 R 2 R 1 Ϫ Ϫ ϩ ϩ Figure 2.6 A closed-loop circuit 36 Understandable electric circuits of voltages in a closed-loop is always equal to zero, i.e. SV ¼ 0. The voltage in KVL includes voltage rising from the voltage sources (E) and voltage dropping on circuit elements or loads. The algebraic sum used in KVL #1 means that there are voltage polarities existing in a closed-loop circuit. It requires assigning a loop direction and it could be in either clockwise or counter-clockwise directions (usually clockwise). ● Assign a positive sign (þ) for voltage (V or E) in the equation SV ¼ 0, if the voltage reference polarity and the loop direction are the same, i.e. if the voltage reference polarity is from positive to negative and the loop direc- tion is clockwise. ● Assign a negative sign (7) for voltage (V or E) in the equation SV ¼ 0, if the voltage reference polarity and the loop direction are opposite, i.e. if the voltage reference polarity is fromnegative topositive, andthe loopdirectionis clockwise. Example 2.5a: Verify KVL #1 for the circuit of Figure 2.7. Solution: Applying SV ¼ 0 in Figure 2.7: V 1 þV 2 þV 3 ÀE 2 ÀE 1 ¼ 0 ð2:5 þ2:5 þ2:5 À5 À2:5Þ V ¼ 0 KVL #1 SV ¼ 0 ● Assign a þve sign for V or E if its reference polarity (þ to 7) and loop direction (clockwise) are the same. ● Assign a 7ve sign for V or E if its reference polarity (7 to þ) and loop direction (clockwise) are opposite. V 1 = 2.5 V E 1 = 2.5 V E 2 = 5 V V 3 = 2.5 V V 2 = 2.5 V ϩ I ϩ ϩ Ϫ Ϫ Ϫ R 3 R 2 R 1 ϩ ϩ Ϫ Ϫ Figure 2.7 Circuit for Example 2.5 Basic laws of electric circuits 37 2.2.3 KVL #2 KVL can also be expressed in another way: the sum of the voltage drops (V) around a closed loop must be equal to the sum of the voltage rises or voltage sources in a closed-loop circuit, i.e. SV ¼ SE. ● Assign a positive sign (þ) for V, if its reference and loop directions are the same; assign a negative sign (7) for V, if its reference and the loop direc- tions are opposite. ● Assign a negative sign (7) for the voltage source (E) in the equation, if its reference polarity and the loop direction are the same, i.e. if its polarity is from þve to –ve and the loop direction is clockwise. Assign a positive sign (þ) for voltage source (E) in the equation if its reference polarity and loop direction are opposite, i.e. if its polarity is from negative to positive and the loop direction is clockwise. Example 2.5b: Verify KVL #2 for the circuit of Figure 2.7. Solution: Applying SV ¼ SE in Figure 2.7: V 1 þV 2 þV 3 ¼ E 2 þE 1 ð2:5 þ2:5 þ2:5ÞV ¼ ð2:5 þ5ÞV 7:5 V ¼ 7:5 V KVL #2 SV ¼ SE ● Assign a þve sign for V if its reference polarity and loop direction are the same; assign a 7ve sign for V if its reference direction and loop direction are opposite. ● Assign a 7ve sign for E if its reference polarity and loop direction are the same; assign a þve sign for E if its polarity and loop direction are opposite. 2.2.4 Experimental circuit of KVL KVL can be approved by an experimental circuit in Figure 2.8. If using a multimeter (voltmeter function) to measure voltages on all resistors and power supply in the circuit of Figure 2.8, the total voltage drops on all the resistors should be equal to the voltage for the DC power supply. KVL #1, SV ¼ 0: (10 þ 10 þ 10 7 30) V ¼ 0 KVL #2, SV ¼ SE: (10 þ 10 þ 10) V ¼ 30 V 38 Understandable electric circuits Example 2.6: Determine resistance R 3 in the circuit of Figure 2.9. Solution: R 3 V 3 I ; V 3 ¼ ? Applying KVL #1, SV ¼ 0: V 1 þ V 2 þ V 3 þ V 4 7E ¼ 0 Therefore: V 1 ¼ I R 1 ¼ (0.25 A)(2.5 O) ¼ 0.625 V V 2 ¼ I R 2 ¼ (0.25 A)(5 O) ¼ 1.25 V V 4 ¼ I R 4 ¼ (0.25 A)(4 O) ¼ 1 V Solve for V 3 from V 1 þ V 2 þ V 3 þ V 4 7E ¼ 0: V 3 ¼ E 7 V 1 7V 2 7 V 4 ¼ (5 70.625 7 1.25 7 1)V ¼ 2.125 V Therefore, R 3 ¼ V 3 I ¼ 2:125 V 0:25 A ¼ 8:5 O V 10 V 10 V 30 V 100 Ω 100 Ω 10 V 100 Ω ϩ ϩ Ϫ Ϫ Ϫ ϩ V V V ϩ Ϫ Figure 2.8 Experimental circuit of KVL E 2 = 5 V R 1 = 2.5 Ω R 2 = 5 Ω R 4 = 4 Ω R 3 = ? I = 0.25 A ϩ ϩ ϩ Ϫ ϩ Ϫ Ϫ Ϫ ϩ Ϫ Figure 2.9 Circuit for Example 2.6 Basic laws of electric circuits 39 2.2.5 KVL extension KVL can be expanded from a closed-loop circuit to any scenario loop in a circuit, because voltage or potential difference in the circuit can exist between any two points in a circuit. V ab in the circuit of Figure 2.10 can be calculated using KVL #2 as follows: SV ¼ SE : V þV ab ¼ E V ab ¼ E ÀV ¼ ð10 À1ÞV ¼ 9 V Example 2.7: Determine the voltage across points a to b (V ab ) in the circuit of Figure 2.11. Solution: V ab can be solved by two methods as follows: Method 1: SV ¼ 0: V 1 þ V ab þ V 4 7E ¼ 0, where V ab ¼ E 7 V 1 7 V 4 ¼ (5 7 1.5 7 1)V ¼ 2.5 V. Method 2: SV ¼ 0: V 2 þ V 3 7 V ab ¼ 0, where V ab ¼ V 2 þ V 3 ¼ (2 þ 0.5)V ¼ 2.5 V. E = 10 V R a I V = 1 V V ab b ϩ ϩ Ϫ Ϫ ϩ Ϫ Figure 2.10 KVL extension V 1 = 1.5 V E = 5 V V 3 = 0.5 V V 2 = 2 V V 4 = 1 V ϩ ϩ ϩ ϩ Ϫ Ϫ Ϫ Ϫ b a ϩ Ϫ Figure 2.11 Circuit for Example 2.7 40 Understandable electric circuits 2.2.6 The physical property of KVL The results from Example 2.7 show that voltage across two points a and b is the same, and it does not matter which path or branch is used to solve for voltage between these two points, the result should be the same. Therefore, the physical property of KVL is that voltage does not depend on the path. 2.3 Kirchhoff’s current law (KCL) 2.3.1 KCL #1 KCL #1 states that the algebraic sum of the total currents entering and exiting a node or junction of the circuit is equal to zero, i.e. SI ¼ 0. ● Assign a positive sign (þ) to the current in the equation if current is entering the node. ● Assign a negative sign (7) to the current in the equation if current is exiting the node. A node or junction is the intersectional point of two or more current paths where current has several possible paths to flow. A branch is a current path between two nodes with one or more circuit components in series. For instance, point A is a node in Figure 2.12, and it has six branches. I 1 , I 2 and I 3 are the currents flowing into node A; I 4 , I 5 and I 6 are the currents exiting the node A. Applying KCL #1: I 1 þ I 2 þ I 3 7I 4 7 I 5 7I 6 ¼ 0 KCL #1 SI ¼ 0 ● Assign a þve sign for current in KCL if I is entering the node. ● Assign a 7ve sign for current in KCL if I is exiting the node. 2.3.2 KCL #2 KCL can also be expressed in another way: the total current flowing into a node is equal to the total current flowing out of the node, i.e. SI in ¼ SI out A I 3 I 4 I 5 I 6 I 1 I 2 Figure 2.12 Nodes and branches Basic laws of electric circuits 41 ● Assign a positive sign (þ) to current I in in the equation if current is entering the node; assign a negative sign (7) for I in if current is exiting the node. ● Assign a positive sign (þ) to current I out in the equation if current is exiting the node; assign a negative sign (7) for I out if current is entering the node. Example 2.8: Verify KVL #1 and #2 for the circuit of Figure 2.13. Solution: KCL #2: SI in ¼ SI out : I 1 þ I 2 ¼ I 3 þ I 4 þ I 5 Substituting I with its respective values, we get (15 þ 10)A ¼ (7 þ 8 þ 10)A KCL #1: SI ¼ 0: I 1 þ I 2 7I 3 7 I 4 7I 5 ¼ 0 Substituting I with its respective values, we get (15 þ10 77 78 710)A¼0 Example 2.9: Determine the current I 1 at node A and B in Figure 2.14. Solution: Node A: SI ¼ 0 : I 1 ÀI 2 ÀI 3 ÀI 4 ¼ 0 SI in ¼ SI out : I 1 ¼ I 2 þI 3 þI 4 Node B: SI ¼ 0 : I 2 þI 3 þI 4 ÀI 1 ¼ 0 SI in ¼ SI out : I 2 þI 3 þI 4 ¼ I 1 a I 2 = 10 A I 3 = 7 A I 4 = 8 A I 5 = 10 A I 1 = 15 A Figure 2.13 Circuit for Example 2.8 A I 1 B I 4 I 3 I 2 ϩ Ϫ Figure 2.14 Circuit for Example 2.9 42 Understandable electric circuits KCL #2 SI in ¼ SI out ● Assign a þve sign for I in if current is entering the node; assign a 7ve sign for I in , if current is exiting the node. ● Assign a þve sign for I out if current is exiting the node; assign a 7ve sign for I out , if current is entering the node. Water flowing in a pipe can be analogized as current flowing in a con- ducting wire with KCL. Water flowing into a pipe should be equal to the water flowing out of the pipe. For example, in Figure 2.15, water flows in the three upstream creeks A, B and C merging together to a converging point and forms the main water flow out of the converging point to the downstream creek. Example 2.10: Determine current I 3 (you may calculate it by using one KCL, and prove it by using another one). A B C Figure 2.15 Creeks I 4 = ? I 1 I 2 I 3 10 A 20 A 5 A Figure 2.16 Circuit for Example 2.10 Basic laws of electric circuits 43 Solution: SI ¼ 0 : I 1 ÀI 2 þI 3 þI 4 ¼ 0; I 4 ¼ I 2 ÀI 1 ÀI 3 ¼ ð20 À10 À5ÞA ¼ 5 A SI in ¼ SI out : I 1 þI 3 þI 4 ¼ I 2 ð10 þ5 þ5ÞA ¼ 20A 20A ¼ 20A ðhence provedÞ The KCL can be proved by an experimental circuit in Figure 2.17. Measure branch currents I 1 and I 2 (entering) using two multimeters (ammeter function), and they are equal to the source branch current I 3 (exit- ing), I 3 ¼ I 1 þ I 2 ¼ 0.25 A. 2.3.3 Physical property of KCL The physical property of KCL is that charges cannot accumulate in a node; what arrives at a node is what leaves that node. This results from conservation of charges, i.e. charges can neither be created nor destroyed or the amount of charges that enter the node equals the amount of charges that exit the node. Another property of KCL is the continuity of current (or charges), which is similar to the continuity of flowing water, i.e. the water or current will never discontinue at any moment in a pipe or conductor. 2.3.4 Procedure to solve a complicated problem It does not matter which field of natural science the problems belong to or how complicated they are, the procedure for analysing and solving them are all similar. The following steps outline the procedure: 1. Start from the unknown value in the problem and find the right equation that can solve this unknown. E = 10 V A A A I 1 = 0.1 A I 2 = 0.15 A I 3 = 0.25 A ϩ Ϫ Figure 2.17 Experimental circuit for KCL 44 Understandable electric circuits 2. Determine the new unknown of the equation in step 1 and find the equa- tion to solve this unknown. 3. Repeat steps 1 and 2 until there are no more unknowns in the equation. 4. Substitute the solution from the last step into the previous equation, and solve the unknown. Repeat until the unknown in the original problem is solved. Now let’s try to use this method to solve I 1 in Example 2.11. Example 2.11: I 1 ¼ ? Solution: The unknown in this problem is I 1 . Find the right equation to solve I 1 . ● At node C: I 1 þ4 A ¼ I 2 þ3 A I 2 ¼ ? ð2:1Þ (Besides I 1 , the unknown in this equation is I 2 .) ● Find the right equation to solve I 2 . At node B: I 2 þ3 A ¼ 4 AþI 3 þ2 A I 3 ¼ ? ð2:2Þ (Besides I 2 , the unknown in this equation is I 3. ) ● Find the right equation to solve I 3 . At node A: I 3 þ 4 A ¼ (5 þ 1)A, solve for I 3 : I 3 ¼ 2 A (there are no more unknown elements in this equation except for I 3 ). ● Substitute I 3 ¼ 2 A into (2.2) and solve for I 2 : I 2 þ 3 A ¼ (4 þ 2 þ 2)A, so I 2 ¼ 5 A. ● Substitute I 2 ¼ 5 A into (2.1) and solve for I 1 : I 1 þ 4 A ¼ (5 þ 3)A, therefore, I 1 ¼ 4 A. 4 A I 2 I 3 I 1 A B C 3 A 1 A 5 A 4 A 2 A Figure 2.18 Circuit for Example 2.11 Basic laws of electric circuits 45 2.3.5 Supernode The concept of the node can be extended to a circuit that contains several nodes and branches, and this circuit can be treated as a supernode. The circuit between nodes a and b in Figure 2.19 within the dashed circle can be treated as an extended node or supernode A; KCL can be applied to it: SI in ¼ SI out or I 1 ¼ I 2 Example 2.12: Determine the magnitudes and directions of I 3 , I 4 and I 7 in the circuit of Figure 2.20. Solution: Treat the circuit between the nodes A and D (inside of the circle) as a supernode, and current entering the node A should be equal to current exiting the node D, therefore, I 7 ¼ I 1 ¼ 5 A. ● At node A: Since current entering node A is I 1 ¼ 5 A, and current leaving node A is I 2 ¼ 6 A, so I 2 4I 1 . I 3 must be current entering node A to satisfy SI in ¼ SI out , i.e. I 1 þ I 3 ¼ I 2 or 5 A þ I 3 ¼ 6 A, therefore, I 3 ¼ 1 A. I 1 I 2 A a b Figure 2.19 Supernode A I 1 = 5 A I 2 = 6 A I 5 = 4 A I 6 = 1 A I 7 = ? I 3 I 4 B C D Figure 2.20 Circuit for Example 2.12 46 Understandable electric circuits ● At node B: Since current entering node B is I 2 ¼ 6 A and currents exiting node B is I 5 ¼ 4 A, so I 2 4I 5 . I 4 must be current exiting node B to satisfy SI in ¼ SI out , i.e. I 2 ¼ I 4 þ I 5 or 6 A ¼ I 4 þ 4 A, therefore, I 4 ¼ 2 A. ● Prove it at node C: I 4 ¼ I 3 þ I 6 , 2 A ¼ 1 A þ 1 A, 2 A ¼ 2 A (proved). 2.3.6 Several important circuit terminologies ● Node: The intersectional point of two or more current paths where current has several possible paths to flow. ● Branch: A current path between two nodes where one or more circuit components is in series. ● Loop: A complete current path where current flows back to the start. ● Mesh: A loop in the circuit that does not contain any other loops (non- redundant loop). Note: A mesh is always a loop, but a loop is not necessary a mesh. A mesh can be analogized as a windowpane, and a loop may include several such windowpanes. Example 2.13: List nodes, branches, meshes and loops in Figure 2.21. Solution: ● Node: four nodes – A, B, C and D ● Branch: six branches – AB, BD, AC, BC, CD and AD ● Mesh: 1, 2 and 3 ● Loop: 1, 2, 3, A–B–D–C–A, A–B–D–A, etc. 2.4 Voltage source and current source A power supply is a circuit device that provides electrical energy to drive the system, and it is a source that can provide EMF (electromotive force) and current to operate the circuit. The power supply can be classified into two categories: voltage source and current source. A B C D 1 2 3 Figure 2.21 Illustration for Example 2.13 Basic laws of electric circuits 47 2.4.1 Voltage source 2.4.1.1 Ideal voltage source An ideal voltage source is a two-terminal circuit device that can provide a con- stant output voltage, V ab , across its terminals, and is shown in Figure 2.22(a). Voltage of the ideal voltage source, V S , will not change even if an external circuit such as a load, R L , is connected to it as shown in Figure 2.22(b), so it is an independent voltage source. This means that the voltage of the ideal voltage source is independent of variations in its external circuit or load. The ideal vol- tage source has a zero internal resistance (R S ¼ 0), and it can provide maximum current to the load. Current in the ideal voltage source is dependant on variations in its exter- nal circuit, so when the load resistance R L changes, the current in the ideal voltage source also changes since I ¼ V/R L . The characteristic curve of an ideal voltage source is shown in Figure 2.22(c). The terminal voltage V ab for an ideal voltage source is a constant, and same as the source voltage (V ab ¼ V S ), no matter what its load resistance R L is. Ideal voltage source ● It can provide a constant terminal voltage that is independent of the variations in its external circuit, V ab ¼ V S . ● Its internal resistance, R S ¼ 0. Its current depends on variations in its external circuit. 2.4.1.2 Real voltage source Usually a real-life application of a voltage source, such as a battery, DC gen- erator or DC power supply, etc., will not reach a perfect constant output vol- tage after it is connected to an external circuit or load, since nothing is perfect. The real voltage sources all have a non-zero internal resistance R S (R S 6¼ 0). The real voltage source (or voltage source) can be represented as an ideal voltage source V S in series with an internal resistor R S as shown in Figure 2.23(a). Once a load resistor R L is connected to the voltage source (Figure 2.23(b)), the terminal voltage of the source V ab will change if the load resistance R L changes. Since the internal resistance R S is usually very small, V ab will be a little bit lower than the source voltage V S (V ab ¼ V S 7IR S ). V S V S V S R L a b a b I V ϩ Ϫ 0 I or t V (a) Ideal voltage source (b) Ideal voltage source with a load (c) Characteristic curve ϩ Ϫ ϩ Ϫ Figure 2.22 Ideal voltage source 48 Understandable electric circuits A smaller internal resistance can also provide a higher current through the external circuit of the real voltage source because I ¼ V S /(R S þ R L ). Once the load resistance R L changes, current I in this circuit will change, and the term- inal voltage V ab also changes. This is why the terminal voltage of the real voltage source is not possible to keep at an ideal constant level (V ab 6¼ V S ). The internal resistance of a real voltage source usually is much smaller than the load resistance, i.e. R S (R L , so the voltage drop on the internal resistance (IR S ) is also very small, and therefore, the terminal voltage of the real voltage source (V ab ) is approximately stable: V ab ¼ V S ÀIR S % V S When a battery is used as a real voltage source, the older battery will have a higher internal resistance R S and a lower terminal voltage V ab . Real voltage source (voltage source) It has a series internal resistance R S , and R S (R L . The terminal voltage of the real voltage source is: V ab ¼ V S 7IR S . Example 2.14: Determine the terminal voltages of the circuit in Figure 2.24(a and b). 0 I or t V V S ↓ ↑ VR S = IR S a b I ϩ Ϫ R S R L V S V ab ϩ Ϫ a b R S V S ϩ Ϫ (a) Real voltage source (b) Real voltage source with a load (c) Characteristic curve Figure 2.23 Real voltage source (b) R S = 25 Ω (a) R S = 0.005 Ω a R S V S R L 3 V 0.005 Ω 5 Ω b ϩ Ϫ I R S V S R L 5 Ω a b 3 V 50 Ω ϩ Ϫ Figure 2.24 Circuits for Example 2.14 Basic laws of electric circuits 49 When R S ¼ 0:005 O; I ¼ V S ðR S þR L Þ ¼ 3 V ð0:005 O þ5 OÞ % 0:5994 A V ab ¼ IR L ¼ ð0:5994 AÞð5 OÞ ¼ 2:997 V When R S ¼ 50 O; I ¼ V S ðR S þR L Þ ¼ 3 V ð50 O þ5 OÞ % 0:055 A V ab ¼ IR L ¼ ð0:055 AÞð5 OÞ ¼ 0:275 V The above example indicates that the internal resistance has a great impact on the terminal voltage and current of the voltage source. Only when the internal resistance is very small, can the terminal voltage of the source be kept approximately stable, such as when R S ¼ 0.005 O, V ab ¼ 2.997 V % V S ¼ 3 V. In this case, the terminal voltage V ab is very close to the source voltage V S . But when R S ¼ 50 O , V ab ¼ 0.275 V (V S ¼ 3 V, i.e. the terminal voltage V ab is much less than the source voltage V S . A real voltage source has three possible working conditions: ● When an external load R L is connected to a voltage source (Figure 2.25(a)): V ab ¼ V S ÀI R S ; I ¼ ðV S =R S þR L Þ. ● Open circuit: when there is no external load R L connected to a voltage source (Figure 2.25(b)): V ab ¼ V S ; I ¼ 0: ● Short circuit: when a jump wire is connected to the two terminals of a voltage source (Figure 2.25(c)): V ab ¼ 0; I ¼ ðV S =R S Þ: 2.4.2 Current source The current source is a circuit device that can provide a stable current to the external circuit. A transistor, an electronic element you may have heard, can be approximated as an example of a current source. 2.4.2.1 Ideal current source An ideal current source is a two-terminal circuit device that can provide a constant output current I S through its external circuit. Current of the ideal V S V S V S R L V ab V ab I = 0 a R S R S R S b a b I ϩ Ϫ ϩ Ϫ a b V ab = 0 (a) With a load (b) Open circuit (c) Short circuit ϩϩϩ ϩ Ϫ Ϫ ϩ Ϫ ϩ Figure 2.25 Three states of a voltage source 50 Understandable electric circuits current source will not change even an external circuit (load R L ) is connected to it, so it is an independent current source. This means the current of the ideal voltage source is independent of variations in its external circuit or load. The ideal current source has an infinite internal resistance (R S ¼ ?), so it can provide a maximum current to the load. Its two-terminal voltage is determined by the external circuit or load. The symbol of an ideal current source is shown in Figure 2.26(a), and its characteristic curve is shown in Figure 2.26(b). I S represents the current for current source, and the direction of the arrow is the current direction of the source. Ideal current source ● It can provide a constant output current I S that does not depend on the variations in its external circuit. ● Its internal resistance R S ¼ ?. ● Its voltage depends on variations in its external circuit. V ab ¼ I S R L . Example 2.15: The load resistance R L is 1 000 and 50 O, respectively, in Figure 2.27. Determine the terminal voltage V ab for the ideal current source in the circuit. I S I S I V or t (a) Symbol of an ideal current source (b) Characteristic curve Figure 2.26 Ideal current source I S R L b a 0.02 A V ab Ϫ ϩ Figure 2.27 Circuit for Example 2.15 Basic laws of electric circuits 51 Solution: When R L ¼ 1000 O; V ab ¼ I S R L ¼ ð0:02AÞ ð1000 OÞ ¼ 20 V When R L ¼ 50 O; V ab ¼ I S R L ¼ ð0:02AÞ ð50 OÞ ¼ 1 V The conditions of open circuit and short circuit of an ideal current source are as follows: ● Open circuit, V ab ¼ ?, I ¼ 0, as shown in Figure 2.28(a). ● Short circuit, V ab ¼ 0, I ¼ I S , as shown in Figure 2.28(b). 2.4.2.2 Real current source Usually a real-life application of current source will not reach a perfect con- stant output current after it is connected to an external circuit or load, as the real current sources all have a non-infinite internal resistance R S . The real current source (or current source) can be represented as an ideal current source I S in parallel with an internal resistor R S . Once a load resistor R L is connected to the current source as shown in Figure 2.29, the current of the source will change if the load resistance R L changes. Since the internal resistance R S of the current source usually is very large, the load current I will be a little bit lower than the source current I S . I S R S R L a b I Figure 2.29 A real current source (a) Open circuit I S Ϫ ϩ a b I V ab ϭ 0 I S a V ab ϭ ∞ I ϭ 0 b ϩ Ϫ (b) Short circuit Figure 2.28 Open circuit and short circuit of an ideal current source 52 Understandable electric circuits Once the load resistance R L changes, the current in the load will also changes. This is why the current of the real current source is not possible to keep at an ideal constant level. A higher internal resistance R S can provide a higher current through the external circuit of the real current source. The internal resistance of a real current source usually is much greater than the load resistance (R S )R L ), and therefore the output current of the real current source is approximately stable. Real current source (current source) ● It has an internal resistance R S (R S ) R L ). ● R S is in parallel with the current source. 2.5 International units for circuit quantities 2.5.1 International system of units (SI) The international system of units (SI) was developed at the General Conference of the International Weight and Measures, which is the international authority that ensures dissemination and modifications of the SI units to reflect the latest advances in science and commerce. SI originates from the French ‘Le Syste`me International d’Unite´s’, which means the international system of units or the metric system to most people. SI system is the world’s most widely used modern metric system of mea- surement. Each physical quantity has a SI unit. There are seven basic units of the SI system and they are listed in Table 2.1. SI units International system of units (SI) is the world’s most widely used modern metric system of measurement. There are seven base units of the SI system. Table 2.1 SI basic units Quantity Quantity symbol Unit Unit symbol Length l Metre m Mass M Kilogram kg Time t Second s Current I Ampere A Temperature T Kelvin K Amount of substance m Mole mol Intensity of light I Candela cd Basic laws of electric circuits 53 All other metric units can be derived from the seven SI basic units that are called ‘derived quantities’. Some derived SI units for circuit quantities are given in Table 2.2. As you study the circuit theory more in-depth, you may use and add more circuit quantities and their derived SI units in this table. 2.5.2 Metric prefixes (SI prefixes) Some time there are very large or small numbers when doing circuit analysis and calculation. A metric prefix (or SI prefix) is often used in the circuit cal- culation to reduce the number of zeroes. Large and small numbers are made by adding SI prefixes. A metric prefix is a modifier on the root unit that is in multiples of 10. In general science, the most common metric prefixes, such as milli, centi and kilo are used. In circuit analysis, more metric prefixes, such as nano and pico are used. Table 2.3 contains a complete list of metric prefixes. Example 2.16: ðaÞ 47000 O ¼ ð?Þ kO ðbÞ 0:0505 A ¼ ð?Þ mA ðcÞ 0:0005 V ¼ ð?Þ mV ðdÞ 15000000000 C ¼ ð?Þ GC Solution: ðaÞ 47000 O ¼ 47 Â10 3 O ¼ 47 kO ðbÞ 0:0505 A ¼ 50:5 Â10 À3 A ¼ 50:5 mA ðcÞ 0:0005 V ¼ 500 Â10 À6 V ¼ 500 mV ðdÞ 15000000000 C ¼ 15 Â10 9 ¼ 15 GC Note: ● If a number is a whole number, move the decimal point to the left, and multiply the positive exponent of 10 (moving the decimal point three places each time). In Example 2.16(a), 47 000O ¼ 47 610 3 O (moving the decimal point three places to the left). Table 2.2 Some circuit quantities and their SI units Quantity Quantity symbol Unit Unit symbol Voltage V Volt V Resistance R Ohm S Charge Q Coulomb C Power P Watt W Energy W Joule J Electromotive force E or V S Volt V Conductance G Siemens S Resistivity r Ohm Á metre O Á m 54 Understandable electric circuits ● If a number is a decimal number, move the decimal point to the right, and multiply the negative exponent of 10. In Example 2.16(b), 0.0505 A ¼ 50.5 610 73 A (moving the decimal point three places to the right). ● If numbers have different prefixes, convert them to the same prefix first, then do the calculation. Example 2.17: Determine the result of 30mA+2000mA. Solution: 30 mAþ2000 mA ¼ 30 Â10 À3 Aþ2000 Â10 À6 A ¼ 30 Â10 À3 Aþ2 Â10 À3 A ¼ ð30 þ2ÞmA ¼ 32 mA Summary Basic concepts ● Power: the speed of energy conversion, or work done per unit of time, P ¼ W/t. Table 2.3 Metric prefix table Prefix Symbol (abbreviation) Exponential (power of 10) Multiple value (in full) Yotta Y 10 24 1 000 000 000 000 000 000 000 000 Zetta Z 10 21 1 000 000 000 000 000 000 000 Exa E 10 18 1 000 000 000 000 000 000 Peta P 10 15 1 000 000 000 000 000 Tera T 10 12 1 000 000 000 000 Giga G 10 9 1 000 000 000 Mega M 10 6 1 000 000 myria my 10 4 10 000 kilo k 10 3 1 000 hecto h 10 2 100 deka da 10 10 deci d 10 71 0.1 centi c 10 72 0.01 milli m 10 73 0.001 micro m (mu) 10 76 0.000 001 nano n 10 79 0.000 000 001 pico p 10 712 0.000 000 000 001 femto f 10 715 0.000 000 000 000 001 atto a 10 718 0.000 000 000 000 000 001 zepto z 10 721 0.000 000 000 000 000 000 001 yocto y 10 724 0.000 000 000 000 000 000 000 001 Note: The most commonly used prefixes are shown in bold. Basic laws of electric circuits 55 ● Energy: the ability to do work. ● The reference direction of power: ● If a circuit has mutually related reference polarity of current and vol- tage: P 40 (absorption energy). ● If a circuit has non-mutually related reference polarity of current and voltage: P 50 (releasing energy). ● Branch: a current path between two nodes where one or more circuit components in series. ● Node: the intersectional point of two or more current paths where current has several possible paths to flow. ● Supernode: a part of the circuit that contains several nodes and branches. ● Loop: a complete current path where current flows back to the start. ● Mesh: a loop in the circuit that does not contain any other loops. ● Ideal voltage source: can provide a constant terminal voltage that does not depend on the variables in its external circuit. Its current depends on variables in its external circuit, V ab ¼ V S , R S ¼ 0. ● Real voltage source: with a series internal resistance R S (R S ( R L ), the terminal voltage of the real voltage source is: V ab ¼ V S 7IR S . ● Ideal current source: can provide a constant output current Is that does not depend on the variations in its external circuit, R S ¼ ?. Its voltage depends on variations in its external circuit. ● Real current source: with an internal resistance R S in parallel with the ideal current source, R S )R L . Formulas ● Work: W ¼ FS ● Power: P ¼ W/t ● Electrical power: P ¼ IV ¼ I 2 R ¼ V 2 /R ● KVL #1: SV ¼ 0 ● Assign a þve sign for V or E if its reference polarity and loop direction are the same. ● Assign a 7ve sign for V or E if its reference polarity and loop direc- tion are opposite. ● KVL #2: SV ¼ SE ● Assign a þve sign for V if its reference polarity and loop direction are the same; assign a 7ve sign for V if its reference direction and loop direction are opposite. ● Assign a 7ve sign for E if its reference polarity and loop direction are the same; assign a þve sign for E if its polarity and loop direction are opposite. ● KCL #1: SI in ¼ 0 ● Assign a þve sign for I if current is entering the node. ● Assign a 7ve sign for I if current is exiting the node. 56 Understandable electric circuits ● KCL #2: SI in ¼ I out ● Assign a þve sign for I in if current is entering the node; assign a 7ve sign for I in if current is exiting the node. ● Assign a þve sign for I out if current is exiting the node; assign a 7ve sign for I out if current is entering the node. ● Some circuit quantities and their SI units ● The commonly used metric prefixes Experiment 2: KVL and KCL Objectives ● Construct and analyse series and parallel circuits. ● Apply Ohm’s law and plot I–V characteristics. ● Experimentally verify KVL. ● Experimentally verify KCL. ● Analyse experimental data, circuit behaviour and performance, and com- pare them to theoretical equivalents. Background information ● Ohm’s law: V ¼ IR Quantity Quantity symbol Unit Unit symbol Voltage V Volt V Resistance R Ohm O Charge Q Coulomb C Power P Watt W Energy W Joule J Electro motive force E or V S Volt V Conductance G Siemens S Resistivity r Ohm Á metre O Á m Prefix Symbol (abbreviation) Exponential (power of 10) pico p 10 712 nano n 10 79 micro m 10 76 milli m 10 73 kilo K 10 3 mega M 10 6 giga G 10 9 tera T 10 12 Basic laws of electric circuits 57 ● I–V characteristics: ● KVL: SV ¼ 0, or SV ¼ SE ● KCL: SI ¼ 0, or SI in ¼ SI out Equipment and components ● Digital multimeter (DMM) ● Breadboard ● DC power supply ● Switch ● Resistors: 240 O, 2.4 kO, 91 O, 2.7 kO, 3.9 kO and 910 O ● Some alligator clips ● Some wires and leads with banana-plug ends Notes: (apply these notes to all experiments in this book) ● The ammeter (function) of the multimeter should be connected to the circuit after the power supply has been turned off. ● The voltmeter (function) of the multimeter should be connected in parallel with the component to measure voltage, and ammeter (function) should be connected in series with the component to measure current. ● Turn off the power supply before doing any circuit rearrangement, other- wise it will damage or harm experimental devices and components. ● Connect the negative terminal of the power supply to the ground using the black wire, and connect the positive terminal of the power supply to the component using the red wire. Connect other circuit components using different colour wires other than red and black. ● Use the actual resistance values to do the calculation. Multimeter guide ● Recall: A multimeter is an electrical and electronic measuring instrument that combines functions of ammeter, voltmeter, ohmmeter, etc. ● Method for measuring voltage with a digital multimeter (voltmeter function): ● Turn on the power supply after components have been connected in the circuit. V I R 0 58 Understandable electric circuits ● Insert the multimeter’s leads into the sockets labelled COM and V/O and turn on the multimeter (Figure L2.1). ● Turn the central selector switch pointing to the voltage ranges with the DCV sign (DCV is for measuring DC voltage, and ACV is for mea- suring AC voltage), and where the estimated voltage value should be less than the maximum range. ● Make the measurement by connecting the component in parallel with the two leads of the multimeter. Connect or touch the red lead from the multimeter to terminal of the component, which is expected to have the more positive voltage, and connect or touch the black lead to the other terminal of the component. ● Read the displayed voltage value on the scale. ● Turn off the multimeter after the measurement. ● Method for measuring current with a digital multimeter (ammeter function): ● Insert the multimeter’s leads into the sockets labelled COM and A. ● Connect the multimeter in series with the resistor branch that is going to make a measurement. Figure L2.1 Multimeter Basic laws of electric circuits 59 ● Turn the central selector switch pointing to the current ranges (with mA or 10 A sign) where the estimated current value is closed to the maximum range. ● Turn on the power supply. ● Read the displayed current value on the scale. ● Turn off the multimeter after the measurement. Procedure Part I: Kirchhoff’s voltage law (KVL) 1. Use the resistor colour code to choose three resistors with resistor values listed in Table L2.1. Measure each resistor using the multimeter (ohmmeter function). Record the values in Table L2.1. 2. Construct the series circuit shown in Figure L2.2 on the breadboard. 3. Calculate circuit current and voltages cross each resistor in Figure L2.2 (assuming the switch is turned on). Record the values in Table L2.2. Table L2.1 Resistance R 1 R 2 R 3 Colour code resistor value 240 O 2.4 kO 91 O Measured value R 1 = 240 R 3 = 91 Ω E 9 V R 2 = 2.4 kΩ ϩ Ϫ Figure L2.2 A series circuit Table L2.2 I V R 1 V R 2 V R 3 V T Formula for calculations Calculated value Measured value 60 Understandable electric circuits 4. Set the power supply to 9 V, then turn on the switch, connect the multi- meter (voltmeter function) in parallel with each resistor and power supply, and measure voltages across each resistor and power supply. Record the values in Table L2.2. 5. Use the direct method or indirect method to measure the circuit current. Record the value in Table L2.2. ● Direct method: Connect the multimeter (ammeter function) in series with the circuit components, then turn on the switch and measure circuit current directly. ● Indirect method: Apply Ohm’s law to calculate the current with mea- sured voltage and resistance. 6. Use measured values to plot I–V characteristics for 240 O resistor. 7. Substitute the measured voltage values from Table L2.2 into KVL equa- tions to verify SV ¼ 0 and SV ¼ SE. Part II: Kirchhoff’s current law (KCL) 1. Construct a parallel circuit as shown in Figure L2.3 to the breadboard. 2. Calculate each branch current and total current in the circuit of Figure L2.3 (assuming the switch is turned on). Record the values in Table L2.3. 3. Set the power supply to 10 V, turn on the switch, measure each branch current and total current in the circuit by using direct or indirect methods (get the multimeter to function as an ammeter). Record the values in Table L2.3. 4. Substitute the measured current values from Table L2.3 into KCL equa- tions to verify SI ¼ 0 and I in ¼ SI out . 2.7 kΩ 3.9 kΩ 910 Ω 10 V Ϫ ϩ Figure L2.3 A parallel circuit Table L2.3 Current I 1 I 2 I 3 I T Formula for calculation Calculated values Measured values Basic laws of electric circuits 61 Conclusion Write your conclusions below: 62 Understandable electric circuits Chapter 3 Series–parallel resistive circuits Objectives After completing this chapter, you will be able to: ● identify series circuits, parallel circuits and series–parallel circuits ● know how to determine the equivalent resistance for series, parallel and series–parallel resistive circuits ● calculate the resistance, voltage, current and power for series, parallel and series–parallel resistive circuits ● understand and apply the voltage-divider (VDR) and current-divider (CDR) rules ● identify the wye (Y) and delta (D) circuits ● know the method of wye (Y) and delta (D) conversions ● apply the method of D–Y conversions to simplify bridge circuits ● understand the method for measuring the unknown resistance of a balanced bridge circuit 3.1 Series resistive circuits and voltage-divider rule Series, parallel and series–parallel resistive circuits are very often used electrical or electronic circuits. It is very important to construct electric circuits in dif- ferent ways as to make practical use of them. 3.1.1 Series resistive circuits A series circuit is the simplest circuit. It has all its elements connected in one loop of wire. It can be analogized by water flowing in a series of tanks con- nected by a pipe. The water flows through the pipe from tank to tank. The same amount of water will flow in each tank. The same is true of an electrical circuit. There is only one pathway by which charges can travel in a series cir- cuit. The same amount of charges will flow in each component of the circuit, such as a light bulb, i.e. the current flow is the same throughout the circuit, so it has just one current in the series circuit. Series circuit ● The components are connected one after the other. ● There is only one current path. ● The current flow through each component is always the same. For example, Figure 3.1 illustrates an electrical circuit with three light bulbs (resistors) connected in series. If an ammeter is connected behind each light bulb, once the power turns on, the same current reading will be read on each ammeter. If we swap the position of the ammeter and the light bulb, the ammeter will still read the same current value. A practical example of a series circuit is a string of old Christmas lights. Many practical series circuits may not be as easily identifiable as Figures 3.1 and 3.2(a). Figure 3.2(b and c) are also series circuits but drawn in different ways. As long as the circuit elements are connected one after the other, and there E Figure 3.1 Series circuit . . . (a) (d) (b) (c) R 1 R 2 R 3 R n ... E V 1 V 2 V 3 V n + − + − − + + − + - Figure 3.2 (a–c) Series resistive circuits and (d) series circuit 64 Understandable electric circuits is only one current path for the circuit, it is said that they are connected in series. It does not matter if there is a different arrangement of the elements. 3.1.1.1 Total series voltage The voltage across the source or power supply (total voltage) is equal to the sum of the voltage that drops across each resistor in a series circuit, i.e. the source voltage shared by each resistor. The terminal of the resistor connecting to the positive side of the voltage source is positive, and the terminal of the resistor connecting to the negative side of the voltage source is negative. The total voltage V T in the circuit of Figure 3.2(d) can be determined by Kirchhoff’s voltage law (KVL) and Ohm’s law. For n resistors connected in series, the total voltage will be as follows: Total series voltage (V T or E) V T ¼ E ¼ V 1 þV 2 þ Á Á Á þV n V T ¼ IR 1 þIR 2 þ Á Á Á þIR n ¼ IR T ð3:1Þ For a series resistive circuit, (3.1) of the total voltage gives V T ¼ IðR 1 þR 2 þ Á Á Á þR n Þ and R T ¼ R 1 þR 2 þ Á Á Á þR n R T is the mathematical equation for computing the total resistance (or equivalent resistance R eq ) of a series resistive circuit. 3.1.1.2 Total series resistance (or equivalent resistance) The total resistance (R T ) of a series resistive circuit is the sum of all resistances in the circuit. It is also called the equivalent resistance (R eq ) because this resistance is equivalent to the sum of all resistances when you look through the two terminals of the series resistive circuit. The equivalent resistance of a series resistive circuit is the amount of resistance that a single resistor would need to equal the overall effect of the all resistors that are present in the circuit. The total resistance of a series resistive circuit is always greater than any single resistance in that circuit. Total series resistance (R T ) or equivalent resistance (R eq ) R T ¼ R 1 þ R 2 þ Á Á Á þ R n Series–parallel resistive circuits 65 3.1.1.3 Series current From the definition of a series circuit we know that there is only one current path in a series circuit, that the current flowing through each element is always the same and that the current is always the same at any point in a series circuit. The current I flowing in a series resistive circuit, such as the one in Figure 3.2(d), can be determined from Ohm’s law as follows: Series current (I) I ¼ V T R T ¼ E R T ¼ V 1 R 1 ¼ V 2 R 2 ¼ Á Á Á ¼ V n R n 3.1.1.4 Series power Each of the resistors in a series circuit consumes power, which is dissipated in the form of heat. The total power (P T ) consumed by a series circuit is the sum of power dissipated by the individual resistor. Since this power must come from the source, the total power is actually the power supplied by the source. Mul- tiply the current I on both sides of the total voltage equation V T ¼ E ¼ V 1 þ V 2 þ Á Á Á þ V n , to get the total power P T ¼ IE ¼ IV 1 þ IV 2 þ Á Á Á þ IV n . Therefore, the total power in a series resistive circuit can be expressed as follows: Total series power (P T ) P T ¼ P 1 þ P 2 þ Á Á Á þ P n or P T ¼ IE ¼ I 2 R T ¼ (E 2 /R T ) The power dissipated by the individual resistor in a series resistive circuit is as follows: P 1 ¼ I 2 R 1 ¼ IV 1 ¼ V 2 1 R 1 P 2 ¼ I 2 R 2 ¼ IV 2 ¼ V 2 2 R 2 Á Á Á P n ¼ I 2 R n ¼ IV n ¼ V 2 n R n 66 Understandable electric circuits Example 3.1: A series resistive circuit is shown in Figure 3.3. Determine the following: (a) Total resistance R T (b) Current I in the circuit (c) Voltage across the resistor R 1 (d) Total voltage V T (e) Total power P T Solution: (a) R T ¼ R 1 þ R 2 þ R 3 ¼ (10 þ 20 þ 30)kO ¼ 60 kO (b) I ¼ (E/R T ) ¼ (60 V/60 kO) ¼ 1 mA (c) V 1 ¼ IR 1 ¼ (1 mA)(10 kO) ¼ 10 V (d) V T ¼ IR T ¼ (1 mA)(60 kO) ¼ 60 V, V T ¼ E ¼ 60 V (checked) (e) P T ¼ IE ¼ (1 mA)(60 V) ¼ 60 mW or P T ¼ I 2 R T ¼ (1 mA) 2 (60 kO) ¼ 60 mW (checked) 3.1.2 Voltage-divider rule (VDR) The VDR can be exhibited by using a pot (short for potentiometer). A pot is a variable resistor whose resistance across its terminals can be varied by turning a knob. A pot is connected to a voltage source, as shown in Figure 3.4(a). Using a voltmeter to measure the voltage across the pot, the voltage relative to the negative side of the 100 V voltage source is ½ E ¼ 50 V when the arrow (knob) is at the middle of the potentiometer. The voltage will increase when the arrow moves up, and the voltage will decrease when the arrow moves down. This is the principle of the voltage divider, i.e. the voltage divider is a design technique used to create different output voltages that is proportional R 2 = 20 kΩ E 60 V R 3 = 30 kΩ R 1 = 10 kΩ Figure 3.3 Circuit for Example 3.1 Series–parallel resistive circuits 67 to the input voltage. Actually, a potentiometer itself is an adjustable voltage divider. The circuit in Figure 3.4(a) is equivalent to (b) since R ¼ R 1 þ R 2 ¼ 100 kO. Therefore, the voltage divider means that the source voltage E or total voltage V T is divided according to the value of the resistors in the series circuit. The output voltage from the divider changes when any of the resistor values change. In Figure 3.4(b), I ¼ E R 1 þR 2 V 1 ¼ IR 1 ¼ E R 1 þR 2 R 1 ¼ E R 1 R 1 þR 2 ¼ 100 V 50 kO ð50 þ 50ÞkO ¼ 50 V V 2 ¼ IR 2 ¼ E R 1 þR 2 R 2 ¼ E R 2 R 1 þR 2 ¼ 100 V 50 kO ð50 þ 50ÞkO ¼ 50 V The above two equations are the VDRs for a series circuit of two resistors. When there are n resistors in series, using the same method we can obtain the general form of the VDR as follows: I ¼ E R 1 þR 2 þ Á Á Á þR n ¼ E R T V X ¼ IR X ¼ E R T R X ¼ E R X R T or V X ¼ V T R X R T where R X and V X are the unknown resistance and voltage, and R T and V T are the total resistance and voltage in the series circuit. E (a) (b) 100 V ← R = 100 kΩ E 100 V R 1 = 50 kΩ R 2 = 50 kΩ I Figure 3.4 Voltage divider 68 Understandable electric circuits VDR ● General form: V X ¼ V T R X R T or V X ¼ E R X R T ðX ¼ 1; 2; . . . ; nÞ ● When there are only two resistors in series: V 1 ¼ V T R 1 R 1 þR 2 ; V 2 ¼ V T R 2 R 1 þR 2 Note: The numerator of the VDR is always the unknown resistance (this is worth memorizing). Example 3.2: Use the VDR to determine the voltage drops across resistors R 2 and R 3 in the circuit of Figure 3.5. Solution: Use the general form of the VDR V X ¼ E(R X /R T ) V 2 ¼ E R 2 R T ¼ E R 2 R 1 þR 2 þR 3 ¼ 60 V 20 kO ð10 þ 20 þ 10Þ kO ¼ 30 V V 3 ¼ E R 3 R T ¼ E R 3 R 1 þR 2 þR 3 ¼ 60 V 10 kO ð10 þ 20 þ 10Þ kO ¼ 15 V The practical application of the voltage divider can be the volume control of audio equipment. The knob of the pot in the circuit will eventually let you adjust the volume of the audio equipment. E 60 V R 1 = 10 kΩ R 2 = 20 kΩ R 3 = 10 kΩ Figure 3.5 Circuit for Example 3.2 Series–parallel resistive circuits 69 3.1.3 Circuit ground Usually, there is a ground for each electric circuit. Electrical circuit grounding is important because it is always at zero potential (0 V), and provides a reference voltage level in which all other voltages in a circuit are measured. There are two types of circuit grounds: one is the earth ground and another is the common ground (or chassis ground). Since an equal number of negative and positive charges are distributed throughout the earth at any given time, the earth is an electrically neutral body. So the earth is always at zero potential (0 V) and measurements can be made by using earth as a reference. An earth ground usually consists of a ground rod or a conductive pipe driven into the soil. A chassis ground is a connection to the main chassis of a piece of electronic or electrical equipment, such as a metal plate. Chassis ground is also called common ground. All chassis grounds should lead to earth ground, so that it also provides a point that has zero voltage. The neutral point in the alternating circuit (AC) is an example of the common ground. The difference between these two grounds can be summarized as follows: ● Earth ground: Connecting one terminal of the voltage source to the earth. The symbol for it is: ● Common ground or chassis ground: The common point for all elements in the circuit. All the common points are electrically connected together through metal plates or wires. The symbol for the common point is: In a circuit, the voltage with the single-subscript notation (such as V b ) is the voltage drop from the point b with respect to ground. And the voltage with the double-subscripts notation (such as V bc ) is the voltage drop across the two points b and c (each point is represented by a subscript). Example 3.3: Determine V bc , V be and V b in the circuit of Figure 3.6. R 1 = 10 kΩ R 2 = 20 kΩ R 3 = 20 kΩ R 4 = 10 kΩ E 120 V a b e d c Figure 3.6 Circuit for Example 3.3 70 Understandable electric circuits Solution: V bc ¼V R 2 ¼ E R 2 R T ¼ E R 2 R 1 þR 2 þR 3 þR 4 ¼120 V 20 kO ð10 þ 20 þ 20 þ 10ÞkO ¼ 120 V 20 kO 60 kO ¼ 40 V V be ¼ E R 2 þR 3 þR 4 R T ¼ 120V ð20 þ 20 þ 10ÞkO 60 kO ¼ 100 V (Use the general form of the VDR V X ¼ E (R X /R T ). There the unknown voltage V X ¼ V be , and the unknown resistance R X ¼ R 2 þ R 3 þ R 4 ). V b ¼ V be ¼ 100 V Ground and voltage subscript notation ● Earth ground: connects to the earth (V ¼ 0). ● Common ground (chassis ground): the common point for all com- ponents in the circuit (V ¼ 0). ● Single-subscript notation: the voltage from the subscript with respect to ground. ● Double-subscript notation: the voltage across the two subscripts. 3.2 Parallel resistive circuits and the current-divider rule 3.2.1 Parallel resistive circuits If any one of the light bulbs or resistors burns out or is removed in a series resistive circuit, the circuit is broken, no charges or current will move through the circuit and the entire circuit would stop operating. This is because there is only one current path in a series circuit. Old style Christmas lights were often wired in series, so if one light bulb burned out, the whole string of lights went off. This is the main disadvantage of a series circuit. A parallel resistive circuit is composed of two or more series circuits con- nected to the same power source. Thus, it has more than one path for charges or current to follow. An obvious advantage of the parallel circuit is that burn out or removal of one bulb does not affect the other bulbs. They continue to operate because there is still a separate, independent path from the source to each of the other bulbs, so the other bulbs will stay lit. Parallel resistive circuits also can be analogized by flowing water. When water flowing in a river across small islands, the one water path will be divided by the islands and split into many more water paths. When the water has passed the islands, it will become a single water path again. This scenario is Series–parallel resistive circuits 71 illustrated in Figure 3.7(a). The circuit in Figure 3.7(b) is parallel resistive cir- cuit equivalent to Figure 3.7(a). In the parallel resistive circuit, the source current divides among the available resistive branches in different paths. The total current I as illustrated in Figure 3.7(b) leaves the positive terminal of the voltage source and flows to node a (or supernode – chapter 2), which is a connecting point for the four branches. At node a, the total current I divides into three currents I 1 , I 2 and I 3 . These three currents flow through their resistors and rejoin at node b. The total current then flows from b back to the negative terminal of the source. Note that the node does not have to physically be one single point. As long as several branches are connected together, then that part of the circuit is considered to be a node. From the parallel circuit in Figure 3.7(b), we can see that circuit elements (resistors) are connected in parallel if the ends of one element are connected directly to the corresponding ends of the other. Many practical series circuits may not be as easily identifiable as Figure 3.7(b), and circuits in Figure 3.8 are also parallel circuits but drawn in different ways. As long as the circuit elements are connected end to end and there are at least two current paths in the circuit, it is said that they are connected in par- allel. It does not matter if there are different arrangements of the elements. R 2 R 1 R 3 E (a) (b) I b I 1 I 2 I 3 a Figure 3.7 Parallel circuit R 1 R 4 R 2 R 3 R 2 R 3 R 1 R 2 R 1 Figure 3.8 Parallel resistive circuits 72 Understandable electric circuits A parallel circuit has two main advantages when compared with series circuits. The first is that a failure of one element does not lead to the failure of the other elements. The other is that more elements may be added in parallel without the need for increasing voltage. Parallel circuit ● The components are connected end to end. ● There are at least two current paths in the circuit. ● The voltage across each component is the same. 3.2.1.1 Parallel voltage Since all resistors in a parallel resistive circuit are connected between the two notes, the voltage between these two notes must be the same. In this case, the voltage drop across each resistor must be the same. All these must be equal to the supply voltage E for the parallel resistive circuit shown in Figure 3.9. If all the resistors are light bulbs and have the same resistances as in Figure 3.9, they will glow at the same brightness as they each receive the same voltage. The voltage drop across each resistor must equal the voltage of the source in a parallel resistive circuit. This can be expressed in the following mathema- tical equation: Parallel voltage V ¼ E ¼ V 1 ¼ V 2 ¼ Á Á Á ¼ V n 3.2.1.2 Parallel current If the parallel circuit was a river, the total volume of water in the river would be the sum of water in each branch (Figure 3.7(a)). This is the same with the current in the parallel resistive circuit. The total current is equal to the sum of currents in each resistive branch, and the total current entering and exiting parallel resistive circuit is the same. V 2 R 1 R n E I 1 I n … … I T I 2 V 1 V R 2 V n Figure 3.9 V and I in a parallel circuit Series–parallel resistive circuits 73 If resistances are different in each branch of a parallel circuit, the branch currents will be different. The branch currents can be determined by Ohm’s law, and the characteristics of the total current in a parallel circuit can be expressed in the following equations: Parallel current ● Branch currents: I 1 ¼ V R 1 ; I 2 ¼ V R 2 ; . . . ; I n ¼ V R n ● Total current: I T ¼ V R eq ¼ I 1 þI 2 þ Á Á Á þI n 3.2.1.3 Equivalent parallel resistance How much current is flowing through each branch in the parallel resistive circuit in Figure 3.9? It depends on the amount of resistance in each branch. The total resistance in a parallel circuit can be found by applying Ohm’s law to the equation of the total current: V R eq ¼ I 1 þI 2 þ Á Á Á þI n ¼ V R 1 þ V R 2 þ Á Á Á þ V R n ¼ V 1 R 1 þ 1 R 2 þ Á Á Á þ 1 R n Dividing the voltages on both sides of the equal sign in the above equation gives 1 R eq ¼ 1 R 1 þ 1 R 2 þ Á Á Á þ 1 R n Solving for R eq from the above equation will give the equivalent resistance for the parallel circuit: R eq ¼ 1 ð1=R 1 Þ þ ð1=R 2 Þ þ Á Á Á þ ð1=R n Þ So the total resistance of a set of resistors in a parallel resistive circuit is found by adding up the reciprocals of the resistance values and then taking the reci- procal of the total. It will be more convenient to use the conductance than the resistance in the parallel circuits. Since the conductance G ¼ 1/R, therefore, G eq ¼ 1 R eq ¼ 1 R 1 þ 1 R 2 þ Á Á Á þ 1 R n ¼ G 1 þG 2 þ Á Á Á þG n 74 Understandable electric circuits When there are only two resistors in parallel: R eq ¼ 1 ð1=R 1 Þ þ ð1=R 2 Þ ¼ 1 ðR 1 þR 2 Þ=ðR 1 R 2 Þ ¼ R 1 R 2 R 1 þR 2 Usually parallel can be expressed by a symbol of ‘//’ such as: R 1 // R 2 // Á Á Á // R n . Equivalent parallel resistance and conductance ● R eq ¼ 1 ð1=R 1 Þ þ ð1=R 2 Þ þ Á Á Á þ ð1=R n Þ ¼ R 1 ==R 2 == Á Á Á ==R n ● G eq ¼ G 1 þ G 2 þ Á Á Á þ G n When n ¼ 2: R eq ¼ R 1 R 2 R 1 þR 2 ¼ R 1 ==R 2 Note: The total resistance of the series resistive circuit is always greater than the individual resistance, and the total resistance of the parallel resistive circuit is always less than the individual resistance. So usually for parallel circuits the equivalent resistance is used instead of the total resistance. 3.2.1.4 Total parallel power The total power is the sum of the power dissipated by the individual resistors in a parallel resistive circuit. By multiplying a voltage V to both sides of the equation for total current I T ¼ I 1 þ I 2 þ Á Á Á þ I n , the total power equation for the parallel circuits is obtained as follows: I T V ¼ I 1 V þI 2 V þ Á Á Á þI n V ¼ P T Total parallel power P T ¼ P 1 þP 2 þ Á Á Á þP n or P T ¼ I T V ¼ I 2 T R eq ¼ V 2 R eq The power consumed by each resistor in a parallel circuit is expressed as: P 1 ¼ I 1 V ¼ I 2 1 R 1 ¼ V 2 R 1 ; P 2 ¼ I 2 V ¼ I 2 2 R 2 ¼ V 2 R 2 ; . . . ; P n ¼ I n V ¼ I 2 n R n ¼ V 2 R n Example 3.4: A parallel circuit is shown in Figure 3.10. Determine (a) R 2 , (b) I T and (c) P 3 , given R eq ¼ 1.25 kO, R 1 ¼ 20 kO, R 3 ¼ 2 kO and I 3 ¼ 18 mA. Series–parallel resistive circuits 75 Solution: (a) Since R 2 ¼ 1/G 2 , determine G 2 first. G eq ¼ G 1 þ G 2 þ G 3 G 2 ¼ G eq ÀG 1 ÀG 3 ¼ 1 R eq À 1 R 1 À 1 R 3 ¼ 1 1:25 kO À 1 20 kO À 1 2 kO ¼ 0:25 mS ; R 2 ¼ 1 G 2 ¼ 1 0:25 mS ¼ 4 kO R eq ¼ 1 ð1=R 1 Þ þ ð1=R 2 Þ þ ð1=R 3 Þ ¼ 1 ð1=20 kOÞ þ ð1=4 kOÞ þ ð1=2 kOÞ ¼1:25 kO ðprovedÞ (b) I T ¼ E R eq ¼ V 3 R eq ¼ I 3 R 3 R eq ¼ ð18 mAÞð2 kOÞ 1:25 kO ¼ 28:8 mA (c) P 3 ¼ I 2 3 R 3 ¼ (18 mA) 2 (2 kO) ¼ 648 mW 3.2.2 Current-divider rule (CDR) The VDR can be used for series circuits, and the CDR can be used for parallel circuits. As previously mentioned, parallel circuits can be analogized by flowing water. It is like a river, the water flowing through the river will be divided by small islands and the flow is split creating more water paths. This was shown in Figure 3.7(a). The equations of the current divider can be derived by the following: A parallel resistive circuit with n resistors was shown in Figure 3.9. In this circuit: I 1 ¼ E R 1 ; I 2 ¼ E R 2 ; . . . ; I n ¼ E R n R 3 R 2 R 1 E I T I 1 I 2 I 3 Figure 3.10 Figure for Example 3.4 76 Understandable electric circuits Inserting E ¼ I T R eq into the above equations gives: I 1 ¼ I T R eq R 1 ; I 2 ¼ I T R eq R 2 ; . . . ; I n ¼ I T R eq R n These are the general form current-divider equations. When there are only two resistors in parallel: I 1 ¼ I T R eq R 1 ¼ I T ðR 1 R 2 =ðR 1 þR 2 ÞÞ R 1 ¼ I T R 2 R 1 þR 2 I 2 ¼ I T R eq R 2 ¼ I T ðR 1 R 2 =ðR 1 þR 2 ÞÞ R 2 ¼ I T R 1 R 1 þR 2 CDR ● General form: I X ¼ I T R eq R X or I X ¼ I T G X G eq ● When there are two resistors in parallel: I 1 ¼ I T R 2 R 1 þR 2 ; I 2 ¼ I T R 1 R 1 þR 2 There I X and R X are unknown current and resistance, and I T is the total current in the parallel resistive circuit. Note: The CDR is similar in form to the VDR. The difference is that the denominator of the general form current divider is the unknown resistance. When there are two resistors in parallel, the numerator is the other resistance (other than the unknown resistance). Recall the VDR: V X ¼ V T R X R T ; V 1 ¼ V T R 1 R 1 þR 2 ; V 2 ¼ V T R 2 R 1 þR 2 Series–parallel resistive circuits 77 Example 3.5: Determine the current I 1 , I 2 and I 3 in the circuit of Figure 3.11. Solution: R eq ¼ R 1 ==R 2 ==R 3 ¼ 1 ð1=R 1 Þ þ ð1=R 2 Þ þ ð1=R 3 Þ ¼ 1 ð1=ð10 OÞÞ þ ð1=ð20 OÞÞ þ ð1=ð30 OÞÞ % 5:455 O I 1 ¼ I T R eq R 1 ¼ 60 mA 5:455 O 10 O ¼ 32:73 mA I 2 ¼ I T R eq R 2 ¼ 60 mA 5:455 O 20 O % 16:37 mA I 3 ¼ I T R eq R 3 ¼ 60 mA 5:455 O 30 O ¼ 10:91 mA The conclusion that can be drawn from the above example is that the greater the branch resistance, the less the current flows through that branch, or the less the share of the total current. R 1 = 10 Ω R 2 = 20 Ω R 3 = 30 Ω E I 1 I 2 I 3 I T = 60 mA Figure 3.11 Circuit for Example 3.5 78 Understandable electric circuits Example 3.6: Determine the resistance R 2 for the circuit in Figure 3.12. Solution: Solve R 2 from the current-divider formula I 2 ¼ I T ½R 1 =ðR 1 þR 2 Þ I 2 ðR 1 þR 2 Þ ¼ I T R 1 I 2 R 2 ¼ I T R 1 ÀI 2 R 1 ; R 2 ¼ R 1 ðI T ÀI 2 Þ I 2 R 2 ¼ 10 Oð30 À 10ÞmA 10 mA ¼ 20 O 3.3 Series–parallel resistive circuits The most practical electric circuits are not simple series or parallel configurations, but combinations of series and parallel circuits, or the series–parallel configura- tions. Many circuits have various combinations of series and parallel components, i.e. circuit elements are series-connected in some parts and parallel in others. The series–parallel configurations have a variety of circuit forms, and some of them may be very complex. However, the same principles and rules or laws that have been introduced in the previous chapters are applied. The key to solving series–parallel circuits is to identify which parts of the circuit are series and which parts are parallel and then simplify them to an equivalent circuit and find an equivalent resistance. Series–parallel circuit The series–parallel circuit is a combination of series and parallel circuits. R 1 = 10 Ω R 2 E I T = 30 mA I 1 I 2 = 10 mA Figure 3.12 Circuit for Example 3.6 Series–parallel resistive circuits 79 3.3.1 Equivalent resistance Method for determining the equivalent resistance of series–parallel circuits: ● Determine the equivalent resistance of the parallel part of the series– parallel circuits. ● Determine the equivalent resistance of the series part of the series–parallel circuits. ● Plot the equivalent circuit if necessary. ● Repeat the above steps until the resistances in the circuit can be simplified to a single equivalent resistance R eq . Note: Determine R eq step by step from the far end of the circuit to the terminals of the R eq . Example 3.7: Analysis of the series–parallel circuit in Figure 3.13. InFigure 3.13(a), the resistor R 5 is inseries withR 6 andinparallel with R 4 and R 3 . This can be expressed by the equivalent circuit in Figure 3.13(b). R 2 is in series with (R 5 þ R 6 ) // R 4 // R 3 and in parallel with R 1 . That is the equivalent resistance R eq for the series–parallel circuit, i.e. R eq ¼ {[(R 5 þ R 6 ) // R 4 // R 3 ] þ R 2 } // R 1 . Example 3.8: Determine the equivalent resistance R eq for the circuit shown in Figure 3.14(a). Solution: R eq ¼ [(R 5 // R 6 þ R 4 ) // (R 7 þ R 8 ) þ R 3 ] // R 2 þ R 1 R 2 R 2 R 3 R 1 R 1 R 4 R 6 R 5 R eq (R 5 + R 6 ) // R 4 // R 3 (a) (b) Figure 3.13 Circuits for Example 3.7 (R 5 // R 6 + R 4 ) // (R 7 + R 8 ) E (a) (b) E R 3 R 2 R 1 R 5 R 4 R 6 R 7 R 8 R 2 R 3 R 1 Figure 3.14 Circuits for Example 3.8 80 Understandable electric circuits Example 3.9: Determine the R eq for the circuit shown in Figure 3.15(a). Solution: R eq ¼ [(R 3 // R 4 // R 5 ) þ R 2 ] // (R 6 þ R 7 ) þ R 1 3.3.2 Method for analysing series–parallel circuits After determining the equivalent resistance of the series–parallel circuit, the total current as well as currents and voltages for each resistor can be deter- mined by using the following steps: ● Apply Ohm’s law with the equivalent resistance solved from the previous section to determine the total current in the equivalent circuit I T ¼ E/R eq . ● Apply the VDR, CDR, Ohm’s law, KCL and KVL to determine the unknown currents and voltages. Example 3.10: Determine the currents and voltages for each resistor in the circuit of Figure 3.16. E (a) (b) E R 3 R 3 R 2 R 1 R 1 R 5 R 4 R 6 R 7 R 4 R 2 R 5 R 6 + R 7 Figure 3.15 Circuits for Example 3.9 E A B C I T I 1 I 3 I 4 I 2 R 3 R 2 R 1 R 5 R 4 Figure 3.16 Circuit for Example 3.10 Series–parallel resistive circuits 81 Solution: ● R eq ¼ (R 1 // R 2 ) þ [(R 4 þ R 5 ) // R 3 ] I T ¼ E R eq ● V R 1 ¼ V R 2 ¼ V AB ¼ I T ðR 1 ==R 2 Þ; I 1 ¼ V AB R 1 ; I 2 ¼ V AB R 2 or I 1 ¼ I T R 2 R 1 þR 2 ; I 2 ¼ I T R 1 R 1 þR 2 ðthe CDRÞ ● V R 3 ¼ V R 4 þV R 5 ¼ V BC ¼ I T ½ðR 4 þR 5 Þ==R 3 I 3 ¼ V BC R 3 ; I 4;5 ¼ V BC R 4 þR 5 Check: I T ¼ I 1 þI 2 or I T ¼ I 3 þI 4;5 ðKCLÞ V AB þV BC ¼ E ðKVLÞ Example 3.11: Determine the current I T in the circuit of Figure 3.17. Solution: I T ¼ I 5 + I 6 The method of analysis: I T ¼ ? I T ¼ I 5 þI 6 ðI 5 ¼ ?; I 6 ¼ ?Þ I 5 ¼ V AB R AB I 5 ¼ V AB R AB ; V AB ¼ E; R AB ¼ ? R AB ¼ ½ðR 1 ==R 2 þR 3 Þ==R 4 þR 5 I 6 ¼ V CD R CD I 6 ¼ V CD R CD ; V CD ¼ E; R CD ¼ ? R CD ¼ ½ðR 8 þR 9 Þ==R 7 þR 6 E A B C D I T I 5 I 6 R 3 R 2 R 1 R 5 R 4 R 6 R 7 R 8 R 9 Figure 3.17 Circuit for Example 3.11 82 Understandable electric circuits 3.4 Wye (Y) and delta (D) configurations and their equivalent conversions 3.4.1 Wye and delta configurations Sometimes the circuit configurations will be neither in series nor in parallel, and the analysis method for series–parallel circuits described in previous chapters may not apply. For example, the configuration of three resistors R a , R b and R c in the circuit of Figure 3.18(a) are neither in series nor in parallel. So how do we determine the equivalent resistance R eq for this circuit? If we convert this to the configuration of resistors R 1 , R 2 and R 3 in the circuit of Figure 3.18(b), the problem can be easily solved, i.e. R eq ¼ [(R 1 + R d ) // (R 2 þ R e )] þ R 3 . The resistors of R a , R b and R c in the circuit of Figure 3.18(a) are said to be in the delta (D) configuration; R 1 , R 2 and R 3 in the circuit of Figure 3.18(b) is called the wye (Y) configuration. The delta and wye designations are from the fact that they look like a triangle and the letter Y, respectively, in electrical drawings. They are also referred to as tee (T) and pi (p) circuits as shown in Figure 3.19. Wye (Y) and delta (D) configurations Y or T configuration: R 2 R 3 R 1 D or p configuration: R a R c R b Wye (Y) and delta (D) configurations are often used in three phase AC circuits. They can also be used in the bridge circuit that will be discussed later. It is very important to know the conversion method of the two circuits and be able to convert back and forth between the wye (Y) and delta (D) configurations. (a) (b) R a R c R b R d R e R eq = ? R d R 1 R 2 R 3 R e R eq = ? Figure 3.18 Delta (D) and wye (Y) conversions Series–parallel resistive circuits 83 3.4.2 Delta to wye conversion (D!Y) There are three terminals in the delta (D) or wye (Y) configurations that can be connected to other circuits. The delta or wye conversion is used to establish equivalence for the circuits with three terminals, meaning that the resistors of the circuits between any two terminals must have the same values for both circuits as shown in Figure 3.20. i:e: R ac ðYÞ ¼ R ac ðDÞ R ab ðYÞ ¼ R ab ðDÞ R bc ðYÞ ¼ R bc ðDÞ The following equations can be obtained from Figure 3.20: R ac ¼ R 1 þR 3 ¼ R b ==ðR a þR c Þ ¼ R b ðR a þR c Þ R b þ ðR a þR c Þ ð3:2Þ R ab ¼ R 1 þR 2 ¼ R c ==ðR a þR b Þ ¼ R c ðR a þR b Þ R c þ ðR a þR b Þ ð3:3Þ R bc ¼ R 2 þR 3 ¼ R a ==ðR b þR c Þ ¼ R a ðR b þR c Þ R a þ ðR b þR c Þ ð3:4Þ Δ ( Delta) R a R c R b π ( Pi ) R a R c R b Y ( Wye) R 3 R 1 R 2 T ( Tee) R 3 R 1 R 2 Figure 3.19 p and T configurations 84 Understandable electric circuits Subtracting equation (3.3) from the sum of equations (3.2) and (3.4) gives ðR 1 þR 3 Þ þ ðR 2 þR 3 Þ À ðR 1 þR 2 Þ ¼ R b R a þR b R c R b þ ðR a þR c Þ þ R a R b þR a R c R a þ ðR b þR c Þ À R c R a þR c R b R c þ ðR a þR b Þ 2R 3 ¼ 2R a R b R a þR b þR c R 3 ¼ R a R b R a þR b þR c ð3:5Þ Similarly, subtracting equation (3.4) from the sum of equations (3.2) and (3.3) gives R 1 ¼ R b R c R a þR b þR c ð3:6Þ And subtracting equation (3.2) from the sum of equations (3.3) and (3.4) gives R 2 ¼ R a R c R a þR b þR c ð3:7Þ The circuit in delta configuration is converted to wye configuration as shown in Figure 3.21. a b c R a R c R b R 3 R 1 R 2 Figure 3.20 Delta and wye configurations Series–parallel resistive circuits 85 Equations for D!Y R 1 ¼ R b R c R a þR b þR c ; R 2 ¼ R a R c R a þR b þR c ; R 3 ¼ R a R b R a þR b þR c 3.4.3 Wye to delta conversion (Y!D) The equations to convert wye to delta configuration can be derived based on (3.5–3.7). Adding the products of equations (3.7) and (3.5), (3.5) and (3.6), and (3.6) and (3.7) gives R 2 R 3 þR 3 R 1 þR 1 R 2 ¼ R a R c R a R b ðR a þR b þR c Þ 2 þ R a R b R b R c ðR a þR b þR c Þ 2 þ R b R c R a R c ðR a þR b þR c Þ 2 R 2 R 3 þR 3 R 1 þR 1 R 2 ¼ R a R b R c ðR a þR b þR c Þ ðR a þR b þR c Þ 2 ¼ R a R b R c R a þR b þR c ð3:8Þ a b c a b c a b c a b c R a R c R b R a R c R b R 1 R 2 R 3 R 1 R 2 R 3 Figure 3.21 Delta converted to wye configuration 86 Understandable electric circuits Using (3.8) to divide (3.6), (3.7) and (3.5) individually will give the equations to convert wye to delta configurations as follows: Equations for Y!D R a ¼ R 1 R 2 þR 2 R 3 þR 3 R 1 R 1 ; R b ¼ R 1 R 2 þR 2 R 3 þR 3 R 1 R 2 ; R c ¼ R 1 R 2 þR 2 R 3 þR 3 R 1 R 3 For example, (3.8) divided by (3.6) gives R 2 R 3 þR 3 R 1 þR 1 R 2 R 1 ¼ R a R b R c =ðR a þR b þR c Þ R b R c =ðR a þR b þR c Þ ¼ R a The circuit in wye configuration is converted to delta as shown in Figure 3.22. 3.4.3.1 R Y and R D If all resistors in the wye (Y) configuration have the same values, i.e. R 1 ¼ R 2 ¼ R 3 ¼ R Y , then all the resistances in the delta (D) configuration will also be the same, i.e. R a ¼ R b ¼ R c ¼ R D . This can be obtained from Y!D or D!Y conversion equations. Use Y!D as an example. R D ¼ R a ¼ R 1 R 2 þR 2 R 3 þR 3 R 1 R 1 ¼ 3R Y R Y R Y ¼ 3R Y R D ¼ R b ¼ R 1 R 2 þR 2 R 3 þR 3 R 1 R 2 ¼ 3R Y R Y R Y ¼ 3R Y R D ¼ R c ¼ R 1 R 2 þR 2 R 3 þR 3 R 1 R 3 ¼ 3R Y R Y R Y ¼ 3R Y c a (a) (b) b a b c R a R c R b R 1 R 2 R 3 Figure 3.22 Wye converted to delta configuration Series–parallel resistive circuits 87 So all delta resistances (R D ) have the same values. R a ¼ R b ¼ R c ¼ R D ¼ 3R Y In the above condition, the delta resistance R D and wye resistance R Y has the following relationship: If R a ¼ R b ¼ R c ¼ R D ; R 1 ¼ R 2 ¼ R 3 ¼ R Y R Y ¼ 1 3 R D or R D ¼ 3R Y Example 3.12: Convert D to Y in the circuit of Figure 3.22, then Y to D to prove the accuracy of the equations. There the delta resistances R a ¼ 30 O, R b ¼ 20 O and R c ¼ 10 O. Solution: D ! Y: R 3 ¼ R a R b R a þR b þR c ¼ ð30 OÞð20 OÞ 30 O þ 20 O þ 10 O ¼ 10 O R 2 ¼ R a R c R a þR b þR c ¼ ð30 OÞð10 OÞ 30 O þ 20 O þ 10 O ¼ 5 O R 1 ¼ R b R c R a þR b þR c ¼ ð20 OÞð10 OÞ 30 O þ 20 O þ 10 O % 3:33 O Y ! D: R a ¼ R 1 R 2 þR 2 R 3 þR 3 R 1 R 1 ¼ ½ð3:33Þð5Þ þ ð5Þð10Þ þ ð10Þð3:33Þ O 2 3:33 O ¼ 99:95O 2 3:33O % 30 O R b ¼ R 1 R 2 þR 2 R 3 þR 3 R 1 R 2 ¼ 99:95 O 2 5 O % 20 O R c ¼ R 1 R 2 þR 2 R 3 þR 3 R 1 R 3 ¼ 99:95 O 2 10 O % 10 O The calculated delta resistances R a ¼ 30 O, R b ¼ 20 O and R c ¼ 10 O are the same with the resistances that were given (proved). 88 Understandable electric circuits 3.4.4 Using D ! Y conversion to simplify bridge circuits Sir Charles Wheatstone (1802–1875), a British physicist and inventor, is most famous for the Wheatstone bridge circuit. He was the first person who imple- mented the bridge circuit when he ‘found’ the description of the device. The bridge was invented by Samuel Hunter Christie (1784–1865), a British scientist. The Wheatstone bridge circuit can be used to measure unknown resistors. A basic Wheatstone bridge circuit is illustrated in Figure 3.23(a). Example 3.13: Determine the equations to calculate the total current I T and branch current I 4 for the bridge circuit in Figure 3.23(a). To determine the equation for the total current I T of the bridge circuit, the equivalent resistance R eq has to be determined first. The normal series–parallel analysis methods cannot be used to determine this resistance. However, Figure 3.23(a) can be converted to Figure 3.23(b) using the D ! Y equivalent con- version, and R 1 , R 2 and R 3 in Figure 3.23(b) can be determined by the equa- tions of D ! Y conversion. R 1 ¼ R b R c R a þR b þR c ; R 2 ¼ R a R c R a þR b þR c ; R 3 ¼ R a R b R a þR b þR c The equivalent resistance R eq of the bridge can be determined as follows: R eq ¼ R 3 þ ½ðR 1 þR 4 Þ==ðR 2 þR 5 Þ The total current can be solved as I T ¼ E/R eq . (a) (b) B A E C R a R c R b R 4 I 4 I T R 5 B A E C I T R 4 I 4 R 1 R 2 R 3 R 5 Figure 3.23 Wheatstone bridge circuit. (a) Delta (D) and (b) wye (Y) Series–parallel resistive circuits 89 The branch current I 4 ¼ I T ððR 2 þR 5 Þ=½ðR 1 þR 4 Þ þ ðR 2 þR 5 ÞÞ (the CDR). If the wire between A and B in the circuit of Figure 3.23(a) is open, the equivalent resistance R eq will be R eq ¼ ðR b þR 4 Þ==ðR a þR 5 Þ 3.4.5 Balanced bridge When the voltage across points A and B in a bridge circuit shown in Figure 3.24 is zero, i.e. V AB ¼ 0, the Wheatstone bridge is said to be balanced. A balanced Wheatstone bridge circuit can accurately measure an unknown resistor. First determine the voltage V AB in the points A and B. The voltage V AB is the voltage from point A to ground (V A ) and then from ground to point B, i.e. V AB ¼ V A þ ðÀV B Þ V AB ¼ E R 2 R 1 þR 2 ÀE R 4 R 3 þR 4 ðthe VDRÞ V AB ¼ E R 2 ðR 3 þR 4 Þ ÀR 4 ðR 1 þR 2 Þ ðR 1 þR 2 ÞðR 3 þR 4 Þ ¼ E R 2 R 3 ÀR 4 R 1 ðR 1 þR 2 ÞðR 3 þR 4 Þ B A E R 1 R 2 R 3 R 4 I T Figure 3.24 A balanced bridge 90 Understandable electric circuits When V AB ¼ 0, or when the bridge is balanced, the numerator of the above equation will be zero, i.e. R 2 R 3 ÀR 4 R 1 ¼ 0; this gives: R 2 R 3 ¼ R 4 R 1 Balanced bridge When V AB ¼ 0, R 2 R 3 ¼ R 4 R 1 3.4.6 Measure unknown resistors using the balanced bridge The method of using the balanced bridge to measure an unknown resistor is as follows: If the unknown resistor is in the position of R 4 in the circuit of Figure 3.24, using a variable (adjustable) resistor to replace R 2 and connecting a Galvan- ometer in between terminals A and B can measure the small current I G in terminals A and B as shown in Figure 3.25. (A galvanometer is a type of ammeter that can measure small current accurately.) Adjust R 2 until the current I G measured by the Galvanometer or current in the A and B branch is zero (I G ¼ 0). This means V AB ¼ 0, or the bridge is balanced. R X ¼ R 4 at this time can be determined by the equation of the balanced bridge as follows: From: R 2 R 3 ¼ R 4 R 1 Solving for R 4 : R X ¼ R 4 ¼ R 2 R 3 R 1 B A E R 1 R 2 R 3 R X I T I G Figure 3.25 Measure an unknown R using a balanced bridge Series–parallel resistive circuits 91 So the unknown resistor value can be determined from the ratios of the resistances in a balanced Wheatstone bridge. Example 3.14: R 1 ¼ 100 O, R 2 ¼ 330 O and R 3 ¼ 470 O in a balanced bridge circuit as shown in Figure 3.25. Determine the unknown resistance R X . Solution: From R 2 R 3 ¼ R 4 R 1 solving for R 4 R X ¼ R 4 R 2 R 3 R 1 ¼ ð330 OÞð470 OÞ 100 O ¼ 1:551 kO Summary Series circuits ● Series circuits: All components are connected one after the other, there is only one circuit path, and the current flow through each component is always the same. ● Total series voltage: V T ¼ E ¼ V 1 þV 2 þ Á Á Á þV n ¼ IR T V T ¼ IR 1 þIR 2 þ Á Á Á þIR n ¼ IR T ● Total series resistance (equivalent resistance R eq ): R T ¼ R 1 þ R 2 þ Á Á Á þ R n ● Series current: I ¼ V T R T ¼ E R T ¼ V 1 R 1 ¼ V 2 R 2 ¼ Á Á Á ¼ V n R n ● Total series power: P T ¼ P 1 þP 2 þ Á Á Á þP n ¼ IE ¼ I 2 R T ¼ E 2 R T ● The VDR General form: V X ¼ V T R X R T or V X ¼ E R X R T ðX ¼ 1; 2; . . . ; nÞ When there are only two resistors in series: V 1 ¼ V T R 1 R 1 þR 2 ; V 2 ¼ V T R 2 R 1 þR 2 ● The earth ground: Connects to the earth (V ¼ 0). ● Common ground or chassis ground: the common point for all components in the circuit (V ¼ 0). 92 Understandable electric circuits ● Single-subscript notation: the voltage from the subscript with respect to ground. ● Double-subscript notation: the voltage across the two subscripts. Parallel circuits ● Parallel circuits: The components are connected end to end, there are at least two current paths in the circuit, and the voltage across each compo- nent is the same. ● Parallel voltage: V ¼ E ¼ V 1 ¼ V 2 ¼ Á Á Á ¼ V n ● Parallel currents: I 1 ¼ V R 1 ; I 2 ¼ V R 2 ; . . . ; I n ¼ V R n I T ¼ V R eq ¼ I 1 þI 2 þ Á Á Á þI n ● Equivalent parallel resistance: R eq ¼ 1 ð1=R 1 Þ þ ð1=R 2 Þ þ Á Á Á þ ð1=R n Þ ¼ R 1 ==R 2 == Á Á Á ==R n When n ¼ 2: R eq ¼ R 1 R 2 R 1 þR 2 ¼ R 1 ==R 2 ● Equivalent parallel conductance: G eq ¼ G 1 þ G 2 þ Á Á Á þ G n ● Total parallel power: P T ¼ P 1 þP 2 þ Á Á Á þP n ¼ I 2 T R eq ¼ V 2 R eq ¼ I T V ● The CDR General form: I X ¼ I T R eq R X or I X ¼ I T G X G eq (I X and R X are unknown current and resistance.) When there are two resistors in parallel: I 1 ¼ I T R 2 R 1 þR 2 ; I 2 ¼ I T R 1 R 1 þR 2 Series–parallel resistive circuits 93 Series–parallel circuits ● Series–parallel circuits are a combination of series and parallel circuits. ● Method for determining the equivalent resistance of series–parallel circuits: ● Determine the equivalent resistance of the parallel part of the series– parallel circuits. ● Determine the equivalent resistance of the series part of the series– parallel circuits. ● Plot the equivalent circuit if necessary. ● Repeat the above steps until the resistance in the circuit can be sim- plified to a single equivalent resistance R eq . ● Method for analysing series–parallel circuits: ● Apply Ohm’s law to determine the total current: I T ¼ E R eq ● Apply VDR, CDR, Ohm’s law, KCL and KVL to determine the unknown currents and voltages. Wye and delta configurations and their conversions ● Y or T circuit: R 2 R 3 R 1 , D or p circuit: R a R c R b ● D ! Y: R 1 ¼ R b R c R a þR b þR c ; R 2 ¼ R a R c R a þR b þR c ; R 3 ¼ R a R b R a þR b þR c ● Y ! D: R a ¼ R 1 R 2 þR 2 R 3 þR 3 R 1 R 1 ; R b ¼ R 1 R 2 þR 2 R 3 þR 3 R 1 R 2 ; R c ¼ R 1 R 2 þR 2 R 3 þR 3 R 1 R 3 ● If R a ¼ R b ¼ R c ¼ R D and R 1 ¼ R 2 ¼ R 3 ¼ R Y : R Y 1 3 R D or R D ¼ 3R Y ● The balanced bridge: When V AB ¼ 0, R 2 R 3 ¼ R 4 R 1 94 Understandable electric circuits Experiment 3: Series–parallel resistive circuits Objectives ● Review series and parallel resistive circuits. ● Construct and analyse series–parallel resistive circuits. ● Measure voltages and currents for series–parallel resistive circuits. ● Review the applications of KCL and KVL. ● Verify the theoretical analysis, and compare the experimental results with theory calculations. ● Apply the CDR to circuit analysis. ● Design and test a voltage divider. ● Measure unknown resistors using a Wheatstone bridge circuit. Background information ● Equivalent (or total) series resistance: R eq ¼ R T ¼ R 1 þ R 2 þ Á Á Á þ R n ● Equivalent parallel resistance: R eq ¼ 1 ð1=R 1 Þ þ ð1=R 2 Þ þ Á Á Á þ ð1=R n Þ ¼ R 1 ==R 2 == Á Á Á ==R n ● When there are only two resistors in parallel: R eq ¼ R 1 R 2 R 1 þR 2 ¼ R 1 ==R 2 ● For a balanced Wheatstone bridge: when V AB ¼ 0, R 2 R 3 ¼ R 4 R 1 ● CDR: I X ¼ I T R eq R X When there are only two resistors in parallel: I 1 ¼ I T R 2 R 1 þR 2 ; I 2 ¼ I T R 1 R 1 þR 2 ● VDR: V X ¼ V T R X R T or V X ¼ E R X R T When there are only two resistors in parallel: V 1 ¼ V T R 1 R 1 þR 2 ; V 2 ¼ V T R 2 R 1 þR 2 Series–parallel resistive circuits 95 Equipment and components ● Digital multimeter ● Breadboard ● DC power supply ● Switch ● Resistors: ● four resistors with any values, ● one 10 kO variable resistor, ● 360 O, 510 O, 5.1 kO, 750 O, 1.2 kO, 2.4 kO, 5.1 kO, 910 O, 2.4 kO, 6.2 and 9.1 kO each and ● two 1.1 kO. Procedure Part I: Equivalent series and parallel resistance 1. Take four unknown resistors and record their colour code resistor values in Table L3.1. 2. Get the multimeter to function as an ohmmeter and measure the values of these four resistors. Record the values in Table L3.1. 3. Connect four resistors in series with the DC power supply as shown in Figure L3.1. Adjust the source voltage to the suitable value according to the value of the resistors, and then connect the DC power supply to the circuit in Figure L3.1 (consult your instructor before you turn on the switch). 4. Calculate the equivalent series resistance R eq , voltage across each resistor V AB, , V BC , V CD , V DE , current I, and record in Table L3.2. Use direct method or indirect method to measure current I. Table L3.1 Resistance R 1 R 2 R 3 R 4 Colour code resistor value Measured value R 1 R 2 R 3 R 4 A B C D V S E Figure L3.1 A series circuit 96 Understandable electric circuits Recall: ● Direct method: Connect the multimeter (ammeter function) in series with the circuit components, turn on the switch and measure circuit current directly. ● Indirect method: Applying Ohm’s law to calculate current by using the measured voltage and resistance. 5. Turn on the switch for the circuit in Figure L3.1, get the multimeter function as an ohmmeter, voltmeter and ammeter, respectively, and mea- sure R eq , V AB, , V BC , V CD , V DE , current I. Record the values in Table L3.2. 6. Connect four resistors in parallel with the DC power supply as shown in Figure L3.2. 7. Calculate the equivalent parallel resistance R eq , currents I R 1 ; I R 2 ; I R 3 ; I R 4 and I T . Record the values in Table L3.3. 8. Turn on the switch for the circuit in Figure L3.2, measure R eq , I R 1 ; I R 2 ; I R 3 ; I R 4 and I T using the multimeter (ohmmeter and ammeter func- tions). Record the values in Table L3.3. Table L3.2 R eq V AB V BC V CD V DE I Formula for calculations Calculated value Measured value V S R 1 R 2 R 3 R 4 Figure L3.2 A parallel circuit Table L3.3 R eq I T I R 1 I R 2 I R 3 I R 4 Formula for calculations Calculated value Measured value Series–parallel resistive circuits 97 Part II: Series–parallel resistive circuit 1. Connect a series–parallel circuit as shown in Figure L3.3 on the breadboard. 2. Calculate the equivalent resistance R eq , currents I T , I 5.1 kO (current flowing through the branch of 5.1 kO resistor), and voltages V AB , V BC and V CD for the circuit in Figure L3.3. Record the values in Table L3.4. 3. Turn on the switch for the circuit in Figure L3.3, measure R eq , I T , I 5.1 kO , V AB , V BC and V CD . Record the values in Table L3.4. Part III: Voltage divider 1. Design and construct a voltage divider as shown in Figure L3.4. When E ¼ 12 V, V A ¼ 6 V (voltage across the resistor R 2 ), and I ¼ 6 mA, calculate resistance R 1 and R 2 . Record the values in Table L3.5. 2. Measure resistance R 1 and R 2 using the multimeter (ohmmeter function) for the circuit in Figure L3.4. Record the values in Table L3.5. 10 V 2.4 kΩ 1.2 kΩ 5.1 kΩ C B 510 Ω 360 Ω 750 Ω D A Figure L3.3 A series–parallel circuit Table L3.4 R eq I T V AB V BC V CD I 5.1 kO Formula for calculations Calculated value Measured value E 12 V R 1 R 2 A I = 6 mA Figure L3.4 Voltage divider circuit 98 Understandable electric circuits 3. Calculate voltages V R 1 and V R 2 for the circuit in Figure L3.4. Record the values in Table L3.5. 4. Measure voltages V R 1 and V R 2 using the multimeter (voltmeter function) for the circuit in Figure L3.4. Record the values in Table L3.5. Part IV: Wheatstone bridge 1. Measure the value of each resistor of R X using the multimeter (ohmmeter function) in Table L3.6. Record the values in Table L3.6. 2. Construct a bridge circuit as shown in Figure L3.5 on the breadboard, and connect the 910 O R X resistor (R 4 ¼ R X ) to the circuit. Table L3.5 R 1 R 2 V R 1 V R 2 Formula for calculations Calculated value Measured value Table L3.6 Colour code value for R X 910 O 2.4 kO 6.2 kO 9.1 kO Multimeter measured R X value Formula to calculate R 3 Multimeter measured R 3 value Bridge measured R X value B A E = 10 V R 1 = 1.1 kΩ R 2 = 1.1 kΩ R 3 R 4 = R X V Figure L3.5 Bridge circuit Series–parallel resistive circuits 99 3. Calculate the value of the variable resistor R 3 for the balanced bridge circuit (when R X ¼ 910 O, and V AB ¼ 0) in Figure L3.5. Record the value in Table L3.6. 4. Turn on the switch, and carefully adjust the variable resistor R 3 when using the multimeter (voltmeter function) to measure the voltage across term- inals A and B until the multimeter voltage is approximately zero. 5. Turn off the switch, use the multimeter (ohmmeter function) to measure the value of the variable resistor R 3 . Record the value in Table L3.6. 6. Calculate the value of R X when V AB ¼ 0 using the formula of the balanced bridge (use measured R 3 value). Record the values in Table L3.6. 7. Turn off the switch for the circuit in Figure L3.5, then replace the other three R X resistors listed in Table L3.6 one by one to the circuit, and repeat steps 3 to 6. 8. Compare the measured and calculated R X values. Are there any significant differences? If so, explain the reasons. Conclusion Write your conclusions below: 100 Understandable electric circuits Chapter 4 Methods of DC circuit analysis Objectives After completing this chapter, you will be able to: ● convert voltage source to an equivalent current source and vice versa ● know the methods of voltage sources in series and parallel ● know the methods of current sources in series and parallel ● understand the branch current analysis method and apply it to circuit analysis ● understand the mesh analysis method and apply it to circuit analysis ● understand the node voltage analysis method and apply it to circuit analysis 4.1 Voltage source, current source and their equivalent conversions 4.1.1 Source equivalent conversion It is sometimes easier to convert a current source to an equivalent voltage source or vice versa to analyse and calculate the circuits. The source equivalent conversion means that if loads are connected to both the terminals of the two sources after conversion, the load voltage V L and current I L of the two sources should be the same (Figure 4.1). So the source equivalent conversion actually means that the source terminals are equivalent, though the internal character- istics of each source circuit are not equivalent. If the internal resistance R S in Figure 4.1(a and b) is equal, the source vol- tage is E ¼ I S R S in Figure 4.1(a) and the source current is I S ¼ E/R S in Figure 4.1(b), then the current source and voltage source can be equivalently converted. When performing the source equivalent conversion, we need to pay attention to the polarities of the sources. The reference polarities of voltage and current of the sources should be the same before and after the conversion as shown in Figures 4.1 and 4.2 (notice the polarities of sources E and I S in the two figures). Source equivalent conversion ● Voltage source ! Current source R S ¼ R S ; I S ¼ E=R S ● Current source ! Voltage source R S ¼ R S ; E ¼ I S =R S The following procedure can verify that the load voltage V L and load current I L in two circuits of Figure 4.1(a and b) are equal after connecting a load resistor R L to the two terminals of these circuits. E R S R L a b (a) (b) I L + – R S R L a b V L + – V L I S I L Figure 4.1 Sources equivalent conversion R S a b (a) (b) R S a b I S E S Figure 4.2 Polarity of conversion 102 Understandable electric circuits ● The voltage source in Figure 4.1(a): I L ¼ E R S þR L V L ¼ E R L R S þR L ¼ I S R S R L R S þR L (Applying the voltage-divider rule and E ¼ I S R S ) ● The current source in Figure 4.1(b): I L ¼ I S R S R S þR L ¼ E R S þR L V L ¼ I L R L ¼ I S R S R S þR L R L (Applying the current-divider rule and E ¼ I S R S ) So the load voltages and currents in the two circuits of Figure 4.1(a and b) are the same, and the source conversion equations have been proved. Example 4.1: Convert the voltage source in Figure 4.3(a) to an equivalent current source and calculate the load current I L for the circuit in Figure 4.3(a and b). Solution: The equivalent current source after the source conversion is shown in Figure 4.3(b); R S is still 2 O in Figure 4.3(b). For Figure 4.3(b): I S ¼ E R S ¼ 6 V 2 O ¼ 3 A I L ¼ I S R S R S þR L ¼ 3 A 2 O ð2 þ 10ÞO ¼ 0:5 A E = 6 V a b (a) (b) I L R S R L = 10 Ω R L = 10 Ω R S = 2 Ω a b I S I L Figure 4.3 Circuit for Example 4.1 Methods of DC circuit analysis 103 For Figure 4.3(a): I L ¼ E R S þR L ¼ 6 V ð2 þ 10ÞO ¼ 0:5 A Example 4.2: Convert the current source in Figure 4.4(a) to an equivalent voltage source, and determine the voltage source E S and internal resistance R S in Figure 4.4(b). Solution: R S ¼ 10 O E S ¼ I S R S ¼ ð5 AÞð10 OÞ ¼ 50 V 4.1.2 Sources in series and parallel 4.1.2.1 Voltage sources in series A circuit of voltage sources in series and its equivalent circuit are shown in Figure 4.5. Voltage sources connected in series are similar with the resistors connected in series, that is the equivalent internal resistance R S for series vol- tage sources is the sum of the individual internal resistances: R S ¼ R S1 þR S2 þ Á Á Á þR Sn and the equivalent voltage E or V S for series voltage sources is the algebraic sum of the individual voltage sources: E ¼ E 1 þE 2 þ Á Á Á þE n or V S ¼ V 1 þV 2 þ Á Á Á þV n R 2 = 10 Ω R 1 = 5 Ω I S = 5 A R S = 10 Ω E 1 = 20 V E 1 = 20 V R 1 = 5 Ω R S = 10 Ω E 2 = 50 V (a) (b) R 2 = 10 Ω Figure 4.4 Circuit for Example 4.2 104 Understandable electric circuits Assign a positive sign (+) if the individual voltage has the same polarity as the equivalent voltage E (or V S ); assign a negative sign (7) if the individual voltage has a different polarity from the equivalent voltage E (or V S ) as shown in Figure 4.5. A flashlight is an example of voltage sources in series, where batteries are connected in series to increase the total equivalent voltage. Voltage sources in series R S ¼ R S1 þR S2 þ Á Á Á þR Sn E ¼ E 1 þE 2 þ Á Á Á þE n or V S ¼ V 1 þV 2 þ Á Á Á þV n ● Assign a +ve sign if E n has same polarity as E (or V S ) ● Assign a 7ve sign if E n has different polarity from E (or V S ) 4.1.2.2 Voltage sources in parallel A circuit of voltage sources in parallel and its equivalent circuit are shown in Figure 4.6. The equivalent voltage for the parallel voltage sources is the same as the voltage for each individual voltage source: E ¼ E 1 ¼ E 2 ¼ Á Á Á ¼ E n or V S ¼ V S1 ¼ V S2 ¼ Á Á Á ¼ V Sn and the equivalent internal resistance R S is the individual internal resistances in parallel: R S ¼ R S1 == R S2 == Á Á Á == R Sn Note: Only voltage sources that have the same values and polarities can be connected in parallel by using the method mentioned above. If voltage sources R S1 E n E 1 R Sn . . . R S2 R S = R S1 + R S2 + ... + R Sn E = E 1 – E 2 + ... + E n E 2 (a) (b) Figure 4.5 Voltage sources in series Methods of DC circuit analysis 105 having different values and polarities are connected in parallel, it can be solved by using Millman’s theory, which will be discussed in chapter 5 (section 5.4). An example of an application for voltage sources connected in parallel is for boosting (or jump starting) a ‘dead’ vehicle. You may have experienced using jumper cables by connecting the dead battery in parallel with a good car battery or with a booster (battery charger) to recharge the dead battery. It is the process of using the power from a charged battery to supplement the power of a discharged battery. It can provide twice the amount of current to the battery of the ‘dead’ vehicle and successfully start the engine. Voltage sources in parallel R S þR S1 == R S2 == Á Á Á == R Sn E ¼ E 1 ¼ E 2 ¼ Á Á Á ¼ E n or V S ¼ V S1 ¼ V S2 ¼ Á Á Á ¼ V S n Only voltage sources that have the same values and polarities can be in parallel. 4.1.2.3 Current sources in parallel A circuit of current sources in parallel and its equivalent circuit are shown in Figure 4.7. Current sources connected in parallel can be replaced by a single equivalent resistance R S in parallel with a single equivalent current I S . R S1 R Sn R S2 I S = I S1 – I S2 +...+ I Sn ... I S1 I S2 I Sn R S = R S1 // R S2 // ... R Sn (a) (b) Figure 4.7 Current sources in parallel R S1 E 2 E n E 1 R Sn R S2 R S = R S1 // R S2 // ... // R Sn E = E 1 = E 2 = ... = E n ... Figure 4.6 Voltage sources in parallel 106 Understandable electric circuits The equivalent resistance R S is the individual internal resistances in parallel: R S ¼ R S1 == R S2 == Á Á Á == R Sn The equivalent current I S is the algebraic sum of the individual current sources: I S ¼ I S1 þI S2 þ Á Á Á þI Sn Assign a positive sign (+) if the individual current is in the same direction as the equivalent current I S ; assign a negative sign (7) if the individual current is in a different direction from the equivalent current I S . Current sources in parallel R S ¼ R S1 == R S2 == . . . == R Sn I S ¼ I S1 þI S2 þ Á Á Á þI Sn ● Assign a +ve sign for I Sn if it has the same polarity as I S ● Assign a 7ve sign for I Sn if it has different polarity from I S 4.1.2.4 Current sources in series Only current sources that have the same polarities and same values can be con- nected in series. There is only one current path in a series circuit, so there must be only one current flowing through it. This is the same concept as Kirchhoff’s current law (KCL), otherwise if the current entering point A does not equal the current exiting point A in Figure 4.8, KCL would be violated at point A. Current sources in series Only current sources that have the same polarities and values can be connected in series. 3 A 7 A ? A Figure 4.8 KCL is violated at point A Methods of DC circuit analysis 107 Example 4.3: Determine the load voltage V L in Figure 4.9. Solution: The process of source equivalent conversion is shown in the circuit of Figure 4.9. Determine V L by using the voltage-divider rule as follows: V L ¼ V ab ¼ IR L ¼ E R T R L ¼ ðÀ4 þ 12 À 2ÞV ð2 þ 2 þ 4 þ 2ÞO ð2 OÞ ¼ 1:2 V 4.2 Branch current analysis The methods of analysis stated in chapter 3 are limited to an electric circuit that has a single power source. If an electric circuit or network has more than one source, it can be solved by the circuit analysis techniques that are discussed in chapters 4 and 5. The branch current analysis is one of several basic methods for analysing electric circuits. The branch current analysis is a circuit analysis method that writes and solves a system of equations in which the unknowns are the branch currents. This method applies Kirchhoff’s laws and Ohm’s law to the circuit and solves the branch currents from simultaneous equations. Once the branch currents 4 Ω 2 Ω 8 Ω 8 Ω 2 A 4 A 4 Ω 4 V a b + – + – V L = ? 2 V 2 V 2 + 4 = 6 A 2 Ω 4 V 8//8 = 4 Ω 2 V 4//4 = 2 Ω R L = 2 Ω R L = 2 Ω V L 2 Ω 4 V + – 2 Ω (2 Ω) (6 A) = 12 V 4 Ω 2 V R L = 2 Ω a a b b V L Figure 4.9 Circuit for Example 4.3 108 Understandable electric circuits have been solved, other circuit quantities such as voltages and powers can also be determined. The branch current analysis technique will use the terms node, branch and independent loop (or mesh); let us review the definitions of these terms. ● Node: The intersectional point of two or more current paths where current has several possible paths to flow. ● Branch: A current path between two nodes where one or more circuit components is in series. ● Loop: A complete current path that allows current to flow back to the start. ● Mesh: A loop in the circuit that does not contain any other loop (it can be analysed as a windowpane). The circuits in Figure 4.10 have three meshes (or independent loops) and different number of nodes (the dark dots). Branch current analysis A circuit analysis method that writes and solves a system of Kirchhoff’s current law (KCL) and voltage law (KVL) equations in which the unknowns are the branch currents (it can be used for a circuit that has more than one source). 4.2.1 Procedure for applying the branch circuit analysis 1. Label the circuit. ● Label all the nodes. ● Assign an arbitrary reference direction for each branch current. ● Assign loop direction for each mesh (choose clockwise direction). 2. Apply KCL to numbers of independent nodes (n 71), where n is the number of nodes. 3. Apply KVL to each mesh (or windowpane), and the number of KVL equations should be equal to the number of meshes, or Equation # ¼ branch # – (nodes # 71). 1 2 3 1 2 3 1 2 3 Figure 4.10 Nodes and meshes Methods of DC circuit analysis 109 4. Solve the simultaneous equations resulting from steps 2 and 3, using determinant or substitution methods to determine each branch current. 5. Calculate the other circuit unknowns from the branch currents in the problem if necessary. The procedure of applying the branch current analysis method is demon- strated in the following example. Example 4.4: Use the branch current analysis method to determine each branch current, power on resistor R 2 and also the voltage across the resistor R 1 in the circuit of Figure 4.11. Solution: This circuit contains two voltage sources, and cannot be solved by using the methods we have learned in chapter 3; let us try to use the branch current analysis method. 1. Label the circuit as shown in Figure 4.11. ● Label the nodes a and b. ● Assign an arbitrary reference direction for each branch current as shown in Figure 4.11. ● Assign clockwise loop direction for each mesh as shown in Figure 4.11. 2. Apply KCL to n À 1 ð Þ ¼ 2 À 1 ð Þ ¼ 1 number of independent nodes (there are two nodes a and b, and n ¼ 2): I 1 þI 2 ¼ I 3 ð4:1Þ 3. Apply KVL to each mesh (windowpane). The number of KVL equations should be equal to the number of meshes. As there are two meshes in Figure 4.11, we should write two KVL equations. Or Equation # ¼ branch # À ðnodes # À 1Þ ¼ 3 À ð2 À 1Þ ¼ 2 Mesh 1: I 1 R 1 þI 3 R 3 ÀE 1 ¼ 0 ð4:2Þ Mesh 2: ÀI 2 R 2 ÀI 3 R 3 þE 2 ¼ 0 ð4:3Þ R 1 = 2 Ω R 2 = 2 Ω R 3 = 3 Ω I 1 I 3 I 2 E 2 = 5 V E 1 = 10 V 1 2 a b Figure 4.11 Circuit for Example 4.4 110 Understandable electric circuits Recall KVL #1 (SV ¼ 0): Assign a positive sign (+) for E or V ¼ IR if its reference polarity and loop direction are the same; otherwise assign a negative sign (7). 4. Solve the simultaneous equations resulting fromsteps 2 and 3, and determine branch currents I 1 , I 2 and I 3 (three equations can solve three unknowns). ● Rewrite the above three equations in standard form: I 1 þI 2 ÀI 3 ¼ 0 I 1 R 1 þ 0 þI 3 R 3 ¼ E 1 0 ÀI 2 R 2 ÀI 3 R 3 ¼ ÀE 2 ● Substitute the values into equations: I 1 þI 2 ÀI 3 ¼ 0 2I 1 þ 0 þ 3I 3 ¼ 10 V 0 À 2I 2 À 3I 3 ¼ À5 V ● Solve simultaneous equations using determinant method: D ¼ 1 1 À1 2 0 3 0 À2 À3 ¼ ð1Þð0ÞðÀ3Þ þ ð2ÞðÀ2ÞðÀ1Þ þ ð0Þð1Þð3Þ À ðÀ1Þð0Þð0Þ À ð3ÞðÀ2Þð1Þ À ðÀ3Þð2Þð1Þ ¼ 4 À ðÀ6Þ À ðÀ6Þ ¼ 16 I 1 ¼ 0 1 À1 10 0 3 À5 À2 À3 D ¼ ð10ÞðÀ2ÞðÀ1Þ þ ðÀ5Þð3Þð1Þ À ðÀ3Þð10Þð1Þ 16 % 2:19 A I 2 ¼ 1 0 À1 2 10 3 0 À5 À3 D ¼ ð1Þð10ÞðÀ3Þ þ ð2ÞðÀ5ÞðÀ1Þ À ðÀ3ÞðÀ5Þð1Þ 16 % 0:31 A Methods of DC circuit analysis 111 I 3 ¼ 1 1 0 2 0 10 0 À2 À5 D ¼ Àð10ÞðÀ2Þð1Þ À ðÀ5Þð2Þð1Þ 16 % 1:88A I 1 % 2:19 A; I 2 % À0:31 A; I 3 % 1:88 A (Negative sign (7) for I 2 indicates that the actual direction of I 2 is opposite with its assigned reference direction.) 5. Calculate the other circuit unknowns from the branch currents: P 2 ¼ I 2 2 R 2 ¼ ðÀ0:31Þ 2 ð2Þ % 0:19 W V 1 ¼ I 1 R 1 ¼ ð2:19Þð2Þ ¼ 4:38 V Example 4.5: Determine current I 3 in Figure 4.12 using the branch current analysis. Solution: 1. Label the nodes, reference direction for branch currents and loop direc- tions in the circuit as shown in Figure 4.12. 2. Apply KCL to n À 1 ð Þ ¼ 2 À 1 ð Þ ¼ 1 number of independent nodes (there are two nodes or supernodes a and b, so n ¼ 2): ÀI 1 þI 2 þI 3 þI 4 ¼ 0 I 4 ¼ 6 A ð Þ: 3. Apply KVL around each mesh (or windowpanes); there are three meshes in Figure 4.12, so you should write three KVL equations. There is no need to write KVL for mesh 1 since mesh 1 current is already known to be equal to the source current I 1 (I 1 ¼ 5 A); therefore, the number of loop equations can be reduced from 3 to 2: Mesh 1: ÀI 1 R 1 þE 1 þE 2 ÀI 2 R 2 ¼ 0 Mesh 2: I 2 R 2 ÀE 2 ÀI 3 R 3 ¼ 0 R 3 = 0.5 Ω E 1 = 2.5 V 2 3 R 2 = 1.5 Ω I 4 = 6 A 1 a b R 1 = 1 Ω E 2 = 4 V I 1 I 2 I 3 Figure 4.12 Circuit for Example 4.5 112 Understandable electric circuits 4. Solve the simultaneous equations resulting from steps 2 and 3, and deter- mine the branch current I 2 . ÀI 1 À 1:5I 2 ¼ À2:5 À 4 ÀI 1 À 1:5I 2 þ 0I 3 ¼ À6:5 0I 1 þ 1:5I 2 À 0:5I 3 ¼ 4 0I 1 þ 1:5I 2 À 0:5I 3 ¼ 4 ÀI 1 þI 2 þI 3 ¼ À6 ÀI 1 þI 2 þI 3 ¼ À6 Solve the above simultaneous equations using the determinant method: D ¼ À1 À1:5 0 0 1:5 À0:5 À1 1 1 ¼ ðÀ1Þð1:5Þð1Þ þ ðÀ1ÞðÀ0:5ÞðÀ1:5Þ À ðÀ0:5Þð1ÞðÀ1Þ ¼ À2:75 I 2 ¼ À1 À6:5 0 0 4 À0:5 À1 À6 1 D ¼ ðÀ1Þð4Þð1Þ þ ðÀ1ÞðÀ0:5ÞðÀ6:5Þ À ðÀ0:5ÞðÀ6ÞðÀ1Þ À2:75 % 1:55 A I 2 % 1:55 A 4.3 Mesh current analysis The branch current analysis in section 4.2 is a circuit analysis method that writes and solves a system of KCL and KVL equations in which the unknowns are the branch currents. Mesh current analysis is a circuit analysis method that writes and solves a system of KVL equations in which the unknowns are the mesh currents (a current that circulates in the mesh). It can be used for a circuit that has more than one source. The branch current analysis is a fundamental method for understanding mesh current analysis; mesh analysis is more practical and easier to use. Mesh current analysis uses KVL and does not need to use KCL. Applying KVL to get the mesh equations and solve unknowns implies that it will have less unknown variables, less simultaneous equations and therefore less calculation than branch current analysis. After solving mesh currents, the branch currents of the circuit will be easily determined. Mesh current analysis A circuit analysis method that writes and solves a system of KVL equa- tions in which the unknowns are the mesh currents (it can be used for a circuit that has more than one source). Methods of DC circuit analysis 113 4.3.1 Procedure for applying mesh current analysis 1. Identify each mesh, and label all the nodes and reference directions for each mesh current (a current that circulates in the mesh) clockwise. 2. Apply KVL to each mesh of the circuit, and the number of KVL equations should be equal to the number of meshes (windowpanes). Or Equation # ¼ branch # À ðnodes # À 1Þ Assign a positive sign (+) for each self-resistor voltage, and a negative sign (7) for each mutual-resistor voltage in KVL equations. ● Self-resistor: A resistor that is located in a mesh where only one mesh current flows through it. ● Mutual resistor: A resistor that is located in a boundary of two meshes and has two mesh currents flowing through it. 3. Solve the simultaneous equations resulting from step 2 using determinant or substitution methods, and determine each mesh current. 4. Calculate the other circuit unknowns such as branch currents from the mesh currents in problem if necessary (choose the reference direction of branch currents first). Note: ● Convert the current source to the voltage source first in the circuit, if there is any. ● If the circuit has a current source, the source current will be the same as the mesh current, so the number of KVL equations can be reduced. The procedure for applying the mesh current analysis method is demon- strated in the following examples. Example 4.6: Use the mesh current analysis method to determine each mesh current and branch currents I R 1 , I R 2 and I R 3 in the circuit of Figure 4.13. Solution: 1. Label all the reference directions for each mesh current I 1 and I 2 (clockwise) as shown in Figure 4.13. R 2 = 10 Ω R 3 = 20 Ω E 3 = 10 V E 1 = 30 V I R3 I R2 I R1 a b E 2 = 20 V R 1 = 10 Ω I 1 I 2 Figure 4.13 Circuit for Example 4.6 114 Understandable electric circuits 2. Apply KVL around each mesh (windowpane), and the number of KVL equations is equal to the number of meshes (there are two meshes in Figure 4.13). Alternatively, use the number of KVL: [branch # 7(nodes #71)]=3 7 (2 71) ¼ 2. Assign a positive sign (+) for each self-resistor voltage, and a negative sign (7) for each mutual-resistor voltage in KVL (SV ¼ SE). Mesh 1: ðR 1 þR 2 Þ I 1 ÀR 2 I 2 ¼ ÀE 1 þE 2 Mesh 2: ÀR 2 I 1 þ ðR 2 þR 3 Þ I 2 ¼ ÀE 2 ÀE 3 Note: These equations were written by inspection of the circuit (inspection method): First column I 1 Second column I 2 Source E Mesh 1: (Self-resistor)I 1 7 (Mutual resistor)I 2 = 7E 1 þ E 2 Mesh 2: (Mutual resistor)I 1 + (Self-resistor)I 2 = 7E 2 7E 3 3. Solve the simultaneous equations resulting from step 2, and determine the mesh currents I 1 and I 2 : ð10 þ 10ÞI 1 À 10I 2 ¼ À30 þ 20 20I 1 À 10I 2 ¼ À10 ð4:4Þ À10I 1 þ ð10 þ 20ÞI 2 ¼ À20 À 10 À10I 1 þ 30I 2 ¼ À30 ð4:5Þ Solve for I 1 and I 2 using the substitution method as follows: ● Solve for I 1 from (4.4): 20I 1 ¼ À10 þ 10I 2 I 1 ¼ À 1 2 þ 1 2 I 2 ð4:6Þ ● Substitute I 1 into (4.5) and solve for I 2 : À10 À 1 2 þ 1 2 I 2 þ 30I 2 ¼ À30 I 2 ¼ À1:4 A ● Substitute I 2 into (4.6) and solve for I 1 : I 1 ¼ 1 2 þ 1 2 ðÀ1:4Þ I 1 ¼ À0:2 A Methods of DC circuit analysis 115 4. Assuming the reference direction of unknown branch current I R 2 as shown in Figure 4.13, calculate I R 2 from the mesh currents by applying KCL at node a: X I ¼ 0 : ÀI R 1 ÀI R 2 þI R 3 ¼ 0 or I 1 ÀI R 2 ÀI 2 ¼ 0 ðsince I 1 ¼ ÀI R 1 and I 2 ¼ ÀI R 3 Þ I R 2 ¼ I 1 ÀI 2 ¼ À0:2 À ðÀ1:4Þ ¼ 1:2 A I R 1 ¼ ÀI 1 ¼ 0:2 A; I R 3 ¼ ÀI 2 ¼ 1:4 A Example 4.7: Write the mesh equations using the mesh current analysis method for the circuit in Figure 4.14. Solution: Convert the current source to a voltage source as shown in Figure 4.14. 1. Label all the nodes and the reference directions for each mesh current (clockwise), as shown in Figure 4.14(b); 2. Apply KVL for each mesh (windowpane), and the number of KVL equations is equal to the number of meshes (there are three meshes in Figure 4.14(b)) Mesh 1: ðR 1 þR 2 þR 3 ÞI 1 ÀR 3 I 2 ÀR 2 I 3 ¼ E s ÀE Mesh 2: ÀR 3 I 1 þ ðR 3 þR 4 þR 5 ÞI 2 ÀR 4 I 3 ¼ E Mesh 3: ÀR 2 I 1 ÀR 4 I 2 þ ðR 2 þR 4 þR 6 ÞI 3 ¼ 0 4.4 Nodal voltage analysis The node voltage analysis is another method for analysis of an electric circuit with two or more sources. The node voltage analysis is a circuit analysis R 1 E R 2 R 4 E s R 2 R 4 E R 3 R 3 R 6 I s R 5 R 1 R 6 R 5 I 1 I 2 I 3 (a) (b) Figure 4.14 Circuit for Example 4.7 (E S ¼ I S R 1 ) 116 Understandable electric circuits method that writes and solves a set of simultaneous KCL equations in which the unknowns are the node voltages. Recall that node is the intersectional point of two or more current paths. Node voltage is voltage between a node and the reference node. Node voltage analysis A circuit analysis method that writes and solves a set of simultaneous KCL equations in which the unknowns are the node voltages (it can be used for a circuit that has more than one source). 4.4.1 Procedure for applying the node voltage analysis 1. Label the circuit. ● Label all the nodes and choose one of them to be the reference node. Usually ground or the node with the most branch connections should be chosen as the reference node (at which voltage is defined as zero). ● Assign an arbitrary reference direction for each branch current (this step can be skipped if using the inspection method). 2. Apply KCL to all n 71 nodes except for the reference node (n is the number of nodes). ● Method 1: Write KCL equations and apply Ohm’s law to the equations; either resistance or conductance can be used. Assign a positive sign (+) for the self- resistor or self-conductance voltage and a negative sign (7) for the mutual-resistor or mutual-conductor voltage. ● Method 2: Convert voltage sources to current sources and write KCL equations using the inspection method. 3. Solve the simultaneous equations and determine each nodal voltage. 4. Calculate the other circuit unknowns such as branch currents from the nodal voltages in the problem, if necessary. The procedure to apply node voltage analysis method is demonstrated in the following example. Example 4.8: Write the node voltage equations for the circuit shown in Figure 4.15(a) using node voltage analysis method. Solution: 1. Label nodes a, b and c, and choose ground c to be the reference node; assign the reference current directions for each branch as shown in Figure 4.15(a). 2. Apply KCL to n – 1 ¼ 3 – 1 ¼ 2 nodes (nodes a and b). Methods of DC circuit analysis 117 ● Method 1: Write KCL equations and apply Ohm’s law to the equations. Node a: I 1 ÀI 2 ÀI 3 ¼ 0; E 1 ÀV a R 1 À V a R 2 À V a ÀV b R 3 ¼ 0 Node b: I 3 ÀI 4 ÀI 5 ¼ 0; V a ÀV b R 3 À V b R 4 À V b þE 2 R 5 ¼ 0 Or use conductance (G ¼ 1/R) ðE 1 ÀV a ÞG 1 ÀV a G 2 À ðV a ÀV b ÞG 3 ¼ 0 ðV a ÀV b ÞG 3 ÀV b G 4 À ðV b þE 2 ÞG 5 ¼ 0 ● Method 2: Convert two voltage sources to current sources from Figure 4.15(a) to Figure 4.15(b), and write KCL equations by inspection. – Use conductance: First column (V a ) Second column (V b ) Source I S Node a: (G 1 þ G 2 þ G 3 )V a 7 G 3 V b = I 1 Node b: 7G 3 V a + (G 3 þ G 4 þ G 5 )V b = 7I 5 R 4 R 5 E 2 I 4 R 2 I 2 E 1 R 1 R 3 I 1 I 3 I 5 a b c R 2 R 1 R 4 a b c I 1 R 5 I 5 R 3 (a) (b) Figure 4.15 (a) Circuit for Example 4.8. (b) Circuit for method 2 118 Understandable electric circuits – Use resistance: 1 R 1 þ 1 R 2 þ 1 R 3 V a À 1 R 3 V b ¼ I 1 À 1 R 3 V a þ 1 R 3 þ 1 R 4 þ 1 R 5 V b ¼ ÀI 5 Note: The inspection method is similar with the one in mesh current analysis. The difference is that mesh current analysis uses mesh currents in each column, and node voltage analysis uses node voltage in each column. (Assign a positive sign (+) for the self-resistance/conductance voltage and entering node current, and a negative sign (7) for the mutual-conductor or mutual-resistor voltage and exiting node current.) 3. Two equations can solve two unknowns, which are the node voltages V a and V b . Example 4.9: Use the node voltage analysis to calculate resistances R 1 and R 2 , and current I 1 and I 2 for the circuit shown in Figure 4.16(a). R 1 R 1 = 12 Ω E R 2 I s R 2 = 24 Ω I s = 2 A E = 60 V I 2 I 1 c a b c d I 1 = 60/12 = 5 A I 1 ' R 1 = 12 Ω d (a) (b) I 1 I 2 (c) R 1 // R 2 I 2 a b b I 1 = 5 A I s = 2 A R 2 = 24 Ω a R 1 = 12 Ω I 1 = 5 A I 2 I s = 2 A Figure 4.16 Circuits for Example 4.9 Methods of DC circuit analysis 119 Solution: ● Label nodes a and b, and choose b to be the reference node, and assign the reference current direction for each branch as shown in Figure 4.16(b). ● Apply KCL to n 7 1 ¼ 2 7 1 ¼ 1 node (node a): ● Use method 1: Write KCL equations and apply Ohm’s law to the equations: I 1 ÀI 2 þI S ¼ 0; E ÀV a R 1 À V a R 2 þI S ¼ 0 ● Or use conductance: E ÀV a ð ÞG 1 ÀV a G 2 þI s ¼ 0 ● Solve the above equation and determine the node voltage V a : E R 1 À V a R 1 À V a R 2 þI S ¼ 0 E R 1 þI S ¼ V a 1 R 1 þ 1 R 2 V a ¼ ðE=R 1 Þ þI S ð1=R 1 Þ þ ð1=R 2 Þ ¼ ½ð60=12Þ þ 2A ½ð1=12Þ þ ð1=24ÞS ¼ 7 A 0:125 S ¼ 56 V ● Calculate the branch currents from the nodal voltages: I 1 ¼ E ÀV a R 1 ¼ ð60 À 56ÞV 12 O ¼ 0:33 A I 2 ¼ À V a R 2 ¼ À 56 V 24 O % À2:33 A ● Use method 2: Convert voltage source to current source from the circuit of Figure 4.16(a) to the circuit of Figure 4.16(c): I 1 ¼ E R 1 ¼ 60 12 ¼ 5 A R 1 ==R 2 ¼ 12==24 ¼ 8 O Write KCL equation to node a using the inspection method: V a R 1 ==R 2 ¼ I 1 þI S V a ¼ ðI 1 þI s Þ ðR 1 ==R 2 Þ ¼ ð5 A þ 2 AÞ ð8 OÞ ¼ 56 V (V a is the same as that from method 1) Example 4.10: Write node voltage equations with resistances and conductances in the circuit of Figure 4.17 using the inspection method. 120 Understandable electric circuits Solution: 1. Label all nodes a, b, c and d (n ¼ 4) in the circuit as shown in Figure 4.17, and choose d to be the reference node. (The step to assign each branch current with reference direction can be skipped since it is used for the inspection method.) 2. Write KCL equations to n 71 ¼ 4 71 ¼ 3 nodes using the inspection method. ● Use resistance: Node a: 1 R 1 þ 1 R 5 V a À 1 R 1 V b À 1 R 5 V c ¼ I S Node b: À 1 R 1 V a þ 1 R 1 þ 1 R 2 þ 1 R 3 V b À 1 R 3 V C ¼ 0 Node c: À 1 R 5 V a À 1 R 3 V b þ 1 R 3 þ 1 R 4 þ 1 R 5 V C ¼ 0 ● Use conductance: Node a: ðG 1 þG 5 ÞV a ÀG 1 V b ÀG 5 V c ¼ I S Node b: ÀG 1 V a þ ðG 1 þG 2 þG 3 ÞV b ÀG 3 V c ¼ 0 Node c: ÀG 5 V a ÀG 3 V b þ ðG 3 þG 4 þG 5 ÞV c ¼ 0 3. Three equations can solve three unknowns (node voltages V a , V b and V c ) 4.5 Node voltage analysis vs. mesh current analysis The choice between mesh current analysis and node voltage analysis is often made on the basis of the circuit structure: ● The node voltage analysis is preferable for solving a circuit that is a par- allel circuit, with current source(s), less nodes and more branches, and thus it is more convenient to solve the circuit unknowns. R 1 R 3 R 2 R 5 R 4 b c I s a d Figure 4.17 Circuit for Example 4.10 Methods of DC circuit analysis 121 ● The mesh current analysis is preferable for solving a circuit that has fewer meshes, more nodes, with voltage sources and requires solving circuit branch currents. Summary Source equivalent conversions and sources in series and parallel ● Voltage source ! Current source: R S ¼ R S ; I S ¼ E R S ● Current source ! Voltage source: R S ¼ R S ; E ¼ I S R S ● Voltage sources in series: R S ¼ R S1 þR S2 þ . . . þR S n Assign a positive sign (+) if it has the same polarity with E (or V S ), otherwise assign a negative sign (7). ● Voltage sources in parallel: R S ¼ R S1 ==R S2 == Á Á Á ==R Sn E ¼ E 1 ¼ E 2 ¼ Á Á Á E n Only voltage sources that have the same values and polarities can be in parallel. ● Current sources in series: Only current sources that have the same pola- rities and values can be in series. Branch current analysis A circuit analysis method that writes and solves a system of KCL and KVL equations in which the unknowns are the branch currents. The procedure for applying the branch current analysis is given in Section 4.2.1. Mesh current analysis A circuit analysis method that writes and solves a system of KVL equations in which the unknowns are the mesh currents (it can be used for a circuit that has more than one source). The procedure for applying the mesh current analysis is given in Section 4.3.1. Nodal voltage analysis A circuit analysis method that writes and solves a set of simultaneous of KCL equations in which the unknowns are the node voltages. The procedure for applying the nodal voltage analysis is given in Section 4.4.1. Note: The branch current analysis, mesh current analysis and node voltage analysis can be used for a circuit that has more than one source. 122 Understandable electric circuits Experiment 4: Mesh current analysis and nodal voltage analysis Objectives ● Construct circuits with two voltage sources. ● Experimentally verify the methods of solving a circuit with two power supplies. ● Experimentally verify the mesh current analysis method. ● Experimentally verify the node voltage analysis method. ● Analyse the experimental data, circuit behaviour and performance, and compare them to the theoretical equivalents. Equipment and components ● Multimeter ● Breadboard ● Dual-output DC power supply ● Switches (2) ● Resistors: 1.8 kO (2), 3 kO, 8.2 kO, 9.1 kO and 3.9 kO Background information ● Mesh current analysis: A circuit analysis method that writes and solves a system of KVL equations in which the unknowns are the mesh currents. It can be used for a circuit that has more than one source. ● Nodal voltage analysis: A circuit analysis method that writes and solves a set of simultaneous KCL equations in which the unknowns are the node voltages. It can be used for a circuit that has more than one source. Lab procedure Part I: Experimentally verify the mesh current analysis method 1. Construct a circuit as shown in Figure L4.1 on the breadboard. 2. Calculate voltage V R 2 and current I R 2 using the mesh current analysis method (assuming the switches are turned on). Record the values in Table L4.1. a b E 1 = 12 V E 2 = 5 V R 1 = 1.8 kΩ R 2 = 1.8 kΩ R 3 = 3 kΩ I 1 I R2 Figure L4.1 Circuit for mesh current analysis Methods of DC circuit analysis 123 3. Set outputs of the dual-output power supply to 12 and 5 V, respectively, and turn on the two switches. Connect the multimeter (voltmeter function) in parallel to resistor R 2 and measure V R 2 . Record the values in Table L4.1. 4. Use direct method or indirect method to measure current I R 2 . Record the value in Table L4.1. 5. Compare the measured values and calculated values; are there any sig- nificant differences? If so, explain the reasons. Part II: Experimentally verify the nodal voltage analysis method 1. Construct a circuit shown in Figure L4.2 on the breadboard. 2. Calculate the nodal voltage V a using the nodal voltage analysis method (assuming the two switches are turned on). Record the value in Table L4.2. 3. Calculate branch currents I 1 , I 2 and I 3 . Record the values in Table L4.2. Table L4.1 Circuit for mesh current analysis Quantity I R 2 V R 2 Formula for calculations Calculated value Measured value a b E 1 = 10 V E 2 = 12 V R 1 = 8.2 kΩ R 2 = 3.9 kΩ R 3 = 9.1 kΩ I 1 I 2 I 3 Figure L4.2 Circuit for nodal voltage analysis Table L4.2 Circuit for nodal voltage analysis V a I 1 I 2 I 3 Formula for calculations Calculated value Measured value 124 Understandable electric circuits 4. Set outputs of the dual-output power supply to 10 and 12 V, respectively, and then turn on the two switches. Connect the multimeter (voltmeter function) in parallel to resistor R 2 and measure V a . Record the values in Table L4.2. 5. Measure branch currents I 1 , I 2 and I 3 using either the direct method or indirect method. Record the values in Table L4.2. 6. Compare the measured values and calculated values; are there any sig- nificant differences? If so, explain the reasons. Conclusion Write your conclusions below: Methods of DC circuit analysis 125 Chapter 5 The network theorems Objectives After completing this chapter, you will be able to: ● understand the concept of linear circuits ● determine currents or voltages of networks using the superposition theorem ● understand Thevenin’s and Norton’s theorems and know how to convert their equivalent circuits ● determine currents or voltages of networks using Thevenin’s and Norton’s theorems ● understand the maximum power transfer theorem, and the method of transferring maximum power to the load ● determine currents or voltages of the parallel voltage source circuits using Millman’s theorem ● determine currents or voltages of networks using the substitution theorem The main methods for analysing series and parallel circuits in chapter 3 are Kirchhoff’s laws. The branch current method, mesh or loop analysis method and node voltage analysis method also use KCL and KVL as the main backbone. When the practical circuits are more and more complex, especially in multi-loop electric circuits, the applications of the above meth- ods solving for currents and voltages can be quite complicated. This is because you need to solve the higher-order mathematic equations when using these methods, i.e. you have to use complex algebra to handle multiple circuit unknowns. The scientists working in the field of electrical engineering have developed more simplified theorems to analyse these kinds of complex circuits (the com- plicated circuit is also called the network). This chapter presents several theo- rems useful for analysing such complex circuits or networks. These theorems include the superposition theorem, Thevenin’s theorem, Norton’s theorem, Millman’s theorem and the substitution theorem. In electrical network analysis, the fundamental rules are still Ohm’s law and Kirchhoff’s laws. Network A network is a complicated circuit. Linearity property The linearity property of a component describes a linear relationship between cause and effect. The pre-requirement of applying some of the above network theorems is that the analysed network must be a linear circuit. The components of a linear circuit are the linear components. An example of linear component is a linear resistor. The voltage and current (input/output) of this linear resistor have a directly proportional (a straight line) relationship. A linear circuit has an output that is directly proportional to its input. The linear circuit can also be defined as follows: as long as the input/ output signal timing does not depend on any characteristic of the input signal, it will be a linear circuit. 5.1 Superposition theorem 5.1.1 Introduction When several power sources are applied to a single circuit or network at the same time, the superposition theorem can be used to separate the original network into several individual circuits for each power source working separately. Then, use series/parallel analysis to determine voltages and currents in the modified circuits. The actual unknown currents and voltages with all power sources can be deter- mined by their algebraic sum; this is the meaning of the theorem’s name – ‘superimposed’. This method can avoid complicated mathematical calculations. Superposition theorem The unknown voltages or currents in a network are the sum of the vol- tages or currents of the individual contributions from each single power supply, by setting the other inactive sources to zero. 5.1.2 Steps to apply the superposition theorem 1. Turn off all power sources except one, i.e. replace the voltage source with the short circuit (placing a jump wire), and replace the current source with an open circuit. Redraw the original circuit with a single source. 2. Analyse and calculate this circuit by using the single source series–parallel analysis method. 128 Understandable electric circuits 3. Repeat steps 1 and 2 for the other power sources in the circuit. 4. Determine the total contribution by calculating the algebraic sum of all contributions due to single sources. (The result should be positive when the reference polarity of the unknown in the single source circuit is the same as the reference polarity of the unknown in the original circuit; otherwise it should be negative.) Note: The superposition theorem can be applied to the linear network to determine only the unknown currents and voltages. It cannot calculate power, since power is a nonlinear variable. Power can be calculated by the voltages and currents that have been determined by the superposition theorem. Example 5.1: Determine the branch current I c in the circuit of Figure 5.1(a) by using the superposition theorem. Solution: 1. Choose E 1 to apply to the circuit first and use a jump wire to replace E 2 as shown in Figure 5.1(b). = (a) (b) (c) I a I b I c E 1 = 48 V E 2 = 24 V R 1 = 8 Ω R 3 = 8 Ω R 4 = 8 Ω R 2 = 8 Ω R 5 = 16 Ω I a ′ I b ′ I c ′ E 1 = 48 V R 3 = 8 Ω R 4 = 8 Ω R 5 = 16 Ω R 1 = 8 Ω R 2 = 8 Ω I a ′′ I b ′′ I c ′′ E 2 = 24 V R 3 = 8 Ω R 4 = 8 Ω R 5 = 16 Ω R 1 = 8 Ω R 2 = 8 Ω + Figure 5.1 Circuit for Example 5.1 2. Determine I c 0 in the circuit of Figure 5.1(b): R 0 eq ¼ R 5 ==ðR 3 þ R 4 Þ þ ðR 1 þ R 2 Þ ¼ 16  ð8 þ 8Þ 16 þ ð8 þ 8Þ þ ð8 þ 8Þ O ¼ 24 O The network theorems 129 (View from the E 1 branch in the circuit of Figure 5.1(b) to determine R eq 0 .) I a 0 ¼ E 1 R eq 0 ¼ 48 V 24 O ¼ 2 A I c 0 ¼ I a 0 R 3 þ R 4 R 3 þ R 4 þ R 5 ¼ 2 A ð8 þ 8ÞO ð8 þ 8 þ 16ÞO ¼ 1 A 3. When E 2 is applied to the circuit, replace E 1 with a short circuit as shown in Figure 5.1(c) and calculate I c 00 : R eq 00 ¼ R 5 ==ðR 1 þ R 2 Þ þ ðR 3 þ R 4 Þ ¼ 16  ð8 þ 8Þ 16 þ ð8 þ 8Þ þ ð8 þ 8Þ O ¼ 24 O (View from the E 2 branch in the circuit of Figure 5.1(c) to determine R eq 00 .) I b 00 ¼ E 2 R 00 eq ¼ 24 V 24 O ¼ 1 A I c ¼ I b R 1 þ R 2 R 1 þ R 2 þ R 5 ¼ 1 A ð8 þ 8ÞO ð8 þ 8 þ 16ÞO ¼ 0:5 A 4. Calculate the sum of currents I c 0 and I c 00 : I c ¼ I 0 c þ I c ¼ ð1 þ 0:5ÞA ¼ 1:5 A Example 5.2: Determine the branch current I 2 and power P 2 of the circuit in Figure 5.2(a) by using the superposition theorem. Solution: 1. When E is applied only to the circuit (using an open circuit to replace the current source I 1 ), calculate I 2 0 by assuming the reference direction of I 2 0 as shown in Figure 5.2(b). 2. Determine I 2 0 in the circuit of Figure 5.2(b): I 2 0 ¼ E ðR 2 ==R 3 Þ þ R 1 ¼ 25 V ð100  100Þ=ð100 þ 100Þ ð Þ þ 50 ½ O ¼ 0:25 A ¼ 250 mA 130 Understandable electric circuits 3. When the current source I 1 is applied only to the circuit (the voltage source E is replaced by a jump wire), the circuit is as shown in Figure 5.2(c). Calculate I 2 0 by assuming the reference direction of I 2 00 as shown in the circuit of Figure 5.2(c): I 2 00 ¼ I 1 R 1 R 1 þ R 2 ==R 3 ¼ 50 mA 50 O 50 þ ð100  100Þ=ð100 þ 100Þ ½ O ¼ 25 mA (Apply the current divider rule to the branches R 1 and R 2 // R 3 .) 4. Calculate the sum of currents I 2 0 and I 2 00 : I 2 ¼ ÀI 2 0 þ I 2 00 ¼ À250 þ 25 ð ÞmA ¼ À225 mA ¼ À0:225 A I 2 0 is negative as its reference direction in Figure 5.2(b) is opposite to that of I 2 in the original circuit of Figure 5.2(a). The negative I 2 implies that the actual direction of I 2 in Figure 5.2(a) is opposite to its reference direction. Determine the power P 2 : P 2 ¼ I 2 2 R 2 ¼ ðÀ0:225 AÞ 2 ð100 OÞ % 5:06 W Example 5.3: Determine the branch current I 3 in the circuit of Figure 5.3(a) using the superposition theorem. = + R 3 = 100 Ω R 3 = 100 Ω R 3 = 100 Ω R 1 = 50 Ω E = 25 V I 1 = 50 mA 100 Ω R 2 I 2 R 1 = 50 Ω 100 Ω E = 25 V R 2 I 2 ′ I 1 = 50 mA R 1 = 50 Ω 100 Ω R 2 I 2 ′′ (a) (b) (c) Figure 5.2 Circuit for Example 5.2 The network theorems 131 Solution: 1. Choose E 1 to apply to the circuit first and use a jump wire to replace E 2 and an open circuit to replace the current source I as shown in Figure 5.3(b). 2. Use the circuit in Figure 5.3(b) to determine I 3 0 : I 3 0 ¼ I 1 0 R 2 R 2 þ ðR 3 þ R 4 Þ ¼ ð19:84 mAÞ 0:55 kO 0:55 kO þ ð0:375 þ 0:45ÞkO % 7:94 mA (Apply the current divider rule to the branches R 2 and (R 3 þ R 4 ).) There I 1 0 ¼ E 1 R eq 0 ¼ E 1 ðR 3 þ R 4 Þ==R 2 þ R 1 ¼ 12:5 V ½ðð0:375 þ 0:45Þ Â 0:55Þ=ðð0:375 þ 0:45Þ þ 0:55Þ þ 0:3kO % 19:84 mA R 1 R 3 R 2 R 1 R 3 R 4 + R 1 R 3 + E 1 R 4 E 2 R 2 R 4 I R 2 R 4 = 0.45 kΩ R 1 = 0.3 kΩ R 2 = 0.55 kΩ R 3 = 0.375 kΩ E 1 = 12.5 V E 2 = 5 V I 3 I = 5 mA = (a) (b) (c) (d) I 3 ′ I 1 ′ I 3 ′′ I 2 ′′ I 3 ′′′ Figure 5.3 Circuit for Example 5.3 132 Understandable electric circuits 3. . Use the circuit in Figure 5.3(c) to determine I 3 00 : I 3 00 ¼ I 2 00 R 1 R 1 þ ðR 3 þ R 4 Þ ¼ ð6:49 mAÞ 0:3 kO ½0:3 þ ð0:375 þ 0:45ÞkO % 1:73 mA (Apply the current divider rule to the branches R 1 and (R 3 þ R 4 ).) There I 2 00 ¼ E 2 R eq 00 ¼ E 2 ðR 3 þ R 4 Þ==R 1 þ R 2 ¼ 5 V ðð0:375 þ 0:45Þð0:3ÞÞ=ðð0:375 þ 0:45Þ þ 0:3Þ þ 0:55 ½ kO % 6:49 mA . Use the circuit in Figure 5.3(d) to determine I 3 000 : I 3 000 ¼ I R 4 ðR 1 ==R 2 þ R 3 Þ þ R 4 ¼ ð5 mAÞ 0:45 kO ðð0:3  0:55Þ=ð0:3 þ 0:55Þ þ 0:375Þ þ 0:45 ½ kO % 2:21 mA (Apply the current divider rule to the branches R 4 and (R 1 // R 2 þ R 3 ).) 4. Calculate the sum of currents I 3 0 , I 3 00 and I 3 000 : I 3 ¼ I 3 0 þ I 3 00 À I 3 000 ¼ 7:94 þ 1:73 À 2:21 ð ÞmA ¼ 7:46 mA I 3 000 is negative since its reference direction is opposite to that of I 3 in the original circuit of Figure 5.3(a). 5.2 Thevenin’s and Norton’s theorems 5.2.1 Introduction Thevenin’s and Norton’s theorems are two of the most widely used theorems to simplify the linear circuit for ease of network analysis. In 1883, French tele- graph engineer M. L. Thevenin published his theorem of network analysis method. Forty-three years later, American engineer E. L. Norton in Bell Tel- ephone laboratory published a similar theorem, but he used the current source to replace the voltage source in the equivalent circuit. These two theorems state that any complicated linear two-terminal network with power supplies can be The network theorems 133 simplified to an equivalent circuit that includes an actual voltage source (Thevenin’s theorem) or an actual current source (Norton’s theorem). Here, ‘the linear two-terminal network with power supplies’ means: ● network: the relatively complicated circuit ● linear network: the circuits in the network are the linear circuits ● two-terminal network: the network with two terminals that can be con- nected to the external circuits ● network with the power supplies: network includes the power supplies No matter how complex the inside construction of any two-terminal net- work with power supplies is, they can all be illustrated in Figure 5.4(a). According to Thevenin’s and Norton’s theorems, we can draw the following conclusion: any linear two-terminal network with power supplies can be replaced by an equivalent circuit as shown in Figure 5.4(b or c). The equivalent means that any load resistor branch (or unknown current or voltage branch) connected between the terminals of Thevenin’s or Norton’s equivalent circuit will have the same current and voltage as if it were connected to the terminals of the original circuit. Thevenin’s and Norton’s theorems allow for analysis of the performance of a circuit from its terminal properties only. R TH V TH I N R N a b a b a R L R L b R L (a) (b) (c) Figure 5.4 Thevenin’s and Norton’s theorems. (a) Linear two-terminal network with the power supply. (b) Thevenin’s theorem. (c) Norton’s theorem 134 Understandable electric circuits Any linear two-terminal network with power supplies can be replaced by a simple equivalent circuit, whichhas a single power source anda single resistor. ● Thevenin’s theorem: Thevenin’s equivalent circuit is a voltage source – with an equivalent resistance R TH in series with an equivalent voltage source V TH . ● Norton’s theorem: Norton’s equivalent circuit is a current source – with an equivalent resistance R N in parallel with an equivalent cur- rent source I N . In short we can conclude that any combination of power supplies and resistors with two terminals can be replaced by a single voltage source and a single series resistor for Thevenin’s theorem, and replaced by a single current source and a single parallel resistor for Norton’s theorem. The key to applying these two theorems is to determine the equivalent resistance R TH and the equivalent voltage V TH for Thevenin’s equivalent cir- cuit, the equivalent resistance R N and the equivalent current I N for Norton’s equivalent circuit. The value of R N in Norton’s equivalent circuit is the same as R TH of Thevenin’s equivalent circuit. Note: The ‘TH’ in V TH and R TH means Thevenin, and the ‘N’ in I N and R N means Norton. These two theorems are used very often to calculate the load (or a branch) current or voltage in practical applications. The load resistor can be varied sometimes (for instance, the wall plug can connect to 60 or 100 W lamps). Once the load is changed, the whole circuit has to be re-analysed or re-calculated. But if Thevenin’s and Norton’s theorems are used, Thevenin’s and Norton’s equivalent circuits will not be changed except for their external load branches. The variation of the load can be determined more conveniently by using Thevenin’s or Norton’s equivalent circuits. 5.2.2 Steps to apply Thevenin’s and Norton’s theorems 1. Open and remove the load branch (or any unknown current or voltage branch) in the network, and mark the letters a and b on the two terminals. 2. Determine the equivalent resistance R TH or R N . It equals the equivalent resistance, looking at it from the a and b terminals when all sources are turned off or equal to zero in the network. (A voltage source should be replaced by a short circuit, and a current source should be replaced by an open circuit.) That is R TH ¼ R N ¼ R ab The network theorems 135 3. Determine Thevenin’s equivalent voltage V TH . It equals the open-circuit voltage from the original linear two-terminal network of a and b, i.e. V TH ¼ V ab 4. Determine Norton’s equivalent current I N . It equals the short-circuit cur- rent from the original linear two-terminal network of a and b, i.e. I N ¼ I sc ðwhere ‘sc’ means the short circuitÞ 5. Plot Thevenin’s or Norton’s equivalent circuit, and connect the load branch (or unknown current or voltage branch) to a and b terminals of the equivalent circuit. Then the load (or unknown) voltage or current can be determined. The above procedure for analysing circuits by using Thevenin’s and Norton’s theorems is illustrated in the circuits of Figure 5.5. Example 5.4: Determine the load current I L in the circuit of Figure 5.6(a) by using Thevenin’s and Norton’s theorems. R TH V TH I N R N a b a a R L R L b R L b V TH = V ab R TH = R N = R ab a b E = 0 I s = 0 I N = I sc a b a b Figure 5.5 The procedure for applying Thevenin’s and Norton’s theorems I L E = 48 V R 1 = 12 Ω R 3 = 12 Ω R 2 = 12 Ω R L = 8 Ω Figure 5.6(a) Circuit for Example 5.4 136 Understandable electric circuits Solution: 1. Open and remove the load branch R L , and mark a and b on the terminals of the load branch as shown in the circuit of Figure 5.6(b). 2. Determine Thevenin’s and Norton’s equivalent resistances R TH and R N (the voltage source is replaced by a short circuit) in the circuit of Figure 5.6(c). R TH ¼ R N ¼ R ab ¼ R 1 ==R 2 ==R 3 ¼ ð12==12==12ÞO ¼ 4 O 3. . Determine Thevenin’s equivalent voltage V TH : Use the circuit in Fig- ure 5.6(d) to calculate the open-circuit voltage across terminals a and b. V TH ¼ V ab ¼ E R 2 ==R 3 R 1 þ R 2 ==R 3 ¼ 48 V ð12==12ÞO ð12 þ 12==12ÞO ¼ 16 V (Apply the voltage divider rule to the resistors R 2 //R 3 and R 1 .) . Determine Norton’s equivalent current I N : Use the circuit in Figure 5.6(e) to calculate the short-circuit current in terminals a and b. E = 48 V R 1 = 12 Ω R 3 = 12 Ω R 2 = 12 Ω a b Figure 5.6(b) Circuit for Example 5.4 R 1 = 12 Ω R 2 = 12 Ω a b Figure 5.6(c) E = 48 V R 1 = 12 Ω R 3 = 12 Ω R 2 = 12 Ω a b Figure 5.6(d) The network theorems 137 I N ¼ I sc ¼ E R 1 ¼ 48 V 12 O ¼ 4 A Since the current in the branch E and R 1 will go through a short cut without resistance – through the branches a and b – and will not go through the branches R 2 and R 3 that have resistances, in this case I N ¼ E=R 1 . 4. Plot Thevenin’s and Norton’s equivalent circuits as shown in Figure 5.6 (f and g). Connect the load R L to a and b terminals of the equivalent circuits and determine the load current I L . ● Use Thevenin’s equivalent circuit in Figure 5.6(f) to determine I L . I L ¼ V TH R TH þ R L ¼ 16 V 4 þ 8 ð ÞO % 1:33 A ● Use Norton’s equivalent circuit in Figure 5.6(g) to calculate I L . E = 48V R 1 = 12 Ω R 3 = 12Ω R 2 = 12Ω a b I = I sc N Figure 5.6(e) R TH = 4 Ω V TH = 16 V b a R L = 8 Ω I L Figure 5.6(f) I N = 4 A R N = 4 Ω a R L = 8 Ω b I L Figure 5.6(g) 138 Understandable electric circuits I L ¼ I N R N R N þ R L ¼ 4 A 4 O 4 þ 8 ð ÞO % 1:33 A (current divider rule) 5.2.3 Viewpoints of the theorems One important way to apply Thevenin’s and Norton’s theorems for analysing any network is to determine the viewpoints of Thevenin’s and Norton’s equivalent circuits. The load branch (or any unknown current or voltage branch) belongs to the external circuit of the linear two-terminal network with power sources. The opening two terminals of the branch are the viewpoints for Thevenin’s and Norton’s equivalent circuits. There could be different viewpoints for the bridge circuit as shown in Figure 5.7(a). If we want to determine the branch current I 3 , we use A and B as viewpoints; if we want to determine the branch current I 2 , we use D and C as viewpoints, etc. Different equivalent circuits and results will be obtained from using different viewpoints. Example 5.5: For the circuit in Figure 5.7(a): 1. Plot Thevenin’s equivalent circuit for calculating the current I 3 . 2. Determine Norton’s equivalent circuit for the viewpoints B–C. 3. Determine Thevenin’s equivalent circuit for the viewpoints D–B. Solution: (a) The viewpoints for calculating I 3 should be A–B (Figure 5.7(a)). 1. Open and remove R 3 in the branch A–B of Figure 5.7(a) and mark the letters a and b, as shown in the circuit of Figure 5.7(b). E A R 3 R 2 R 4 R 1 B C D R 1 R 2 R 4 E A B C D I 3 R 3 I 2 I 4 Figure 5.7(a) Viewpoints for the theorem R 4 R 2 R 1 b a A B E Figure 5.7(b) Circuit for Example 5.5 The network theorems 139 2. Determine R TH and R ab . Replace the voltage source E with a short circuit, as shown in the circuit of Figure 5.7(c). R TH ¼ R ab ¼ ðR 2 ==R 4 Þ==R 1 3. Determine V TH using the circuit in Figure 5.7(d). V TH ¼ V ab ¼ E R 1 R 1 þ R 2 ==R 4 (voltage divider rule) 4. Plot Thevenin’s equivalent circuit as shown in the circuit of Figure 5.7(e). Connect R 3 to a and b terminals of the equivalent circuit and determine the current I 3 : I 3 ¼ V TH R TH þ R 3 R 4 R 2 R 1 b a A B R TH = R ab R 4 R 1 R 2 a b R TH = R ab Figure 5.7(c) R 4 R 2 R 1 a E V TH = V ab b Figure 5.7(d) R 3 R TH a V TH b I 3 A B Figure 5.7(e) 140 Understandable electric circuits (b) Norton’s equivalent circuit for the viewpoints B–C: 1. Open and remove R 4 in the branch B–C of Figure 5.7(a), and mark the letters a and b on the two terminals as shown in the circuit of Figure 5.7(f). 2. Determine R N . Replace the voltage source with a short circuit as shown in the circuit of Figure 5.7(g). 3. Determine I N using the circuit in Figure 5.7(h). Since the current will go through the short cut without resistance – the branch a and b – and will not go through the branch with resistance R 2 , I N ¼ I T : I N ¼ I T ¼ E R 1 ==R 3 R 3 R 2 R 1 a C B E b Figure 5.7(f) R 3 R 2 R 1 a C B b R TH = R ab Figure 5.7(g) R 3 R 2 R 1 a E b I T I N = I sc Figure 5.7(h) The network theorems 141 4. Plot Norton’s equivalent circuit as shown in the circuit of Figure 5.7(i). (c) Thevenin’s equivalent circuit for the viewpoint D–B: 1. Open branch D–B (Figure 5.7(a)) and mark the letters a and b on the two terminals as shown in the circuit of Figure 5.7(j). 2. Determine R TH . Replace the voltage source with a short circuit as shown in the circuits of Figure 5.7(k). R TH ¼ R ab ¼ ðR 1 ==R 2 Þ þ ðR 3 ==R 4 Þ R 4 R N a b C B I s Figure 5.7(i) R 4 b a D B R 3 R 1 R 2 E Figure 5.7(j) R 4 b a D B R 3 R 1 R 2 R 2 R 1 R 3 R 4 a b Figure 5.7(k) 142 Understandable electric circuits 3. Determine V TH using the circuit in Figure 5.7(j): V TH ¼ V ab ¼ V a þ ðÀV b Þ ¼ E R 2 R 1 þ R 2 À E R 4 R 3 þ R 4 4. Plot Thevenin’s equivalent circuit as shown in the circuit of Figure 5.7(l). Example 5.6: Determine current I L in the circuit of Figure 5.8(a) by using Norton’s theorem. Solution: 1. Open and remove R L in the load branch (Figure 5.8(b)) and mark the letters a and b on its two terminals, as shown in the circuit of Figure 5.8(b). 2. Determine R N . Replace the current source with an open circuit as shown in the circuit of Figure 5.8(c). a b R TH V TH D B Figure 5.7(l) R 1 = 200Ω I L R 2 = 200Ω R = 400Ω I = 1mA 3 R L =100Ω R 4 = 500Ω Figure 5.8(a) Circuit for Example 5.6 R 1 = 200Ω R 2 = 200Ω R 3 = 400Ω I = 1mA a b R 4 = 500Ω Figure 5.8(b) The network theorems 143 R N ¼ R ab ¼ ðR 1 ==R 2 þ R 3 Þ==R 4 ¼ ½ð200==200 þ 400Þ==500O ¼ 250 O 3. Calculate I N using the circuit of Figure 5.8(d). I N ¼ I ¼ 1 mA Since the current I will flow through the short cut without resistance – the branch a and b – and will not go through the branch with resistance, I N ¼ I. 4. Plot Norton’s equivalent circuit as shown in the circuit of Figure 5.8(e). Connect R L to the a and b terminals of the equivalent circuit, and calculate the current I L . I L ¼ I N R N R L þ R N ¼ 1 mA 250 O ð250 þ 100ÞO % 0:71 mA R 1 = 200 Ω R 2 = 200 Ω a b R 3 = 400 Ω R 4 = 500 Ω R N = R ab Figure 5.8(c) I N = I sc R 2 = 200Ω R 3 = 400Ω R 4 = 500Ω R 1 = 200Ω I = 1mA a b Figure 5.8(d) R N = 250 Ω R L = 100 Ω a I L I N = 1 mA b Figure 5.8(e) 144 Understandable electric circuits When applying Thevenin’s and Norton’s theorems to analyse networks, it is often necessary to combine theorems that we have learned in the previous chapters. This is explained in the following example. Example 5.7: Determine Norton’s equivalent circuit for the left part of the terminals a and b in the circuit of Figure 5.9(a) and determine the current I L . Solution: 1. Open and remove the current source part on the right side of the circuit from the terminals a and b (Figure 5.9(a)), as shown in the circuit of Figure 5.9(b). 2. Determine R N . Replace the voltage source with a short circuit, and the current source with an open circuit, as shown in the circuit of Figure 5.9(c). R L = 42 Ω I 1 = 45 mA I 2 = 140 mA E = 6 V R 1 = 30 Ω I L a b R 2 = 70 Ω Figure 5.9(a) Circuit for Example 5.7 I 1 = 45 mA E = 6 V R 1 = 30 Ω a b R 2 = 70 Ω Figure 5.9(b) R 1 = 30 Ω R 2 = 70 Ω R N = Rab a b Figure 5.9(c) The network theorems 145 R N ¼ R ab ¼ R 1 ==R 2 ¼ ð30  70ÞO ð30 þ 70ÞO ¼ 21 O 3. Determine I N using the circuit in Figure 5.9(d). Since there are two power supplies in this circuit, it is necessary to apply the network analysing method for this complex circuit. Let us try to use the superposition theo- rem to determine I N . ● When the single voltage source E is applied only to the circuit, the circuit is shown in Figure 5.9(e). Since R 2 is short circuited by the I N 0 (recall that current always goes through the short cut without resistance): ; I N 0 ¼ E R 1 ¼ 6 V 30 O ¼ 0:2 A ¼ 200 mA ● When the single current source I 1 is applied only to the circuit, the circuit is shown in Figure 5.9(f). R 1 = 30 Ω R 2 = 70 Ω E = 6 V a b I N I 1 = 45 mA Figure 5.9(d) R 1 = 30 Ω R 2 = 70 Ω E = 6 V a b I N ′ Figure 5.9(e) R 1 = 30 Ω R 2 = 70 Ω a b I ′′ N I 1 = 45 mA Figure 5.9(f) 146 Understandable electric circuits Since R 1 and R 2 are short circuited by I N 00 : I N 00 ¼ I 1 ¼ 45 mA ● Determine I N : I N ¼ I N 0 ÀI N 00 ¼ ð200 À 45Þ mA ¼ 155 mA 4. Plot Norton’s equivalent circuit. Connect the right side of the a and b terminals of the current source (Figure 5.9(a)) to the a and b terminals of Norton’s equivalent circuit, as shown in the circuits of Figure 5.9(g). Determine the current I L . I L ¼ ðI N þ I 2 Þ R N R L þ R N ¼ ð155 þ 140ÞmA 21 O ð42 þ 21ÞO % 98:33 mA 5.3 Maximum power transfer Practical circuits are usually designed to provide power to the load. When working in electrical or electronic engineering fields you are sometimes asked to design a circuit that will transfer the maximum power from a given source to a load. The maximum power transfer theorem can be used to solve this kind of problem. The maximum power transfer theorem states that when the load resistance is equal to the source’s internal resistance, the maximum power will be transferred to the load. From the last section, we have learned that any linear two-terminal net- work with power supply can be equally substituted by Thevenin’s or Norton’s equivalent circuits. Therefore, the maximum power transfer theorem implies that when the load resistance (R L ) of a circuit is equal to the internal resistance (R S ) of the source or the equivalent resistance of Thevenin’s or Norton’s equivalent circuits (R TH or R N ), maximum power will be dissipated in the load. This concept is illustrated in the circuits of Figure 5.10. R N = 21 Ω R N = 21 Ω I 2 = 140 mA I N = 155 mA a b R L = 42 Ω I L I N + I 2 a b R L = 42 Ω I L Figure 5.9(g) The network theorems 147 The maximum power transfer theorem When the load resistance is equal to the internal resistance of the source (R L ¼ R S ); or when the load resistance is equal to the Thevenin’s/ Norton’s equivalent resistance of the network (R L ¼ R TH ¼ R N ), the maximum power can be transferred to the load. The maximum power transfer theorem is used very often in radios, recor- ders, stereos, CDs, etc. If the load component is a speaker and the circuit that drives the speaker is a power amplifier, when the resistance of the speaker R L is equal to the internal resistance R S of the amplifier equivalent circuit, the amplifier can transfer the maximum power to the speaker, i.e. the maximum volume can be delivered by the speaker. Using the equivalent circuit in Figure 5.10(d) to calculate the power consumed by the load resistor R L gives P L ¼ I L 2 R L ¼ V S R S þ R L 2 R L ð5:1Þ When R L ¼ R S , the maximum power that can be transferred to the load is P L ¼ V S 2 ð2R S Þ 2 R S ¼ V S 2 4R S R TH V TH I N R N a b a b a R L R L b R L R L = R TH R L = R N R L V S R s a b R L = R S (a) (b) (c) (d) I L Figure 5.10 The maximum power transfer 148 Understandable electric circuits The maximum load power P L ¼ V S 2 4R S If V S ¼ 10 V; R S ¼ 30O; and R L ¼ 30O Then P L ¼ V S 2 4R S ¼ 10 2 V 4ð30 OÞ % 833 mW The maximum power transfer theorem can be proved by using an experi- ment circuit as shown in Figure 5.11. When the variable resistor R L is adjusted, it will change the value of the load resistor. Replacing the load resistance R L with different values in (5.1) gives different load power P L , as shown in Table 5.1. Such as when R L ¼ 10 O: P L ¼ I 2 R L ¼ V S R S þ R L 2 R L ¼ 10 V ð30 þ 10ÞO 2 10 O ¼ 0:625 W R s = 30 Ω a b R L V s = 10 V Figure 5.11 The experiment circuit Table 5.1 The load power R L (O) P L (W) 10 0.625 20 0.8 30 0.833 40 0.816 50 0.781 The network theorems 149 When R L ¼ 20 O: P L ¼ I 2 R L ¼ V S R S þ R L 2 R L ¼ 10 V ð30 þ 20ÞO 2 20 O ¼ 0:8 W The R L and P L curves can be plotted from Table 5.1 as shown in Figure 5.12. Table 5.1 and Figure 5.12 shows that only when R L ¼ R S (30 O), the power for the resistor R L reaches the maximum point A (0.833 W). Proof of the maximum power transfer equation: The maximum power transfer equation R L ¼ R S can be derived by the method of getting the max- imum value in calculus (skip this part if you haven’t learned calculus yet). Take the derivative for R L in (5.1), and let its derivative equal to zero, giving the following: dP dR L ¼ d ðV S =ðR S þ R L ÞÞ 2 R L h i dR L ¼ 0 V S 2 ðR S þ R L Þ 2 À 2R L ðR S þ R L Þ ðR S þ R L Þ 4 ¼ 0 V S 2 R S 2 þ 2R S R L þ R L 2 À 2R L R S À 2R L 2 ðR S þ R L Þ 4 ¼ 0 P (W) L R (Ω) L 10 20 30 40 50 0 A R L = R S 0.2 0.4 0.6 0.8 1 Figure 5.12 R L –P L curve 150 Understandable electric circuits V S 2 R S 2 À R L 2 ðR S þ R L Þ 4 ¼ 0 V S 2 ðR S þ R L ÞðR S À R L Þ ðR S þ R L Þ 4 ¼ 0 V S 2 R S À R L ðR S þ R L Þ 3 ¼ 0 i:e: R S À R L ¼ 0 or R L ¼ R S hence proved: 5.4 Millman’s and substitution theorems 5.4.1 Millman’s theorem Millman’s theorem is named after the Russian electrical engineering professor Jacob Millman (1911–1991) who proved this theorem. A similar method, known as Tank’s method, had already been used before Millman’s proof. The method using series–parallel power sources was stated in chapter 4. However, the series–parallel method can only be used in power sources that have the same polarities and values. Millman’s theorem in this chapter can be used to analyse circuits of parallel voltage sources that have different polarities and values. This can be shown in the circuits of Figure 5.13. Millman’s theorem states that for a circuit of parallel branches, with each branch consisting of a resistor or a voltage/current source, this circuit can be replaced by a single voltage source with voltage V m in series with a resistor R m as shown in Figure 5.13. Millman’s theorem, therefore, can determine the voltage across the parallel branches of a circuit. R n E 1 R 1 a b E 2 R 2 E n a b V m R m … Figure 5.13 Millman’s theorem The network theorems 151 Millman’s theorem When several voltage sources or branches consisting of a resistor are in parallel, they can be replaced by a single voltage source. V m ¼ R m I m ¼ R m E 1 R 1 þ E 2 R 2 þ ::: þ E n R n R m ¼ R 1 ==R 2 == . . . ==R n Note: V m is the algebraic sum for all the individual terms in the equation. It will be positive if E n and V m have the same polarities, otherwise it will be negative. The letter m in V m and R m means Millman. Example 5.8: Determine the load voltage V L in the circuit of Figure 5.14 using Millman’s theorem. Solution: R m ¼ R 1 ==R 2 ==R 3 ==R 4 ¼ ð100==100==100==100ÞO ¼ 25 O V m ¼ R m I m ¼ R m E 1 R 1 þ E 2 R 2 À E 3 R 3 À E 4 R 4 ¼ ð25 OÞ 40 V 100 O þ 30 V 100 O À 20 V 100 O À 10 V 100 O ¼ 10 V V L ¼ V m R L R L þ R m ¼ ð10 VÞ 30 O ð30 þ 25ÞO % 5:455 V 5.4.2 Substitution theorem Substitution theorem A branch in a network that consists of any component can be replaced by an equivalent branch that consists of any combination of components, as long as the currents and voltages on that branch do not change after the substitution. E 1 = 40V R 1 = 100Ω a b a b E 2 = 30V R 4 = 100Ω E 3 = 20V R m = 25Ω V m = 10V R L = 30Ω V L R 2 = 100Ω R 3 = 100Ω E 4 = 10V R L = 30Ω Figure 5.14 Circuit for Example 5.8 152 Understandable electric circuits This theorem can be illustrated in the circuits of Figures 5.15 and 5.16. The current and voltage of branch a–b in the circuit of Figure 5.15 can be determined as follows: The voltage across branch a–b: V 2 ¼ E R 2 R 1 þ R 2 ¼ 20 V 6kO ð2 þ 6ÞkO ¼ 15 V The current in branch a–b: I ¼ E R 1 þ R 2 ¼ 20V ð2 þ 6ÞkO ¼ 2:5 mA According to the definition of the substitution theorem, any branch in the circuit of Figure 5.16 can replace the a–b branch in the circuit of Figure 5.15, since their voltages and currents are the same as the voltages and currents in branch a–b in the circuit of Figure 5.15. Example 5.9: Use a current source with a 30 O internal resistor to replace the a–b branch in the circuit of Figure 5.17(a). a b R 1 = 2 kΩ E = 20 V R 2 = 6 kΩ V 2 = 15 V I = 2.5 mA Figure 5.15 Circuit 1 of the substitution theorem a R = 6 kΩ I = 2.5 mA 15 V + - 2.5 mA a E = 15 V I = 2.5 mA a + - 15 V 5 V a R = 2 kΩ E = 10 V + - I = 2.5 mA Figure 5.16 Circuit 2 of the substitution theorem R 3 = 7.5 Ω E = 2.5 V R 1 = 2 Ω a b R 2 = 5 Ω I Figure 5.17(a) Circuit for Example 5.9 The network theorems 153 Solution: ● Figure 5.17(b) shows the resultant circuit after the current source with a 30 O internal resistor replaced the a–b branch in the circuit of Figure 5.17(a). ● Determine the voltage and current in the a–b branch of the circuit in Figure 5.17(a). V ab ¼ E R 2 ==R 3 R 1 þ R 2 ==R 3 ¼ 2:5 V ðð5  7:5Þ=ð5 þ 7:5ÞÞO ð2 þ ð5  7:5Þ=ð5 þ 7:5ÞÞO ¼ 1:5 V ðvoltage divider ruleÞ I ¼ V ab R 3 ¼ 1:5 V 7:5 O ¼ 0:2 A ¼ 200 mA ● Determine the currents in the substituted branch and the current source branch using the circuit in Figure 5.17(c). I 3 0 ¼ V ab R 3 0 ¼ 1:5 V 30 O ¼ 50 mA Therefore, to maintain the terminal voltage V ab ¼ 1.5 V in the original branch, the current I 3 0 in the R 3 0 branch should be 50 mA. Using KCL we can get the current I 0 in the current source branch: I 0 ¼ IÀI 3 0 ¼ ð200 À 50ÞmA ¼ 150 mA R 3 ′ = 30 Ω E = 2.5 V R 1 = 2 Ω a b R 2 = 5 Ω I ′ Figure 5.17(b) R 3 ′ = 30 Ω I ′ = 50 mA I 3 ′ = 50 mA a b I = 200 mA + - V ab = 1.5 V Figure 5.17(c) 154 Understandable electric circuits Summary Linear circuit: includes the linear components (such as resistors). Superposition theorem ● The unknown voltages or currents in any linear network are the sum of the voltages or currents of the individual contributions from each single power supply, by setting the other inactive sources to zero. ● Steps to apply superposition theorem: 1. Turn off all power sources except one, i.e. replace the voltage source with the short circuit, and replace the current source with an open circuit. Redraw the original circuit with a single source. 2. Analyse and calculate this circuit by using the single source series– parallel analysis method, and repeat steps 1 and 2 for the other power sources in the circuit. 3. Determine the total contribution by calculating the algebraic sum of all contributions due to single sources. The result should be positive when the reference polarity of the unknown in the single source circuit is the same as the reference polarity of the unknown in the original circuit; otherwise it should be negative. The linear two-terminal network with the sources It is a linear complex circuit that has power sources and two terminals. Thevenin’s and Norton’s theorems ● Any linear two-terminal network with power supplies can be replaced by a simple equivalent circuit that has a single power source and a single resistor. ● Thevenin’s theorem: The equivalent circuit is a voltage source (with an equivalent resistance R TH in series with an equivalent voltage source V TH ). ● Norton’s theorem: The equivalent circuit is a current source (with an equivalent resistance R N in parallel with an equivalent current source I N ). ● Steps to apply Thevenin’s and Norton’s theorems: 1. Open and remove the load branch (or any unknown current or voltage branch) in the network, and mark the letter a and b on the two terminals. 2. Determine the equivalent resistance R TH or R N . It equals the equiva- lent resistance, looking at it from the a and b terminals when all sources are turned off or equal to zero in the network. (A voltage source should be replaced by a short circuit, and a current source should be replaced by an open circuit.) That is R TH ¼ R N ¼ R ab The network theorems 155 3. Determine Thevenin’s equivalent voltage V TH . It equals the open- circuit voltage from the original linear two-terminal network of a and b, i.e. V TH ¼ V ab 4. Determine Norton’s equivalent current I N . It equals to the short-cir- cuit current from the original linear two-terminal network of a and b, i.e. I N ¼ I sc . 5. Plot Thevenin’s or Norton’s equivalent circuit, and connect the load branch (or unknown current or voltage branch) to a and b terminals of the equivalent circuit. Then the load (or unknown) voltage or current can be determined. Maximum power transfer theorem When the load resistance is equal to the internal resistance of the source (R L ¼ R S ); or when the load resistance is equal to the Thevenin’s/Norton’s equivalent resistance of the network (R L =R TH ¼ R N ), maximum power will be trans- ferred to the load. Millman’s theorem When several voltage sources or branches consisting of a resistor are in par- allel, they can be replaced by a branch with a voltage source. V m ¼ R m I m ¼ R m E 1 R 1 þ E 2 R 2 þ Á Á Á þ E n R n ; R m ¼ R 1 == R 2 == ::: == R n Substitution theorem A branch in a network that consists of any component can be replaced by an equivalent branch that consists of any combination of components, as long as the currents and voltages on that branch do not change after the substitution. Experiment 5A: Superposition theorem Objectives ● Understand the superposition theorem through experiment. ● Construct a circuit with two voltage sources, and collect and evaluate experimental data to verify applications of the superposition theorem. ● Analyse experimental data, circuit behaviour and performance, and compare them to theoretical equivalents. Equipment and components ● Breadboard ● Multimeter ● Dual-output DC power supply ● Resistors: 5.1, 7.5, and 11 kO 156 Understandable electric circuits Background information ● Superposition theorem: The unknown voltages or currents in any linear network are the sum of the voltages or currents of the individual con- tributions from each single power supply, by setting the other inactive sources to zero. ● Steps to apply superposition theorem: 1. Turn off all power sources except one, i.e. replace the voltage source with a short circuit (by placing a jump wire), and replace the current source with an open circuit. Redraw the original circuit with a single source. 2. Analyse and calculate this circuit using the single source series–parallel analysis method, and repeat steps 1 and 2. 3. Determine the total contribution by calculating the algebraic sum of all contributions due to single sources. (The result should be positive when the reference polarity of the unknown in the single source circuit is the same with the polarity of the unknown in the original circuit; otherwise it should be negative.) Procedure 1. Measure the values of resistors listed in Table L5.1 using a multimeter (ohmmeter function) and record in Table L5.1. 2. Construct a circuit on the breadboard as shown in Figure L5.1(b). 3. Calculate the equivalent resistance R eq 0 from the terminals a and b in the circuit of Figure L5.1(b). Record the value in Table L5.2. 4. Calculate currents I 1 0 and I 3 0 in the circuit of Figure L5.1(b). Record the values in Table L5.2. 5. Measure the equivalent resistance R eq 0 and currents I 1 0 and I 3 0 using the multimeter (ohmmeter and ammeter functions) in Figure L5.1(b). Record the values in Table L5.2. 6. Reconstruct a circuit as shown in Figure L5.1(c) on the breadboard. 7. Calculate the equivalent resistance R eq 00 from the terminals c and d in the circuit of Figure L5.1(c). Record the values in Table L5.2. 8. Calculate currents I 1 00 and I 3 00 in the circuit of Figure L5.1(c). Record the values in Table L5.2. Table L5.1 Resistance R 1 R 2 R 3 Nominal value 5.1 kO 7.5 kO 11 kO Measured value The network theorems 157 9. Measure the equivalent resistance R eq 00 and currents I 1 00 and I 3 00 in Figure L5.1(c) using a multimeter (ohmmeter and ammeter functions). Record the values in Table L5.2. 10. Calculate I 3 ¼ I 3 0 þ I 3 00 from Table L5.2 using the calculated values. Record the value in Table L5.2. 11. Construct a circuit as shown in Figure L5.1(a) on the breadboard and mea- sure current I 3 in the circuit of Figure L5.1(a). Record the value in Table L5.2. 12. Compare the measured values and calculated values; are there any sig- nificant differences? If so, explain the reasons. Conclusion Write your conclusions below: Table L5.2 Resistance R eq 0 I 1 0 I 3 0 R eq 00 I 1 00 I 3 00 I 3 Formula for calculations Calculated value Measured value E 2 = 12V R 3 = 11kΩ I 3 E 1 = 6V b c d a R 1 = 5.1kΩ R 2 = 7.5kΩ I 1 I 2 (a) + E 1 = 6V R 3 = 11kΩ = a b c d I ′ 1 I ′ 3 I ′ 2 R 1 = 5.1kΩ R 2 = 7.5kΩ R 3 = 11kΩ E 2 = 12V a b c d I ′′ 1 I ′′ 3 I ′′ 2 R 1 = 5.1kΩ R 2 = 7.5kΩ (b) (c) Figure L5.1 Superposition theorem 158 Understandable electric circuits Experiment 5B: Thevenin’s and Norton’s theorems Objectives ● Understand Thevenin’s and Norton’s theorems and the maximum power transfer theorem through this experiment. ● Construct electric circuits, and collect and evaluate experimental data to verify the applications of Thevenin’s and Norton’s theorems. ● Construct electric circuits, and collect and evaluate experimental data to verify the applications of the maximum power transfer theorem. ● Analyse experimental data, circuit behaviour and performance, and compare them to the theoretical equivalents. Equipment and components ● Breadboard ● Multimeter ● DC power supply ● Resistors: 300 O, 750 O, 620 O, 180 O, 1 kO, and one 10 kO variable resistor Background information ● Any linear two-terminal network (complex circuit) with power supplies can be replaced by a simple equivalent circuit that has a single power source and a single resistor. ● Thevenin’s theorem: Thevenin’s equivalent circuit is an actual voltage source that has an equivalent resistance R TH in series with an equivalent voltage source V TH . ● Norton’s theorem: Norton’s equivalent circuit is an actual current source that has an equivalent resistance R N in parallel with an equivalent current source I N . ● The maximum power transfer theorem: When the load resistance is equal to the internal resistance of the source (R L ¼ R S ); or when the load resis- tance is equal to the Thevenin/Norton equivalent resistance of the circuit (R L ¼ R TH ¼ R N ), maximum power will be dissipated in the load. Procedure Part I: Thevenin’s and Norton’s theorems 1. Measure the values of the resistors listed in Table L5.3 using a multimeter (ohmmeter function) and record in Table L5.3. Table L5.3 Resistor R 1 R 2 R 3 R L Colour code value 300 O 620 O 750 O 180 O Measured Value The network theorems 159 2. Calculate R TH and V TH of Thevenin’s equivalent circuit in Figure L5.2(a). Record the values in Table L5.4. 3. Calculate R N and I N of Norton’s equivalent circuit in Figure L5.2(a). Record the values in Table L5.4. 4. Calculate the load current I L in Thevenin’s and Norton’s equivalent circuits of Figure L5.2(b and c). Record the values in Table L5.4. 5. Construct a circuit as shown in Figure L5.3(a) on the breadboard. 6. Measure the open-circuit voltage V ab (V ab ¼ V TH ) on the two terminals a and b in the circuit of Figure L5.3(a) using a multimeter (voltmeter func- tion). Record the value in Table L5.4. = 180 Ω R R L R L R L N I N a b I L = 180 Ω a b E TH I L R TH 3 = 750 Ω E 1 = 12V a b = 180 Ω 1 = 300 Ω R R R 2 = 620 Ω (a) (b) (c) Figure L5.2 Thevenin’s and Norton’s equivalent circuits Table L5.4 R TH V TH (V ab ) R N I N (I SC ) I L (Figure L5.2(b)) I L (Figure L5.2(c)) Formula for calculations Calculated value Measured value 3 = 750Ω E 1 = 12V a b 1 = 300Ω R R R 2 = 620Ω a b R 3 = 750Ω 1 = 300Ω R R 2 = 620Ω (a) (b) Figure L5.3 Thevenin’s and Norton’s circuits 160 Understandable electric circuits 7. Measure short-circuit current I sc (I sc ¼ I N ) from a to b in Figure L5.3(a) using a multimeter (ammeter function). Record the value in Table L5.4. 8. Disconnect the power supply E in Figure L5.3(a), and use a jump wire con- necting the two terminals of E as shown in Figure L5.3(b). Measure the equivalent resistance at terminals of a to b (R ab ¼ R TH ¼ R N ) in Figure L5.3 (b) using a multimeter (ohmmeter function). Record the value in Table L5.4. 9. Construct a circuit as shown in Figure L5.2(b) (with measured V TH and R TH ) on the breadboard, and measure the load current I L using a multi- meter (ammeter function). Record the value in Table L5.4. 10. Compare the measured values and the calculated values. Are there any significant differences? If so, state the reasons. Part II: Maximum power transfer 1. Construct a circuit as shown in Figure L5.4 on the breadboard. R S repre- sents the internal resistance for power supply or the Thevenin’s or Norton’s equivalent resistances. 2. Measure the open-circuit voltage V ab in the circuit of Figure L5.4 (without connecting the load R L ) using a multimeter (voltmeter function). Record the value in Table L5.5. R s = 1 kΩ a b R L V s = 10 V Figure L5.4 Maximum power transfer circuit Table L5.5 Load Resistor R L Measured R L Value Measured V R L Value Calculated P L Value R L1 R L2 R L3 R L4 R L5 1 kO R L6 R L7 R L8 R L9 Open-circuit voltage V ab = The network theorems 161 3. Connect the 10-kO variable resistor R L to the circuit in Figure L5.4 and change the value of the variable resistor nine times from lower to higher values (one of the values should be 1 kO). Measure each R L and V R L using a multimeter (ohmmeter and voltmeter functions). Record the values in Table L5.5. 4. Calculate power dissipated in each load resistor and record the values in Table L5.5. 5. Sketch the R L –P L curve (use P L as vertical axis and R L as horizontal axis). 6. When the load resistance is equal to the internal resistance of the source (R L ¼ R S ¼ 1 kO), is P L at the maximum point on the curve? Why? Conclusion Write your conclusions below: 162 Understandable electric circuits Chapter 6 Capacitors and inductors Objectives After completing this chapter, you will be able to: ● describe the basic structure of the capacitor and inductor ● explain the charging and discharging behaviours of a capacitor ● understand the storing and releasing energy of an inductor ● define capacitance and inductance ● list the factors affecting capacitance and inductance ● understand the relationship between voltage and current in capacitive and inductive circuits ● calculate energy stored in capacitors and inductors ● determine the equivalent capacitance and inductance in series, parallel and series–parallel configurations There are three important fundamental circuit elements: the resistor, capacitor and inductor. The resistor (R) has appeared in circuit analysis in the previous chapters. The other two elements – the capacitor (C) and inductor (L) will be introduced in this chapter. Both of these electric elements can store energy that has been absorbed from the power supply, and release it to the circuit. A capacitor can store energy in the electric field, and an inductor can store energy in the magnetic field. This is different with a resistor that consumes or dis- sipates electric energy. Three basic circuit components ● Resistor (R) ● Capacitor (C) ● Inductor (L) A circuit containing only resistors has limited applications. Practical elec- tric circuits usually combine the above three basic elements and possibilities along with other devices. 6.1 Capacitor 6.1.1 The construction of a capacitor A capacitor has applications in many areas of electrical and electronic circuits, and it extends from households to industry and the business world. For instance, it is used in flash lamps (for flash camera), power systems (power supply smoothing, surge protections), electronic engineering, communications, compu- ters, etc. There are many different types of capacitors, but no matter how differently their shapes and sizes, they all have the same basic construction. A capacitor has two parallel conductive metal plates separated by an iso- lating material (the dielectric). The dielectric can be of insulating material, such as paper, vacuum, air, glass, plastic film, oil, mica, ceramics, etc. The basic construction of a capacitor is shown in Figure 6.1. A capacitor can be represented by a capacitor schematic symbol as its circuit model. Similar to resistors, there are two basic types of capacitors, variable and fixed, and their schematic symbols are shown in Figure 6.2 (a and b). Dielectric Plate Plate Figure 6.1 The basic construction of a capacitor + – + – (a) (b) Figure 6.2 Symbols of capacitor. (a) Fixed: unpolarized and polarized and (b) variable 164 Understandable electric circuits A variable capacitor is a capacitor that possesses a value that may be changed manually or automatically. A fixed capacitor is a capacitor that pos- sesses a fixed value and cannot be adjusted. For a fixed polarized capacitor, connect its positive (þ) lead to the higher voltage point in the circuit, and negative (7) lead to the lower voltage point. For an unpolarized capacitor, it does not matter which lead connects to where. Electrolytic capacitors are usually polarized, and non-electrolytic capaci- tors are unpolarized. Electrolytic capacitors can have higher working voltages and store more charges than non-electrolytic capacitors. Capacitor C An energy storage element that has two parallel conductive metal plates separated by an isolating material (the dielectric). 6.1.2 Charging a capacitor A purely capacitive circuit with an uncharged capacitor (V C ¼ 0), a three- position switch, and a DC (direct current) voltage source (E) is shown in Fig- ure 6.3(a). With the switch at position 0, the circuit is open, and the potential difference between the two metal plates of the capacitor is zero (V C ¼ 0). Two plates of the capacitor have the same size and are made by the same conducting material, so they should have the same number of charges at the initial condition. Once the three-position switch is turned on to position 1 as shown in Figure 6.3(b), the DC voltage source is connected to the two leads of the capacitor. From the rule ‘opposites attract and likes repel’, we know that the positive pole of the voltage source will attract electrons from the positive plate of the capacitor, and the negative pole of the voltage source will attract positive charges from the negative plate of the capacitor; this causes current I to flow in the circuit. Plate A loses electrons and shows positive; plate B loses positive charges and thus shows negative. Thus, the electric field is built up between the two metal plates of the capacitor, and the potential difference (V C ) appears on the E 1 2 C E 1 + + + + - - - - - - + + 2 A B I 0 V c = 0 V c (a) (b) Figure 6.3 Charging a capacitor Capacitors and inductors 165 capacitor with positive (þ) on plate A and negative (7) on plate B, as shown in Figure 6.3(b). Once voltage across the capacitor V C has reached the source voltage V S , i.e. V C ¼ E, there is no more potential difference between the source and capacitor, the charging current cesses to flow (I ¼ 0), and the process of charging the capacitor is completed. This is the process of charging a capacitor. 6.1.3 Energy storage element When the switch is turned off to position 0 in the circuit shown in Figure 6.3(a), the capacitor and power supply will disconnect. If the voltage across the capacitor V C is measured at this time using a multimeter (voltmeter function), V C should still be the same with the source voltage (V C ¼ E) even without a power supply connected to it. This is why a capacitor is called an energy sto- rage element, as it can store charges absorbed from the power supply and store electric energy obtained from charging. Once a capacitor has transferred some charges through charging, an electric field is built up between the two plates of the capacitor, and it can maintain the potential difference across it. The isolating material (dielectric) between the two metal plates isolates the charges between the two plates. Charges will not be able to cross the insulating material from one plate to another. So the energy storage element capacitor will keep its charged voltage V C for a long time (duration will depend on the quality and type of the capacitor). Since the insulating material will not be perfect and a small leakage current may flow through the dielectric, this may eventually slowly dissipate the charges. 6.1.4 Discharging a capacitor When the switch is closed to position 2 as shown in the circuit of Figure 6.4, the capacitor and wires in the circuit forms a closed path. At this time, the capacitor is equivalent to a voltage source, as voltage across the capacitor V C will cause the current to flow in the circuit. Since there is no resistor in this circuit, it is a short circuit, and a high current causes the capacitor to release its charges or stored energy in a short time. This is known as discharging a capacitor. After the capacitor has released all its stored energy, the voltage across the capacitor will be zero (V C ¼ 0), the current in the circuit ceases to flow (I ¼ 0) and the discharge process is completed. E 1 2 C E 1 + + + − − − 2 A B V C I (a) (b) Figure 6.4 Discharging a capacitor 166 Understandable electric circuits The capacitor cannot release energy that is more than it has absorbed and stored, therefore it is a passive component. A passive component is a compo- nent that absorbs (but not produce) energy. The concept of a capacitor may be analogous to a small reservoir. It acts as a reservoir that stores and releases water. The process of charging a capacitor from the power supply is similar to a reservoir storing water. The process of discharging a capacitor is similar to a reservoir releasing water. There is an important characteristic that implies in the charge and dis- charge of a capacitor. That is, the voltage on the capacitor won’t be able to change instantly; it will always take time, i.e. gradually increase (charging) or decrease (discharge). Charging/Discharging a capacitor An electric element that can store and release charges that it absorbed from the power supply. ● Charging: The process of storing energy. ● Discharging: The process of releasing energy. 6.1.5 Capacitance As previously mentioned, once the source voltage is applied to two leads of a capacitor, the capacitor starts to store energy or charges. The charges (Q) that are stored are proportional to the voltage (V) across it. This can be expressed by the following formula: Q ¼ CV or C ¼ Q V This is analogous to a pump pumping water to a reservoir. The higher the pressure, the more water will be pumped into the reservoir. The higher the voltage, the more charges a capacitor can store. The voltage and charge (V–Q) characteristic of a capacitor is shown in Figure 6.5, demonstrating that the capacitor voltage is proportional to the amount of charges a capacitor can store. V Q Figure 6.5 Q–V characteristic of a capacitor Capacitors and inductors 167 C is the capacitance, which is the value of the capacitor and describes the amount of charges stored in the capacitor. Just as a resistor is a component and resistance is the value of a resistor, capacitor is a component and capacitance is the value of a capacitor. Resistor is symbolized by R while resistance is R: capacitor is symbolized by C while capacitance is C. Capacitance C C, the value of the capacitor, is directly proportional to its stored charges, and inversely proportional to the voltage (V) across it. C ¼ Q V A capacitor can store 1 C charge when 1 V of voltage is applied to it. That is, 1 F ¼ 1 C 1 V Farad is a very large unit of measure for most practical capacitors. Microfarad (mF) or picofarad (pF) are more commonly used units for capacitors. Recall 1 mF ¼ 10 À6 F and 1 pF ¼ 10 À12 F Note: m is a Greek letter called ‘mu’ (see Appendix A for a list of Greek letters). Example 6.1: If a 50 mC charge is stored on the plates of a capacitor, determine the voltage across the capacitor if the capacitance of the capacitor is 1000 pF. Solution: Q ¼ 50 mC; C ¼ 1000 pF; V ¼ ? V ¼ Q C ¼ 50 mC 1000 pF ¼ 50  10 À6 C 1000  10 À12 F ¼ 0:05  10 6 V ¼ 50 KV Quantity Quantity symbol Unit Unit symbol Capacitance C Farad F Charge Q Coulomb C Voltage V Volt V 168 Understandable electric circuits 6.1.6 Factors affecting capacitance There are three basic factors affecting the capacitance of a capacitor, and they are determined by the construction of a capacitor as shown below: ● The area of plates (A): A is directly proportional to the charge Q; the larger the plate area, the more electric charges that can be stored. ● The distance between the two plates (d): The shorter the distance between two plates, the stronger the produced electric field that will increase the ability to store charges. Therefore, the distance (d) between the two plates is inversely proportional to the capacitance (C). ● The dielectric constant (k): Different insulating materials (dielectrics) will have a different impact on the capacitance. The dielectric constant (k) is directly proportional to the capacitance (C). The factors affecting the capacitance of a capacitor are illustrated in Figure 6.6. Factors affecting capacitance C ¼ 8:85  10 À12 kA d Dielectric constants for some commonly used capacitor materials are listed in Table 6.1. A d k Figure 6.6 Factors affecting capacitance Quantity Quantity symbol Unit Unit symbol Plates area A Square meter m 2 Distance d Meter m Dielectric constant k No unit Capacitance C Farad F Capacitors and inductors 169 Example 6.2: Determine the capacitance if the area of plates for a capacitor is 0.004 m 2 , the distance between the plates is 0.006 m and the dielectric for this capacitor is mica. Solution: A ¼ 0:004 m 2 ; d ¼ 0:006 m and k ¼ 5 C ¼ 8:85  10 À12 kA d ¼ 8:85  10 À12 5  0:004m 2 0:006m ¼ 29:5 pF 6.1.7 Leakage current The dielectric between two plates of the capacitor is insulating material, and practically no insulating material is perfect (i.e. 100 per cent of the insulation). Once voltage is applied across the capacitor, there may be a very small current through the dielectric, and this is called the leakage current in the capacitor. Although the leakage current is very small, it is always there. That is why the charges or the energy stored on the capacitor plates will eventually leak off. But the leakage current is so small that it can be ignored for the application. (Electrolytic capacitors have higher leakage current.) Leakage current A very small current through the dielectric. 6.1.8 Breakdown voltage As mentioned earlier, a capacitor charging acts as a pump pumping water into a reservoir, or a water tank. The higher the pressure, the more water will be pumped into the tank. If the tank is full and still continues to increase pressure, the tank may break down or become damaged by such high pressure. Table 6.1 Dielectric constants of some insulating materials Material Dielectric constant Vacuum 1 Air 1.0006 Paper (dry) 2.5 Glass (photographic) 7.5 Mica 5 Oil 4 Polystyrene 2.6 Teflon 2.1 170 Understandable electric circuits This is similar to a capacitor. If the voltage across a capacitor is too high and exceeds the capacitor’s working or breakdown voltage, the capacitor’s dielectric will break down, causing current to flow through it. As a result, this may explode or permanently damage the capacitor. Therefore, when using a capacitor, pay attention to the maximum working voltage, which is the max- imum voltage a capacitor can have. The applied voltage of the capacitor can never exceed the capacitor’s breakdown voltage. Breakdown voltage The voltage that causes a capacitor’s dielectric to become electrically conductive. It may explode or permanently damage the capacitor. 6.1.9 Relationship between the current and voltage of a capacitor The relationship between the current and voltage for a resistor is Ohm’s law for a resistor. The relationship between the current and voltage for a capacitor is Ohm’s law for a capacitor. It can be obtained mathematically as follows. A quantity that varies with time (such as a capacitor that takes time to charge/discharge) is called instantaneous quantity, which is the quantity at a specific time. Usually the lowercase letters symbolize instantaneous quantities, and the uppercase letters symbolize the constants or average quantities. The equation Q ¼ CV in terms of instantaneous quantity is q ¼ Cv. Note: If you haven’t learned calculus, just keep in mind that i ¼ C(Dv/Dt) or i C ¼ C(dv C /dt) is Ohm’s law for a capacitor, and skip the following mathematic derivation process, where Dv and Dt or dv and dt are very small changes in voltage and time. Differentiating the equation q ¼ Cv yields dq dt ¼ C dv dt Recall that current is the rate of movement of charges, and has the i ¼ dq/dt notation in calculus. Substitute i ¼ dq/dt into the equation of dq/dt ¼ C(dv/dt) yields i ¼ C dv dt or i ¼ C Dv Dt This is Ohm’s law for a capacitor. The relationship between voltage and cur- rent of a capacitor can be expressed by Figure 6.7(b). The relationship of voltage and current for a capacitor shows that when the applied voltage at two leads of the capacitor changes, the charges (q) stored on the plates of the capacitor will also change. This will cause current to flow in Capacitors and inductors 171 the capacitor circuit. Current and the rate of change of voltage are directly proportional to each other. The reference polarities of capacitor voltage and current should be mutually related. That is, the reference polarities of voltage and current of a capacitor should be consistent, as shown in Figure 6.7(a). Ohm’s law for a capacitor The current of a capacitor i C is directly proportional to the ratio of capacitor voltage dv C /dt (or Dv C /Dt) and capacitance C. i C ¼ C dv C dt or i ¼ C Dv C Dt where dv C and dt or Dv C , and Dt are very small changes in voltage and time. The relationship of voltage and current in a capacitive circuit shows that the faster the voltage changes with time, the greater the amount of capacitive current flows through the circuit. Similarly, the slower the voltage changes with time, the smaller the amount of current, and if voltage does not change with time, the current will be zero. Zero current means that the capacitor acts like an open circuit for DC voltage at this time. Voltage that does not change with time is DC voltage, meaning that current is zero when DC voltage is applied to a capacitor. Therefore, the capacitor may play an important role for blocking the DC current. This is a very important characteristic of a capacitor. DC blocking Current through a capacitor is zero when DC voltage applied to it (open- circuit equivalent). A capacitor can block DC current. Note: Although there is a DC voltage source applied to the capacitive circuit in Figures 6.3 and 6.4, the capacitor charging/discharging happened at the moment i C V + - i dt dv C 0 (a) (b) Figure 6.7 Relationship between v and i of a capacitor 172 Understandable electric circuits when the switch turned to different locations, i.e. when the voltage across the capacitor changes within a moment. When the capacitor charging/dischaging has finished, the capacitor is equivalent to an open circuit for that circuit. 6.1.10 Energy stored by a capacitor As mentioned earlier, a capacitor is an energy storage element. It can store energy that it absorbed from charging and maintain voltage across it. Energy stored by a capacitor in the electric field can be derived as follows. The instantaneous electric power of a capacitor is given by p ¼ ni. Sub- stituting this into the capacitor’s current i ¼ C(dv/dt) yields p ¼ Cv dv dt Since the relationship between power and work is P ¼ W/t (energy is the ability to do work), and, instantaneous power for this expression is p ¼ dw/dt, substituting it into p ¼ Cv(dv/dt) yields dw dt ¼ Cv dv dt Integrating the above expression: ð t 0 dw dt dt ¼ C ð v 0 v dv dt dt gives W ¼ 1 2 Cv 2 Note: If you haven’t learned calculus, just keep in mind that W ¼ ½(Cn 2 ), and skip the above mathematic derivation process. Energy stored by a capacitor W C ¼ 1 2 Cv 2 Quantity Quantity symbol Unit Unit symbol Energy W Joule J Capacitor C Farad F Voltage V Volt V Capacitors and inductors 173 The expression for energy stored by a capacitor demonstrates that the capacitor’s energy depends on the values of the capacitor and voltage across the capacitor. Example 6.3: A 15 V voltage is applied to a 2.2 mF capacitor. Determine the energy this capacitor has stored. Solution: W C ¼ 1 2 Cv 2 ¼ 1 2 ð2:2 mFÞ Â ð15 VÞ 2 ¼247:5 mJ 6.2 Capacitors in series and parallel Same as resistors, capacitors may also be connected in series or parallel to obtain a suitable resultant value that may be either higher or lower than a single capacitor value. The total or equivalent capacitance C eq will decrease for a series capacitive circuit and it will increase for a parallel capacitive circuit. The total or equivalent capacitance has the opposite form with the total or equivalent resistance R eq . 6.2.1 Capacitors in series A circuit of n capacitors is connected in series as shown in Figure 6.8. Applying Kirchhoff’s voltage law (KVL) to the above circuit gives E ¼ V 1 þ V 2 þ Á Á Á þ V n and since V ¼ Q/C, substituting it into the above expression yields Q eq C eq ¼ Q 1 C 1 þ Q 2 C 2 þ Á Á Á þ Q n C n E V 1 C 1 Q 1 Q 2 Q n C 2 C n V 2 V n … Figure 6.8 n capacitors in series 174 Understandable electric circuits where E ¼ Q eq /C eq , Q eq is the equivalent (or total) charges and C eq is the equivalent (or total) capacitance for a series capacitive circuit respectively. Since only one current flows in a series circuit, each capacitor will store the same amount of charges, i.e. Q eq ¼ Q 1 ¼ Q 2 ¼ . . . ¼ Q n ¼ Q therefore, Q C eq ¼ Q C 1 þ Q C 2 þ Á Á Á þ Q C n Dividing by Q on both sides of the above expression gives 1 C eq ¼ 1 C 1 þ 1 C 2 þ Á Á Á þ 1 C n or C eq ¼ 1 ð1=C 1 Þ þ ð1=C 2 Þ þ Á Á Á þ ð1=C n Þ This is the equation for calculating the series equivalent (total) capacitance. This formula has the same form with the formula for calculating equivalent parallel resistance (1/R eq ) ¼ (1/R 1 ) þ (1/R 2 ) þ . . . þ (1/R n ). When there are two capacitors in series, it also has the same form with the formula for cal- culating two resistors in parallel, i.e. C eq ¼ C 1 C 2 /(C 1 þ C 2 ) and R eq ¼ R 1 R 2 / (R 1 þ R 2 ). Equivalent (total) series capacitance ● n capacitors in series: C eq ¼ 1 ð1=C 1 Þþð1=C 2 ÞþÁÁÁþð1=C n Þ ● Two capacitors in series: C eq ¼ C 1 C 2 C 1 þC 2 Example 6.4: Determine the charges Q stored by each capacitor in the circuit of Figure 6.9. Solution: Since Q ¼ CV, or Q ¼ C eq E, solve for C eq first. E = 25 V C 1 C 2 C 3 C 4 100 μF 100 μF 100 μF 100 μF Figure 6.9 Circuit for Example 6.4 Capacitors and inductors 175 C eq ¼ 1 ð1=C 1 Þ þ ð1=C 2 Þ þ ð1=C 3 Þ þ ð1=C 4 Þ ¼ 1 ½ð1=100Þ þ ð1=100Þ þ ð1=100Þ þ ð1=100ÞmF ¼ 25 mF Therefore, Q ¼ C eq E ¼ ð25 mFÞ ð25 VÞ ¼ 625 mC Example 6.4 shows that when capacitors are connected in series, the total or equivalent capacitance C eq (25 mF) is less than any one of the individual capacitances (100 mF). The physical characteristic of the series equivalent capacitance is that the single series equivalent capacitance C eq has the total dielectric (or total distance between the plates) of all the individual capacitors. The formula for factors affecting the capacitance (C ¼ 8.85 6 10 712 kA/d) shows that if the distance between the plates of a capacitor (d) increases, the capacitance (C) will decrease. This is shown in Figure 6.10. 6.2.2 Capacitors in parallel A circuit of n capacitors connected in parallel is shown in Figure 6.11. The charge stored on the individual capacitor in this circuit is Q 1 ¼ C 1 V; Q 2 ¼ C 2 V; Q n ¼ C n V; ðwhere V ¼ EÞ C eq Figure 6.10 The physical characteristic of series C eq E … Q 1 C 1 C 2 C n Q 2 Q n Figure 6.11 n capacitors in parallel 176 Understandable electric circuits The total charge Q eq in this circuit should be the sum of all stored charges on the individual capacitor, i.e. Q eq ¼ Q 1 þ Q 2 þ Á Á Á þ Q n therefore, C eq V ¼ C 1 V þ C 2 V þ Á Á Á þ C n V dividing both sides by V yields C eq ¼ C 1 þ C 2 þ Á Á Á þ C n This is the equation for calculating the parallel equivalent (total) capacitance. As you may have noticed, this equation has the same form with the equation for calculating series resistances (R eq ¼ R 1 þ R 2 þ . . . þ R n ). Equivalent (total) parallel capacitance C eq ¼ C 1 þ C 2 þ . . . þ C n Equation for calculating capacitance is exactly opposite the equations for calculating resistance. Capacitors in series result in parallel form as resistances, and capacitors in parallel result in series form as resistances. Example 6.5: Determine the total charge in all the capacitors in the circuit of Figure 6.12. Solution: Since Q ¼ CV, i.e. Q eq ¼ C eq E and C eq ¼ C 1 þ C 2 þ Á Á Á þ C n ¼ ð100 þ 10 þ 20 þ 320ÞmF ¼ 450 mF therefore, Q eq ¼ C eq E ¼ ð450 mFÞð60 VÞ ¼ 27 000 mC From Example 6.5, we can see that when capacitors are connected in parallel, the total or equivalent capacitance C eq (450 mF) is greater than any one of the E = 60 V C 1 = 100 μF C 2 = 10 μF C 3 = 20 μF C 4 = 320 μF Figure 6.12 Circuit for Example 6.5 Capacitors and inductors 177 individual capacitances (C 1 ¼ 100 mF, C 2 ¼ 10 mF, C 3 ¼ 20 mF and C 4 ¼ 320 mF). The physical characteristic of the equation for calculating the parallel equivalent capacitance is that a single parallel equivalent capacitor C eq has the total area of plates of the individual capacitors. From the formula of factors affecting the capacitance (C ¼ 8.85 610 712 kA/d), we can see that if the area of plates (A) of a capacitor increases, the capacitance will increase. This is shown in Figure 6.13. 6.2.3 Capacitors in series–parallel Similar to resistors, capacitors may also be connected in various combinations. When serial and parallel capacitors are combined together, series–parallel capacitor circuits result and an example is shown in the following. Example 6.6: Determine the equivalent capacitance through two terminals a and b in the circuit of Figure 6.14. Solution: C 4;5 ¼ C 4 C 5 C 4 þ C 5 ¼ ð0:5mFÞð2mFÞ ð0:5 þ 2ÞmF ¼ 0:4 mF C 2;3;4;5 ¼ C 2 þ C 3 þ C 4;5 ¼ ð2 þ 0:6 þ 0:4ÞmF ¼ 3 mF C eq ¼ C 1 C 2;3;4;5 C 1 þ C 2;3;4;5 ¼ ð6mFÞð3mFÞ ð6 þ 3ÞmF ¼ 2 mF C eq Figure 6.13 The physical characteristic of parallel C eq b a C 1 = 6 μF C 2 = 2 μF C 4 = 0.5 μF C 3 = 0.6 μF C 5 = 2 μF Figure 6.14 Circuit for Example 6.6 178 Understandable electric circuits 6.3 Inductor We have learned about two of the three important fundamental passive circuit elements (components that absorb but not produce energy), the resistor and the capacitor. The third element is the inductor (or coil). Inductors have many applications in electrical and electronic devices, including electrical generators, transformers, radios, TVs, radars, motors, etc. As previously mentioned, both capacitors and inductors are energy storage elements. The difference between the two is that a capacitor stores transferred energy in the electric field, and an inductor stores transferred energy in the magnetic field. Since inductors are based on the theory of electromagnetism induction, let us review some concepts of electromagnetism induction you may have learned in physics that will be used in the following section. 6.3.1 Electromagnetism induction 6.3.1.1 Electromagnetic field All stationary electrical charges are surrounded by electric fields, and the move- ment of a charge will produce a magnetic field. When the charge changes its velocity of motion (or when the charge is accelerated), an electromagnetic field is generated. Therefore, whenever a changing current flows through a conductor, the area surrounding the conductor will produce an electromagnetic field. The electromagnetic field can be visualized by inserting a current-carrying conductor (wire) through a hole in a cardboard and sprinkling some iron filings on it. As the changing current flows through the conductor, the iron fillings will align themselves with the circles surrounding the conductor; these are magnetic lines of force. The direction of these lines of force can be deter- mined by the right-hand spiral rule, as shown in Figure 6.15. The area shows that the magnetic characteristics are called the magnetic field, as it is produced by the changing current-carrying conductor, and therefore, it is also called the electromagnetic field. This is the principle of electricity producing magnetism. ● Right-hand spiral rule: Thumb = the direction of current. Four fingers = the direction of magnetic lines of force or direction of the flux (the total magnetic lines of force). E I → → Figure 6.15 Electricity produces magnetism Capacitors and inductors 179 Electromagnetic field The surrounding area of a conductor with a changing current can gen- erate an electromagnetic field. 6.3.1.2 Faraday’s law In 1831, British physicist and chemist Michael Faraday discovered how an electromagnetic field can be induced by a changing magnetic flux. When there is a relative movement between a conductor and a magnetic field (or a chan- ging current through the conductor), it will induce a changing magnetic flux F (the total number of magnetic lines of force) surrounding the conductor, hence an electromagnetic field is generated. This electromagnetic field will produce an induced voltage and current. For example, in Figure 6.16, if a magnet bar is moved back and forth in a coil of wire (conductor), or if the coil is moved back and forth close to the magnet and through the magnetic field, the magnetic lines of flux will be cut and a voltage v L across the coil will be induced (v L can be measured by using a voltmeter.) Or, an electromotive force (emf, e L ) that has an opposite polarity with v L will be induced, and this will result in an induced current in the coil. This is the principle of a magnet producing electricity. Faraday observed that the induced voltage (v L ) is directly proportional to the rate of change of flux (df/dt) and also the number of turns (N) in the coil, and is expressed mathematically as v L ¼ N(df/dt). In other words, the faster the relative movement between the conductor and magnetic fields, or the more the turns the coil has, the higher the voltage will be produced. Faraday’s law ● When there is a relative movement between a conductor and mag- netic field, the changing magnetic flux will induce an electromagnetic field and produce an induced voltage (v L ). ● v L is directly proportional to the rate of change of flux (df/dt) and the number of turns (N) in the coil, v L ¼ N(df/dt). V Figure 6.16 Magnet produces electricity 180 Understandable electric circuits 6.3.1.3 Lenz’s law In 1834, Russian physicist Heinrich Lenz developed a companion result with the Faraday’s law. Lenz defined the polarity of induced effect and stated that an induced effect is always opposed to the cause producing it. When there is a relative movement between a conductor and a magnetic field (or a changing current through the conductor), an induced voltage (v L ) or induced emf (e L ) and also an induced current (i) will be produced. The polarity of the induced emf is always opposite to the change of the original current. When the switch is turned on in the circuit of Figure 6.17, the current (cause) in the circuit will increase, but the induced emf (effect) will try to stop it from increasing. When the switch is turned off, the current i will decrease, but the polarity of induced emf (e L ) changes and will try to stop it from decreasing. This is because an induced current in the circuit flows in a direction that can create a magnetic field that will counteract the change in the original magnetic flux. Mathematically, Lenz’s law can be expressed as follows: If i > 0; di dt > 0; then e L ¼ ÀL di dt ; or v L ¼ L di dt where di/dt is the rate of change of current, and the minus sign for e L is to remind us that the induced emf always acts to oppose the change in magnetic flux that generates the emf and current. The induced voltage (v L ) and induced emf (e L ) have opposite polarities (E ¼ 7V); this emf is also called the counter emf. However, the induced vol- tage (v L ) has the same polarity with the direction of induced current (i). This is similar to the concept of the mutually related reference polarity of voltage and current. Lenz’s law ● When there is a changing current through the conductor, an induced voltage (v L ) or induced emf (e L ) and also an induced current (i) will be produced. E R L e L v L e L v L + - - + i E + - + - Figure 6.17 Lenz’s law Capacitors and inductors 181 ● The polarity of the induced emf (e L ) is always opposite to the change of the original current. e L ¼ ÀL di dt or v L ¼ L di dt The letter L in the above equation is called inductance (or self-inductance), which is discussed below. 6.3.2 Inductor An inductor (L) is made by winding a given length of wire into a loop or coil around a core (centre of the coil). Inductors may be classified as air-core inductors or iron-core inductors. An air-core inductor is simply a coil of wire. But this coil turns out to be a very important electric/electronic element because of its magnetic properties. Iron-core provides a better path for the magnetic lines of force and a stronger magnetic field for the iron-core inductor as compared to the air-core inductor. The schematic symbol for an air-core inductor looks like a coil of wire as shown in Figure 6.18(a). The schematic symbol for an iron-core inductor is shown in Figure 6.18(b). Similar to resistors and capacitors, the inductor can be also classified as fixed and variable. Inductor L An inductor is an energy storage element that is made by winding a given length of wire into a loop or coil around a core. 6.3.3 Self-inductance When current flows through an inductor (coil) that is the same as a current- carrying conductor, a magnetic field will be induced around the inductor. According to the principle of electromagnetic induction, Faraday’s law and Lenz’s law, when there is a relative movement between an inductor and L L L L Air-core Variable air-core Iron-core Variable iron-core (b) (a) Figure 6.18 Schematic symbols for inductors 182 Understandable electric circuits magnetic field or when current changes in the inductor, the changing magnetic flux will induce an electromagnetic field resulting in an induced voltage (v L ), or induced emf (e L ), and also an induced current (i). The measurement of the changing current in an inductor that is able to generate induced voltage is called inductance. The inductor is symbolized by L while inductance is symbolized by L, and the unit of inductance is henry (H). The resistor, capacitor and inductor are circuit components, and the resistance, capacitance and inductance are the value or capacity of these components. So inductance is the capacity to store energy in the magnetic field of an inductor. Inductance L (or self-inductance) The measurement of the changing current in an inductor that is able to gene- rate induced voltage is called inductance that is measured in henries (H). 6.3.4 Relationship between inductor voltage and current Lenz’s law v L ¼ L(di/dt) shows the relationship between current and voltage for an inductor, and it is Ohm’s law for an inductor. There, the inductance (L) and the current rate of change (di/dt) determine the induced voltage (n L ). The induced voltage n L is directly proportional to the inductance L and the current rate of change di/dt. This relationship can be illustrated as in Figure 6.19. Ohm’s law for an inductor An inductor’s voltage n L is directly proportional to the inductance L and the rate of change current di/dt: n L ¼ L(di/dt). Ohm’s law for an inductor n L ¼ L(di/dt) has a similar form as Ohm’s law for a capacitor i C ¼ C(dn C /dt). These two are very important formulas that will be used in future circuits. Quantity Quantity symbol Unit Unit symbol Inductance (or Self-Inductance) L Henry H L + - dt di dt di v L = L i v L L Figure 6.19 Characteristics of an inductor’s voltage and current Capacitors and inductors 183 The larger the inductance, or the greater the change of current, the higher the induced voltage in the coil. When the current does not change with time (DC current), i.e. di/dt ¼ 0, the inductor voltage (v L ) is also zero. Zero voltage means that an inductor acts like a short circuit for DC current. Therefore, the inductor may play an important role for passing the DC current. This is a very important characteristic of an inductor and is opposite to that of a capacitor. Recall that a capacitor can block DC and acts like an open circuit for DC. Passing DC ● Voltage across an inductor is zero when a DC current flows through it (short-circuit equivalent). ● An inductor can pass DC. 6.3.5 Factors affecting inductance There are some basic factors affecting the inductance of an inductor (iron- core). These parameters are determined by the construction of an inductor as shown in the following (if all other factors are equal): ● The number of turns (N) for the coil: More turns for a coil will produce a stronger magnetic field resulting in a higher induced voltage and inductance. ● The length of the core (l): A longer core will make a loosely spaced coil and a longer distance between each turn, and therefore producing a weaker magnetic field resulting in a smaller inductance. ● The cross-section area of the core (A): A larger core area requires more wire to construct a coil, and therefore it can produce a stronger magnetic field resulting in a higher inductance. ● The permeability of the material of the core (m): A core material with higher permeability will produce a stronger magnetic field resulting in a higher inductance. (Permeability of the material of the core determines the ability of material to produce a magnetic field. Different materials have different degrees of permeability.) Factors affecting the inductance of an inductor are illustrated in Figure 6.20. Factors affecting inductance L ¼ N 2 Am l 184 Understandable electric circuits Where inductance is symbolized by L, measured in henries (H); area of the core is symbolized by A, measured in m 2 ; permeability is symbolized by m; number of turns is symbolized by N. From the expression of the factor affecting inductance, we can see that either when the number of turns of a coil increases, or when the cross-section area of the core increases, or when core material with higher permeability is chosen, or when the length of core is reduced, the inductance of an inductor will increase. 6.3.6 The energy stored by an inductor Same as a capacitor, an inductor is also an energy storage element. When voltage is applied to two leads of an inductor, the current flows through the inductor and will generate energy, and this energy is then absorbed by the inductor and stored in the magnetic field as electromagnetic field builds up. The energy stored by an inductor can be derived as follows: The instantaneous electric power of an inductor is given by p ¼ iv L Since the relationship between power and work is P ¼ W=t (energy is the ability to do work), and the instantaneous power for this expression is p ¼ dw/dt. Substituting p ¼ dw/dt and v L ¼ L(di/dt) into the instantaneous power expression p ¼ in L gives dw dt ¼ Li di dt A N L μ Figure 6.20 Factors affecting inductance Capacitors and inductors 185 Integrating both sides: ð t 0 dw dt dt ¼ ð t 0 Li di dt dt Therefore w ¼ L ð t 0 idi; i:e: w ¼ 1 2 Li 2 Note: If you haven’t learned calculus, just keep in mind that W L ¼ (½)Li 2 , and skip the above mathematic derivation process. This equation has a similar form with the energy equation of a capacitor (W C ¼ (½)Cv 2 ). The equation for energy stored by an inductor shows that the inductor’s energy depends on the inductance and the inductor’s current. When current increases, an inductor absorbs energy and stores it in the magnetic field of the inductor. When current decreases, an inductor releases the stored energy to the circuit. Same as a capacitor, an inductor will not be able to release more energy than it has stored, so it is also called a passive element. Energy stored by an inductor W L ¼ 1 2 Li 2 where inductance L is measured in H, energy W is measured in J and current i is measured in A. Example 6.7: Current in a 0.01 H inductor is i(t) ¼ 5e 72t A, determine the energy stored by the inductor and induced voltage n L . Solution: W L ¼ 1 2 Li 2 ¼ 1 2 ð0:01 HÞð5e À2t AÞ 2 ¼ 1 2 ð0:01 HÞð25e À4t AÞ ¼ 0:125e À4t J v L ¼ L di dt ¼ 0:01 d dt ð5e À2t Þ ¼ 0:01 HðÀ2Þð5Þðe À2t AÞ ¼ À0:1e À2t V Note: If you haven’t learned calculus, skip the v L part. 6.3.7 Winding resistor of an inductor When winding a given length of wire into a loop or coil around a core, an inductor is formed. A coil or inductor always has resistance. This is because 186 Understandable electric circuits there is always a certain internal resistance distributed in the wire, and the longer the wire, the more turns of coils there are, and thus the wire will have a significantly higher internal resistance. This is called the winding resistance of a coil (R w ). An inductor circuit with winding resistance is shown in Figure 6.21. Winding resistance R w The internal resistance in the wire of an inductor. Example 6.8: The winding resistance for an inductor in the circuit of Figure 6.22 is 5 O. When the current approaches a steady state (does not change any more), the energy stored by the inductor is 4 J. What is the inductance of the inductor? R w L Figure 6.21 Winding resistance L I E = 20 V R = 45 Ω R w = 5 Ω Figure 6.22 Circuit for Example 6.8 Solution: E ¼ 20 V; R ¼ 45 O; R w ¼ 5 O; and W L ¼ 4 J: L ¼ ? I ¼ E R þ R w ¼ 20 V ð45 þ 5ÞO ¼ 0:4 A Capacitors and inductors 187 From W L = ½(Li 2 ) solving for L: L ¼ 2W L I 2 ¼ 2  4 J ð0:4 AÞ 2 ¼ 50 H (i ¼ I since the current approaches to steady state) 6.4 Inductors in series and parallel Similar to resistors and capacitors, inductors may also be connected in series or in parallel to obtain a suitable resultant value that may be either higher or lower than a single inductor value. The equivalent (total) series or parallel inductance has the same form as the equivalent (total) series or parallel resis- tance. The equivalent inductance will increase if inductors are in series, and the equivalent (total) inductance will decrease if inductors are in parallel. 6.4.1 Inductors in series A circuit of n inductors connected in series is shown in Figure 6.23. Equivalent series inductance L eq ¼ L 1 þ L 2 þ . . . þ L n This is the equation for calculating the equivalent (total) series inductance. As you may have noticed, this formula has the same form as the formula for calculating series resistances (R eq ¼ R 1 þ R 2 þ . . . þ R n ). 6.4.2 Inductors in parallel A circuit of n inductors connected in parallel is shown in Figure 6.24. … L eq L 1 L 2 L n Figure 6.23 Inductors in series L 1 L 2 L n L eq … Figure 6.24 Inductors in parallel 188 Understandable electric circuits Equivalent parallel inductance ● n inductors in parallel: L eq ¼ 1 ð1=L 1 Þþð1=L 2 ÞþÁÁÁþð1=L n Þ ● Two inductors in parallel: L eq ¼ L 1 L 2 L 1 þL 2 These are the equations for calculating the equivalent parallel inductance. As you may have noticed, these equations have the same forms as the equa- tions for calculating parallel resistance R eq ¼ 1/(1/R 1 ) þ (1/R 2 ) þ . . . þ (1/R n ), also R eq ¼ R 1 R 2 /(R 1 þ R 2 ). 6.4.3 Inductors in series–parallel Similar to resistors and capacitors, inductors may also be connected in various combinations of series and parallel. An example of a series–parallel inductive circuit is shown in the following. Example 6.9: Determine the equivalent inductance for the series–parallel inductive circuit shown in Figure 6.25. Solution: L eq ¼ ½ðL 4 þ L 5 Þ==L 3 þ ðL 2 þ L 6 Þ==L 1 L eq ¼ ðð1 þ 1Þ Â 2Þ=ðð1 þ 1Þ þ 2Þ þ 1 þ 1 ½  3 ðð1 þ 1Þ Â 2Þ=ðð1 þ 1Þ þ 2Þ þ 1 þ 1 ½ þ 3 H ¼ 1:5 H Example 6.10: There are three inductors in a series–parallel inductive circuit: 40, 40 and 50 H. If L eq ¼ 70 H, how are these inductors connected? L eq L 2 = 1H L 1 = 3H L 6 = 1H L 5 = 1H L 4 = 1H L 3 = 2H Figure 6.25 Circuit for Example 6.9 Capacitors and inductors 189 Solution: L eq ¼ 50 H þ ð40 HÞ==ð40 HÞ ¼ 70 H or L eq ¼ 50 H þ 40  40 40 þ 40 H ¼ 70 H So two 40 H inductors are in parallel, and then in series with a 50 H inductor. Summary Capacitor ● Capacitor (C): An energy storage element that has two conductive plates separated by an isolating material (the dielectric). ● Capacitor charging: Capacitor stores absorbed energy. ● Capacitor discharging: Capacitor releases energy to the circuit. ● Capacitance (C): The value of the capacitor, C ¼ Q/U. ● Factors affecting capacitance: C ¼ 8:85  10 À12 kA d ● Leakage current: A very small current through the dielectric. ● Breakdown voltage: The voltage that causes a capacitor’s dielectric to become electrically conductive, it can explode or permanently damage the capacitor. ● Blocking DC: A capacitor can block DC current (open-circuit equivalent). Inductor ● Electromagnetic field: The surrounding area of a conductor with a chan- ging current can generate an electromagnetic field. ● Faraday’s law: v L ¼ N df dt ● Lenz’s law : e L ¼ ÀL di dt or v L ¼ L di dt ● Inductor (L): An energy storage element that is made by winding a given length of wire into a loop or coil around a core. ● Inductance (L): The measurement of the changing current in an inductor that produces the ability to generate induced voltage is called inductance. 190 Understandable electric circuits ● Factors affecting inductance: L ¼ N 2 Am l ● Winding resistance (R w ): The internal resistance in the wire of an inductor. The characteristics of the resistor, capacitor and inductor Experiment 6: Capacitors Objectives ● Understand the characteristics of a capacitor through experiment. ● Verify the equations of capacitors in series and parallel through experiment. ● Apply the voltage divider rule in a capacitive circuit. ● Analyse experimental data, circuit behaviour and performance, and com- pare them to the theoretical equivalents. Background information ● Parallel equivalent capacitance: C eq ¼ C 1 þ C 2 þ Á Á Á þ C n ● Series equivalent capacitance: C eq ¼ 1 ð1=C 1 Þ þ ð1=C 2 Þ þ Á Á Á þ ð1=C n Þ Characteristic Resistor Capacitor Inductor Ohm’s law V ¼ IR i C ¼ C(dn C /dt) n L ¼ L(di/dt) Energy W ¼ pt or dw ¼ pdt W C ¼ ½(Cn 2 ) W L ¼ ½(Li 2 ) Series R eq ¼ R 1 þ R 2 þ . . . þ R n C eq ¼ 1/[(1/C 1 ) þ (1/C 2 ) þ . . . þ (1/C n )] Two capacitors: C eq ¼ C 1 C 2 / (C 1 þ C 2 ) L eq ¼ L 1 þ L 2 þ . . . þ L n Parallel R eq ¼ 1/[(1/R 1 ) þ (1/R 2 ) þ . . . þ (1/R n )] Two resistors: R eq ¼ R 1 R 2 / (R 1 þ R 2 ) C eq ¼ C 1 þ C 2 þ . . . þ C n L eq ¼ 1/[(1/L 1 ) þ (1/L 2 ) þ . . . þ (1/L n )] Two inductors: R eq ¼ L 1 L 2 / (L 1 þ L 2 ) Elements in DC Open-circuit equivalent Short-circuit equivalent Capacitors and inductors 191 When n ¼ 2: C eq ¼ C 1 C 2 C 1 þ C 2 ● Voltage divider rule for capacitive circuit: V C 1 ¼ E C eq C 1 ; V C 2 ¼ E C eq C 2 Recall the voltage divider rule for resistive circuit: V R 1 ¼ E R 1 R eq V R 2 ¼ E R 2 R eq Equipment and components ● Multimeter ● Breadboard ● DC power supply ● Z meter or LCZ meter (or any other measuring instruments that can be used to measure capacitance): ● Z meter: A measuring instrument that can be used to measure the values of capacitors. ● LCZ meter: A measuring instrument that can be used to measure the values of capacitors and inductors. ● Capacitors: ● 30 and 470 mF electrolytic capacitors each ● Four non-electrolytic capacitors with any values Note: Electrolytic capacitors are polarized. Connect the positive lead of the capacitor to the positive terminal of the DC power supply, and negative lead to the negative terminal of the DC power supply. Non-electrolytic capacitors are non-polarized, so they can be connected either way in a circuit. Procedure Part I: Capacitors in series and parallel 1. Take four non-electrolytic capacitors, short circuit each capacitor with a bit of wire to release or discharge the stored charges on the capacitor. Record their nominal values in Table L6.1. (A capacitor can hold its stored charges for days or weeks and can shock you even when it is not connected to a circuit.) Table L6.1 Capacitor C 1 C 2 C 3 C 4 Nominal value Measured value 192 Understandable electric circuits 2. Measure the value of each capacitor using a Z meter or LCZ meter and record in Table L6.1. 3. Connect each circuit as shown in Figure L6.1(a, b and c) on the bread- board. Calculate the equivalent capacitance for each circuit. Record the values in Table L6.2. 4. Measure the equivalent capacitance for each circuit in Figure L6.1. Record the values in Table L6.2. 5. Compare the measured values and calculated values, are there any sig- nificant differences? If so, explain the reasons. Part II: Apply voltage divider rule in the capacitive circuit 1. Take two capacitors with the value shown in Table L6.3, short circuit each capacitor with a bit of wire to release or discharge the stored charges on the capacitor. 2. Connect a series capacitive circuit as shown in Figure L6.2 on the breadboard. C 1 C 2 C 3 C 4 C 1 C 2 C 3 C 4 C 1 C 2 C 3 C 4 (a) (b) (c) Figure L6.1 Capacitors in series and parallel Table L6.2 Equivalent capacitance C eq for Figure L6.1(a) C eq for Figure L6.1(b) C eq for Figure L6.1(c) Formula for calculations Calculated value Measured value Table L6.3 Capacitor C 1 C 2 Nominal value 30 mF 470 mF Measured value Capacitors and inductors 193 3. Calculate V C 1 and V C 2 in the circuit of Figure L6.2 using the voltage divider rule. Record the values in Table L6.4. 4. Measure the voltage across each capacitor from Figure L6.2 using the multimeter (voltmeter function). Record the values in Table L6.4. 5. Compare the measured values and calculated values, are there any sig- nificant differences? If so, explain the reasons. Note: Take down the measurement quickly, otherwise the capacitor will dis- charge gradually. Conclusion Write your conclusions below: E = 6 V V C 1 V C 2 C 1 = 30 μF C 2 = 470 μF Figure L6.2 Capacitor in series Table L6.4 V C 1 V C 2 Formula for calculations Calculated value Measured value 194 Understandable electric circuits Chapter 7 Transient analysis of circuits Objectives After completing this chapter, you will be able to: ● understand the first-order circuits and concepts of the step response and source-free response of the circuits ● understand the initial conditions in the switching circuit ● understand the concepts of the transient and steady states of RL and RC circuits ● determine the charging/discharging process in an RC circuit ● determine the energy storing/releasing process in an RL circuit ● understand the concepts of time constants for RL and RC circuits ● plot the voltages and currents verse time curves for RL and RC circuits ● understand the relationship between the time constant and the charging/ discharging in an RC circuit ● understand the relationship between the time constant and the energy storing/releasing in an RL circuit 7.1 Transient response 7.1.1 The first-order circuit and its transient response There are three basic elements in an electric circuit, the resistor R, capacitor C and inductor L. The circuits in this chapter will combine the resistor(s) R with an energy storage element capacitor C or an inductor L to form an RL (resistor–inductor) or RC (resistor–capacitor) circuit. These circuits exhibit the important behaviours that are fundamental to much of analogue electronics, and they are used very often in electric and electronic circuits. Analysis RL or RC circuits still use Kirchhoff’s current law (KCL) and Kirchhoff’s voltage law (KVL). The main difference between these types of cir- cuits and pure resistor circuits is that the pure resistor circuits can be analysed by algebraic methods. Since the relationship of voltages and currents in the capacitor and inductor circuits is expressed by the derivative, differential equations (the equations with the derivative) will be used to describe RC and RL circuits. RL or RC circuits that are described by the first-order differential equa- tions, or the circuits that include resistor(s), and only one single energy storage element (inductor or capacitor), are called the first-order circuits. First-order circuit ● The circuit that contains resistor(s), and a single energy storage ele- ment (L or C). ● RL or RC circuits that are described by the first-order differential equations. We have discussed the concept of charging/discharging behaviour of the energy storage element capacitor C in chapter 6. Another energy storage ele- ment inductor L also has the similarly energy storing/releasing behaviour. The difference is that charging/discharging of a capacitor is in the electric filed, and the energy storing/releasing of an inductor is in the magnetic field. There are two types of circuit states in RL or RC circuit, the transient and steady states. The transient state is the dynamic state that occurs by a sudden change of voltage, current, etc. in a circuit. That means the dynamic state of the circuit has been changed, such as the process of charging/discharging a capacitor or energy storing/releasing for an inductor as the result of the operation of a switch. The steady-state is an equilibrium condition that occurs in a circuit when all transients have finished. It is the stable-circuit state when all the physical quantities in the circuit have stopped changing. For the process of charging/discharging a capacitor or energy storing/releasing for an induc- tor, it is the result of the operation of a switch in the circuit after a certain time interval. Transient state: The dynamic state that occurs when the physical quan- tities have been changed suddenly. Steady state: An equilibrium condition that occurs when all physical quantities have stopped changing and all transients have finished. 7.1.2 Circuit responses A response is the effect of an output resulting from an input. The first-order RL or RC circuit has two responses, one is called the step response, and the other is the source-free response. The step response for a general system states that the time behaviour of the outputs when its inputs change from zero to unity value (1) in a very short time. And the step response for an RC or RL circuit is the circuit responses 196 Understandable electric circuits (outputs) when the initial state of the energy store elements L or C is zero and the input (DC power source) is not zero in a very short time. It is when a DC source voltage is instantly applied to the circuit, the energy store elements L or C hasn’t stored energy yet and the output current or vol- tage generated in this first-order circuit. Or the charging process of the energy storing process of the capacitor or inductor. The step response can be analo- gized as a process to fill up water in a reservoir or a water bottle. Following are some of the basic terms for a step response: ● The initial state: the state when an energy storage element hasn’t stored energy yet. ● Input (excitation): the power supply. ● Output (response): the resultant current and voltage. Step response The circuit response when the initial condition of the energy store ele- ments (L or C) is zero, and the input (DC power source) is not zero in a very short time, i.e. the charging/storing process of the C or L. The source-free response or natural response is opposite to the step response. It is the circuit response when the input is zero, and the initial con- dition of the capacitor or inductor is not zero (the energy has been stored to the capacitor or inductor). It is the discharging or energy releasing process of the capacitor or inductor in an RC or RL circuit. The source-free response can also be analogized as the process to release water in a reservoir or a water bottle. Source-free (or natural) response The circuit response when the input (DC power source) is zero, and the initial condition of the energy store elements (L or C) is not zero, i.e. the discharging/releasing process of the C or L. When an RC or RL circuit that is initially at ‘rest’ with zero initial condi- tion and a DC voltage source is switched on to this circuit instantly, this DC voltage source can be analogized as a unit-step function, since it ‘steps’ from zero to a unit constant value (1). So the step response can be also called the unit-step response. The unit-step response is defined as follows: All initial conditions of the circuit are zero at time less than zero (t 5 0), i.e. at the moment of time before the power turns on. And the response v(t) or output voltage for this condition is obviously also zero. After the power turns on (t 4 0), the response v(t) will Transient analysis of circuits 197 be a constant unit value 1, as shown in the following mathematical expression and also can be illustrated in Figure 7.1(b). vðtÞ ¼ 0; t < 0 1; t > 0 This unit-step function can be expressed as the switch in the circuit of Figure 7.1(a), when t ¼ 0, the switch turns to position 1, a DC power source is con- nected to the RC circuit, and produces a unit-step response to the circuit. 7.1.3 The initial condition of the dynamic circuit The process of charging and discharging of a capacitor needs a switch to connect or disconnect to the DC source in the RC circuit, as shown in the circuit of Figure 7.1(a). The instantly turned on or turned off the switch, or the source input that is switched ‘on’ or ‘off’ in an RC or RL circuit is called the switching circuit. At the moment when the circuit is suddenly switched, the capacitor voltage and inductor current will not change instantly, this concept can be described as t ¼ 07and t ¼ 0þ. ● t ¼ 07is the instant time interval before switching the circuit (turn off the switch). ● t ¼ 0þ is the instant time interval after switching the circuit (turn on the switch). At this switching moment, the non-zero initial capacitor voltage and inductor current can be expressed as follows: v C ð0þ Þ ¼ v C ð0À Þ and i L ð0þÞ ¼ i L ð0ÀÞ And ● v C (07) is the capacitor voltage at the instant time before the switch is closed. ● v C (0þ) is the capacitor voltage at the instant time after the switch is closed. ● v L (07) is the inductor current at the instant time before the switch is closed. ● i L (0þ) is the inductor current at the instant time after the switch is closed. (a) (b) 0 t 1 C 1 2 3 R E = 1 V Figure 7.1 Step function 198 Understandable electric circuits Initial conditions ● Switching circuit: the instantly turned on or turned off switch in the circuit. ● t ¼ 07: the instant time interval before the switch is closed. ● t ¼ 0þ: the instant time interval after the switch is closed. ● At the instant time before/after the switch is closed, v C and i L do not change instantly: v C ð0þÞ ¼ v C ð0ÀÞ and i L ð0þÞ ¼ i L ð0ÀÞ: 7.2 The step response of an RC circuit Chapter 6 has discussed the charging and discharging process of a capacitor. When there are no resistors in the circuit, a pure capacitive circuit will fill with electric charges instantly, or release the stored electric charges instantly. But there is always a small amount of resistance in the practical capacitive circuits. Sometimes a resistor will be connected to a capacitive circuit that is used very often in the different applications of the electronic circuits. Figure 7.2 is a resistor–capacitor series circuit that has a switch connecting to the DC power supply. Such a circuit is generally referred to as an RC circuit. All important concepts of step response (charging) or source-free response (discharging) and transient and steady state of an RC circuit can be analysed by this simple circuit. 7.2.1 The charging process of an RC circuit Assuming the capacitor has not been charged yet in the circuit of Figure 7.2, the switch is in position 2 (middle). What will happen when the switch is turned to position 1, and the DC power source (E) is connected to the RC series circuit as shown in the circuit of Figure 7.3? E C 1 R 2 3 Figure 7.2 An RC circuit Transient analysis of circuits 199 The energy storing element capacitor C will start charging. Since there is a resistor in this circuit, the process of the capacitor’s charging will not finish instantly, the capacitor will gradually store the electric charges. This RC circuit is similar to a reservoir (or a water bottle) filling with water to capacity. If the door of the reservoir opened only to a certain width, the reservoir will need more time to fill up with the water (or the water bottle will need more time to fill up with water if the tab didn’t fully open). The voltage across the capacitor v C is not instantaneously equal to the source voltage E when the switch is closed to 1. The capacitor voltage v is zero at the beginning. It needs time to overcome the resistance R of the circuit to gradually charge to the source voltage E. After this charging time interval or the transient state of the RC circuit, the capacitor can be fully charged; this is shown Figure 7.4. Figure 7.4(b) indicates that capacitor voltage v C increases exponentially from zero to its final value (E). The voltage across the capacitor will be increased until it reaches the source voltage (E), at that time no more charges will flow onto the plates of the capacitor, i.e. the circuit current stops flowing. And the capacitor will reach a state of dynamic equilibrium (steady state). The state of the circuit voltage or current after charging is called the steady state. Once they reach the steady state, the current and voltage in the circuit will not change any more, and at this time, the capacitor voltage is equal to the source voltage, i.e. v C ¼ E. The circuit current is zero, and the capacitor is equivalent to an open circuit as shown in the circuit of Figure 7.4(a). For this E 1 3 R v C = E E v C 0 (a) (b) t Figure 7.4 The charging process of an RC circuit E C 1 v R 3 R v C + − − + i Figure 7.3 RC charging circuit 200 Understandable electric circuits open circuit, the current stops to flow. Therefore, there is no voltage drop across the resistor. The phenomena of the capacitor voltage v C increases dexponentially from zero to its final value E (or a charging process) in an RC circuit can be also analysed by the quantity analysis method as follows. 7.2.2 Quantity analysis for the charging process of the RC circuit The polarities of the capacitor and resistor voltages of the RC circuit are shown in Figure 7.3. Applying KVL to this circuit will result in v R þv C ¼ E ð7:1Þ The voltage drop across the resistor is Ri (Ohm’s law) while the current through this circuit is i ¼ Cðdv C =dtÞ (from chapter 6, section 1.9), i.e. v R ¼ Ri i ¼ C dv C dt Therefore, v R ¼ RC dv C dt ð7:2Þ Substituting (7.2) into (7.1) yields RC dv C dt þv C ¼ E ð7:3Þ ● Determine the capacitor voltage v C Note: If you haven’t learned calculus, then just keep in mind that (7.4) is the equation for the capacitor voltage v C during the discharging process in an RC circuit, and skip the following mathematical derivation process. The first-order differentia (7.3) can be rearranged as v C ÀE ¼ ÀRC dv C dt Divide both sides by 7RC À 1 RC ðv C À EÞ ¼ dv C dt rearrange À dt RC ¼ dv C v C ÀE Transient analysis of circuits 201 Integrating the above equation on both sides yields À 1 RC ð t 0 dt ¼ ð v C 0 dv C v C ÀE À t RC t 0 ¼ ln v C À Ej v C 0 rearrange À t RC ¼ ln v C ÀEj j À ln ÀE j j ln v C À E ÀE ¼ À t RC Taking the natural exponent (e) on both sides results in e ln ðv c ÀEÞ=ðÀEÞ j j ¼ e Àt=RC v C À E ÀE ¼ e Àt=RC Solve for v C v C ¼ Eð1 À e Àt=RC Þ ð7:4Þ The above equation is the capacitor voltage during the charging process in an RC circuit. ● Determine the resistor voltage v R Applying KVL in the circuit of Figure 7.3 v R þv C ¼ E rearrange v R ¼ E Àv C ð7:5Þ Substituting the capacitor voltage v C ¼ Eð1 À e Àt=RC Þ into (7.5) yields v R ¼ E ÀEð1 À e Àt=RC Þ Therefore, the resistor voltage is v R ¼ Ee Àt=RC 202 Understandable electric circuits ● Determine the charging current i Dividing both sides of the equation v R ¼ Ee Àt=RC by R yields v R R ¼ E R e Àt=RC Applying Ohm’s law to the left side of the above equation will result in the charging current i i ¼ E R e Àt=RC Charging equations for an RC circuit ● Capacitor voltage: v C ¼ Eð1 À e Àt=RC Þ ● Resistor voltage: v R ¼ Ee Àt=RC ● Charging current: i ¼ E R e Àt=RC Mathematically, these three equations indicate that capacitor voltage increases exponentially from initial value zero to the final value E; the resistor voltage and the charging current decay exponentially from initial value E and E/R (or I max ) to zero, respectively. And t is the charging time in the equations. According to the above mathematical equations, the curves of v C , v R and i versus time can be plotted as in Figure 7.5. Example 7.1: For the circuit shown in Figure 7.3, if E ¼ 25 V, R ¼2.5 kO and C ¼ 2.5 mF, the charging time t ¼ 37.5 ms. Determine the resistor voltage v R and capacitor voltage v C . Solution: RC ¼ (2.5 kO)(2.5 mF) ¼ 6.25 ms ● v R ¼ Ee Àt=RC ¼ ð25 VÞðe ðÀ37:5=6:25Þms Þ ¼ ð25 VÞðe À6 Þ % 0:062 V v C E t 0 E R E i v R t t 0 0 Figure 7.5 v C , v R and i versus t Transient analysis of circuits 203 ● v C ¼ Eð1 À e Àt=RC Þ ¼ ð25 VÞð1 À e ðÀ37:5=6:25Þms Þ ¼ ð25 VÞð1 À e À6 Þ ¼ 24:938 V These results can be checked by using KVL: v R þ v C ¼ E. Substituting the values into KVL yields v R þv C ¼ E ð0:062 þ 24:938ÞV ¼ 25 V ðcheckedÞ Thus, the sum of the capacitor voltage and resistor voltage must be equal to the source voltage in the RC circuit. 7.3 The source-free response of the RC circuit 7.3.1 The discharging process of the RC circuit Consider a capacitor C that has initially charged to a certain voltage value v 0 (such as the DC source voltage E) through the charging process of the last sec- tion in the circuit of Figure 7.4(a). The voltage across the capacitor is v C ¼ E, whose function will be the same as a voltage source in the right loop of the RC circuit in Figure 7.6. Once the switch turns to position 3 as shown in the circuit of Figure 7.6(b), the capacitor will start discharging, but now it will be different than a pure capacitive circuit that can discharge instantly. The discharging time will increase since there is a resistor in the circuit. It needs some time to overcome the resistance and eventually release all the charges from the capacitor. Once the capacitor has finished the discharge, the capacitor voltage v C will be 0, the discharging curve is shown in Figure 7.7. This is similar to a reservoir that has an opened door to release the water (or the water bottle has an opened lid to pour water). But the releasing door of a reservoir is not open wide enough, so it will need some time to release all the water. v R 3 1 E v C = E + − + − i E C 1 R 2 3 v C = E (a) (b) Figure 7.6 Discharging process of the RC circuit 204 Understandable electric circuits 7.3.2 Quantity analysis of the RC discharging process The equations used to calculate the capacitor voltage v C , resistor voltage v R and discharging current i of the capacitor discharging circuit can be determined by the following mathematical analysis method. Applying KVL to the circuit in Figure 7.6(b) will result in v R Àv C ¼ 0 or v R ¼ v C ð7:6Þ Since v R ¼ iR and i ¼ ÀC dv C dt Substitute i into the equation of v R v R ¼ ÀR C dv C dt ð7:7Þ The negative sign (7) in the above equation is because the current i and vol- tage v C in the circuit of Figure 7.6(b) have opposite polarities. Substitute (7.7) into the left-hand side of (7.6) ÀRC dv C dt ¼ v C Divide both sides of the above equation by 7RC dv C dt ¼ À 1 RC v C ð7:8Þ ● Determine the capacitor voltage v C Note: If you haven’t learned calculus, then just keep in mind that (7.10) is the equation for the capacitor voltage v C during the charging process in an RC circuit, and skip the following mathematical derivation process. Integrating (7.8) on both sides yields ð dv C v C ¼ À 1 RC ð dt t v C E 0 Figure 7.7 Discharge curve of the RC circuit Transient analysis of circuits 205 ln jv C j ¼ À 1 RC t þ ln A (ln A – the constant of the integration) or ln jv C j À ln A ¼ À 1 RC t Rearrange ln v C A ¼ À t ðRCÞ Taking the natural exponent (e) on both sides of the above equation: e ln jv C =Aj ¼ e Àt=RC Therefore, v C A ¼ e Àt=RC or v C ¼ Ae Àt=RC ð7:9Þ As the capacitor has been charged to an initial voltage value v 0 before being connected to the circuit in Figure 7.6(b), the initial condition (initial value) of the capacitor voltage should be v C ð0ÀÞ ¼ V 0 v 0 can be any initial voltage value for the capacitor, such as the source voltage E. Immediately before/after the switch is closed to the position 3 in the circuit of Figure 7.6(b), v C does not change instantly (from the section 7.1.3), therefore, v C ð0 þ Þ ¼ v C ð0 À Þ or v C ¼ V 0 When t ¼ 0, substituting v C ¼ v 0 in (7.9) yields V 0 ¼ Ae À0=RC That is V 0 ¼ A 206 Understandable electric circuits Substitute v 0 ¼ A into (7.9) v C ¼ V 0 e Àt=RC ð7:10Þ This is the equation of the capacitor voltage for the RC discharging circuit. ● Determine the resistor voltage v R According to (7.6) v R ¼ v C Substitute (7.10) into (7.6) yields v R ¼ V 0 e Àt=RC ð7:11Þ ● Determine the discharge current i Since i ¼ v R R ðOhm ‘ s lawÞ Substitute (7.11) into the above Ohm’s law will result in i ¼ V 0 R e Àt=RC Discharging equations for an RC circuit ● Capacitor voltage: v C ¼ V 0 e Àt=RC ● Resistor voltage: v R ¼ V 0 e Àt=RC ● Discharging current: i ¼ V 0 R e Àt=RC In the above equations, t is the discharging time. V 0 is the initial capacitor voltage. These three equations mathematically indicate that capacitor voltage v C and the resistor voltage v R decay exponentially from initial value V 0 to the final value zero; and the discharging current i decays exponentially from the initial value v 0 /R (or I max ) to the final value zero. The curves of v C , v R and i versus time t can be illustrated as shown in Figure 7.8. The discharging voltage and current decay exponentially from the initial value to zero, this means that the capacitor gradually releases the stored energy, and eventually the energy stored in the capacitor will be released to the circuit completely, and it will be received by the resistor and convert to heat energy. Transient analysis of circuits 207 7.3.3 RC time constant t In an RC circuit, the charging and discharging is a gradual process that needs some time. The time rate of this process depends on the values of circuit capacitance C and resistance R. The variation of the R and C will affect rate of the charging and discharging. The product of the R and C is called the RC time constant and it can be expressed as a Greek letter t (tau), i.e. t ¼ RC. Generally speaking, the time constant is the time interval required for a system or circuit to change from one state to another, i.e. the time required from the transient to the steady state or to charge or discharge in an RC circuit. RC time constant t ¼ RC The time constant t represents the time the capacitor voltage reaches (increa- ses) to 63.2% of its final value (steady state) or the time the capacitor voltage decays (decreases) below to 36.8% of its initial value. The higher the R and C values (or when the time constant t increases), the longer the charging or dis- charging time; lesser the v C variation, longer the time to reach the final or initial values. This can be shown in Figure 7.9. Quantity Quantity symbol Unit Unit symbol Resistance R Ohm O Capacitance C Farad F Time constant t Second s t v C , v R V 0 0 t 0 i R V 0 Figure 7.8 The curves of v C , v R and i versus t 208 Understandable electric circuits 7.3.4 The RC time constant and charging/discharging The capacitor charging/discharging voltages when the time is 1 t and 2 t can be determined from the equations of the capacitor voltage in the RC charging/ discharging circuit. That is v C ¼ Eð1 À e Àt= Þ and v C ¼ V 0 e Àt= For example, when v 0 ¼ E ¼ 100 V, ● at t ¼ 1 t ● Capacitor charging voltage: v C ¼ Eð1 À e Àt= Þ ¼ 100 Vð1 À e À1= Þ % 63:2 V ● Capacitor discharging voltage: v C ¼ V 0 e Àt= ¼ 100 Ve À1= % 36:8 V ● at t ¼ 2 t ● Capacitor charging voltage: v C ¼ Eð1 À e Àt= Þ ¼ 100 Vð1 À e À2= Þ % 86:5 V ● Capacitor discharging voltage: v C ¼ V 0 e Àt= ¼ 100 Ve À2= % 13:5 V (a) (b) v C E t 0 τ Increases E v C t 0 τ Increases Figure 7.9 The effect of the time constant t to v C . (a) Charging and (b) discharging Transient analysis of circuits 209 Using the same method as above, the capacitor charging/discharging vol- tages can be determined when the time is 3, 4 and 5 t. These results are sum- marized in Table 7.1 and Figure 7.10. The data in Table 7.1 and graphs in Figure 7.10 mean that when the time constant is 1 t, the capacitor will charge to 63.2% of the final value (source vol- tage), and discharge to 36.8% of the initial value (the initial capacitor voltage). If the final and initial value are 100 V, it will charge to 63.2 V and discharge to 36.8 V. When the time constant is 2 t, the capacitor will charge to 86.5% of the final value, and discharge to 13.5% of the initial value. According to this sequence, if the time constant is 5 t, the capacitor will charge to 99.3% of the final value, and discharge to 0.67% of the initial value. When the time is 5 t, the circuit will reach the steady state, which means that the capacitor will charge approaching to the source voltage E, or discharge approaching to zero. Therefore, when time has passed 4 to 5 t, charging/dis- charging of the capacitor will be almost finished. After 5 t, the transient state of RC circuit will be finished and enter the steady state of the circuit. Time constant t for and charging/discharging ● When t ¼ 1 t: the capacitor charges to 63.2% of the final value and discharges to 36.8% of the initial value. t v C E 0 t 0 63.2% 86.5% 95% 98.2% 99.3% v C E 36.8% 13.5% 5% 1.8% 0.67% 1 2 3 4 5 τ τ τ τ τ 1 2 3 4 5 τ τ τ τ τ Figure 7.10 The charging/discharging curves of the capacitor voltage Table 7.1 The capacitor charging/discharging voltages Charging/discharging time Capacitor charging voltage: v C ¼ Eð1 À e Àt= Þ Capacitor discharging voltage: v C ¼ V 0 e Àt= 1 t 63.2% of E 36.8% of E 2 t 86.5% of E 13.5% of E 3 t 95.0% of E 5% of E 4 t 98.2% of E 1.8% of E 5 t 99.3% of E 0.67% of E 210 Understandable electric circuits ● When t ¼ 5 t: the capacitor charges to 99.3% of the final value and discharges to 0.67% of the initial value. Example 7.2: In the circuit of Figure 7.6(a), the source voltage is 100 V, the resistance is 10 kO, and the capacitance is 0.005 mF. In how much time can the capacitor voltage be discharged to 5 V after the switch is turned to position 3? Solution: E ¼ 100 V; R ¼ 10 kO; C ¼ 0:005 mF; t ¼ ? The time constant t for discharging is: t ¼ RC¼ (10 kO)(0.005 mF) ¼ 50 ms. The capacitor voltage discharging to 5 V is 5% of the initial value E (100 V). Table 7.1 and Figure 7.10 indicate that the time capacitor discharges to 5% of the initial value is 3 t. Therefore, the capacitor discharging time is t ¼ 3 ¼ 3ð50 msÞ ¼ 150 ms Example 7.3: In an RC circuit, R ¼ 5 kO, the transient state has last 1 s in this circuit. Determine the capacitance C. Solution: The transient state in the RC circuit will last 5 t, therefore, 5 ¼ 1 s or ¼ 1 5 ¼ 0:2 s , ¼ RC therefore C ¼ R ¼ ð0:2 sÞ ð5k OÞ ¼ 40 ms 7.4 The step response of an RL circuit Figure 7.11(a) is a resistor and inductor series circuit, it runs through a switch connecting to the DC power supply. Such a circuit is generally referred to as an RL circuit. An RC circuit stores the charges in the electric field, and an RL circuit stores the energy in the magnetic field. We will use the term charging/ discharging in an RC circuit, and the term energy storing/releasing in an RL circuit. All important concepts of the magnetic storing/releasing or transient and steady state of RL circuit can be analysed by a simple circuit as shown in Figure 7.11. The step response (storing) and source-free response (releasing) of an RL circuit is similar to the step response and source-free response of an RC circuit. After understanding the RC circuit, its method of analysis can be used to analyse the RL circuit in a similar fashion. Transient analysis of circuits 211 Figure 7.11(a) is a circuit that can be used to analyse RL step response and source-free response. 7.4.1 Energy storing process of the RL circuit In the circuit of Figure 7.11(a), assuming the energy has not been stored in the inductor yet, the switch is in position 2. What will happen when the switch is turned to position 1, and the DC power source is connected to the RL series circuit as shown in the circuit of Figure 7.11(b)? As it has been mentioned in chapter 6, when the switch in Figure 7.11(a) turns to position 1, the current will flow through this RL circuit, the electromagnetic field will be built up in the inductor L, and will produce the induced voltage V L . The inductor L absorbs the electric energy from the DC source and converts it to magnetic energy. This energy storing process of the inductor in an RL circuit is similar to the electron charging process of the capacitor in an RC circuit. Since there is a resistor R in the circuit of Figure 7.11(b), it will be different as a pure inductor circuit that can store energy instantly. After the switch is turned to position 1, the current needs time to overcome the resistance in this RL circuit. Therefore, the process of the inductor’s energy storing will not finish instantly. The current i L in the RL circuit will reach the final value (maximum value) after a time interval, as shown in Figure 7.12. The phenomenon of the inductor current i L in an RL circuit increases exponentially from zero to its final value (I max ) or from the transient to the steady state can also be analysed by the quantitative analysis method below. i L I max t 0 5τ Figure 7.12 Current versus time curve in the RL circuit E L 1 R 2 3 E V R 1 i L V L + − + − 3 (a) (b) Figure 7.11 RL circuit 212 Understandable electric circuits 7.4.2 Quantitative analysis of the energy storing process in an RL circuit The polarities of the inductor and resistor voltages of an RL circuit are shown in the circuit of Figure 7.11(b). Applying KVL to this circuit will result in v L þ v R ¼ E ð7:12Þ Substituting v L ¼ L(di/dt) (from chapter 6, section 6.3.4) and v R ¼ Ri, where i ¼ i L , with (7.12) yields L di L dt þRi L ¼ E Applying a similar analysis method for the RC charging circuit in section 7.2 will yield the equation of the current in RL circuit during the process of energy storing as given in the following sections. ● Determine the current i L i L ¼ E R ð1 À e Àt=ðL=RÞ Þ ¼ E R ð1 À e Àt= Þ ¼ I max ð1 À e Àt= Þ ð7:13Þ The time constant of RL circuit is ¼ L R The final value for the current is I max ¼ E R ● Determine the resistor voltage v R Applying Ohm’s law v R ¼ Ri ð7:14Þ Keeping in mind that i ¼ i L and substituting i by the current i L in (7.14) yields v R ¼R E R ð1 À e Àt= Þ ¼Eð1 À e Àt= Þ Transient analysis of circuits 213 The final value for the resistor voltage is E ¼ I max R ● Determine the inductor voltage v L According to (7.12) v L þv R ¼ E Substitute v R and solving for v L v L ¼ E Àv R ¼ E ÀEð1 À e Àt= Þ ¼ Ee Àt= Energy storing equations for an RL circuit ● Circuit current: i L ¼ E R ð1 À e Àt= Þ ● Resistor voltage: v R ¼ Eð1 À e Àt= Þ ● Inductor voltage: v L ¼ Ee Àt= In the above equations, t is the energy storing time, and t ¼ L/R is the time constant of the RL circuit. These three equations mathematically indicate that circuit current and resistor voltage increase exponentially from initial value zero to the final value E/R and E, respectively; the inductor voltage decays exponentially from initial value E to zero. According to the above mathematical equations, the curves of i L , v R and v L versus time can be illustrated in Figure 7.13. Example 7.4: The resistor voltage v R ¼ 10(1 7 e 72t )V and circuit current i L ¼ 2(17e 72t ) A in an RL circuit is shown in the circuit of Figure 7.11(b). Determine the time constant t and inductance L in this circuit. i L t 0 v R t t 0 0 v L E E R E Figure 7.13 Curves of i L , v R and v L versus time 214 Understandable electric circuits Solution: The given resistor voltage v R ¼ Eð1 À e Àt= Þ ¼ 10ð1 À e À2t ÞV with E ¼ 10 V and À t ¼ À2t or ¼ 1 2 s ¼ 0:5 s The given current i L ¼ E R ð1 À e Àt= Þ ¼ 2ð1 À e À2t Þ A with E R ¼ 2 A E ¼ 10 V; R ¼ E I ¼ 10 V 2 A ¼ 5 O The time constant ¼ L R Solve for L L ¼ R ¼ ð5 OÞð0:5 sÞ ¼ 2:5 H 7.5 Source-free response of an RL circuit 7.5.1 Energy releasing process of an RL circuit Consider an inductor L that has initially stored energy and has the induced voltage v L through the energy storing process of the last section. If the switch turns to position 3 at this moment (Figure 7.14(b)), the inductor voltage v L has a function just like a voltage source in the right loop of this RL circuit. (a) (b) E L 1 2 3 v L = E v R R + − 3 1 E v L + − + − i L v R R Figure 7.14 RL circuit Transient analysis of circuits 215 Without connecting the resistor R in this circuit, at the instant when the switch turns to position 3, the inductor will release the stored energy immedi- ately. This might produce a spark on the switch and damage the circuit com- ponents. But if there is a resistor R in the circuit, the resistance in the circuit will increase the time required for releasing energy, the current in the circuit will take time to decay from the stored initial value to zero. This means the inductor releases the energy gradually, and the resistor absorbs the energy and converts it to heat energy. The current i L curve of the energy release process in the RL circuit is illustrated in Figure 7.15. 7.5.2 Quantity analysis of the energy release process of an RL circuit The equations to calculate the inductor voltage v L , resistor voltage v R and circuit current i L of the RL energy releasing circuit can be determined by the following mathematical analysis method. Applying KVL to the circuit in Figure 7.14(b) will result in v L þ v R ¼ 0 or v L ¼ Àv R ð7:15Þ Substituting v L ¼ Lðdi L =dtÞ and v R ¼ Ri L into (7.15) yields L di L dt ¼ ÀRi L ð7:16Þ ● Determine the circuit current Note: If you haven’t learned calculus, then just keep in mind that (7.18) is the equation for the current in the RL circuit during the energy releasing, and skip the following mathematical derivation process. Divide L on both sides in (7.16) di L dt ¼ À Ri L L i L t 0 R E Figure 7.15 Energy release curve of the RL circuit 216 Understandable electric circuits Integrating the above equation on both sides yields ð di L i L ¼ À ð R L dt; ln ji L j ¼ À R L t þ ln A Rearrange: lnji L j À ln A ¼ À R L t Taking the natural exponent (e) on both sides results in e lnji L =Aj ¼ e ðÀR=LÞt or i L A ¼ e ðÀR=LÞt Solve for i L i L ¼ Ae ðÀR=LÞt ð7:17Þ Since energy has been stored in the inductor before it is been connected to the circuit in Figure 7.14(b), its initial condition or value should be i L ð0ÀÞ ¼ I 0 (I 0 can be any initial current, such as I 0 ¼ E/R) Since immediately before/after the switch is closed to position 3, i L does not change, therefore, i L ð0þÞ ¼ i L ð0ÀÞ or i L ¼ I 0 When t ¼ 0, substitute i L ¼ I 0 into (7.17) yields I 0 ¼ Ae ðÀR=LÞÂ0 i:e: I 0 ¼A Therefore, i L ¼ I 0 e Àt= ð7:18Þ In the above equation, t ¼ L/R is the time constant for the RL circuit. ● Determine the resistor voltage Keep in mind that i ¼ i L and apply Ohm’s law to (7.18) v R ¼ Ri ¼ RðI 0 e Àt= Þ ¼ RI 0 e Àt= ● Determine the inductor voltage Substituting (7.18) into v L þ v R ¼ 0 (as in (7.15)) results in v L ¼ Àv R ¼ ÀRI 0 e Àt= Transient analysis of circuits 217 Energy releasing equations for an RL circuit ● Circuit current: i L ¼ I 0 e Àt= ● Resistor voltage: v R ¼ I 0 Re Àt= ● Inductor voltage: v L ¼ ÀI 0 Re Àt= In the above equations, t is the energy releasing time, I 0 ¼ E/R is the initial current for the inductor and t ¼ L/R is the time constant for the RL circuit. These three equations mathematically indicate that inductor current, resistor voltage and inductor voltage decay exponentially from initial value I 0 , I 0 R and 7I 0 R, respectively, to the final value zero. The curves of i L , v R and v L versus time can be illustrated in Figure 7.16. 7.5.3 RL time constant t In an RL circuit, the storing and releasing of energy is a gradual process that needs time. The time rate of this process depends on the values of the circuit inductance L and resistance R. The variation of R and L will affect the rate of the energy storing and releasing. The quotient of L and R is called the RL time constant t ¼ L/R. The RL time constant is the time interval required from the transient to the steady state or the energy storing/releasing time in an RL circuit. v R v L t 0 t i L 0 I 0 I 0 R 0 t Ϫ I 0 R Figure 7.16 Curves of i L , v R and v L versus time RL time constant ¼ L R Quantity Quantity symbol Unit Unit symbol Resistance R Ohm O Inductance L Henry H Time constant t Second s 218 Understandable electric circuits The time constant t represents the time the inductor current reaches (increases) to 63.2% of its final value (steady state); the time of the inductor current decays (decreases) below to 36.8% of the its initial value. The higher the value of L, the lower the R (or when the time constant t increases), the longer the storing or releasing time, the lesser the i L variation and the longer the time to reach the final or initial values. This can be shown in Figure 7.17. 7.5.4 The RL time constant and the energy storing and releasing Similar to an RC circuit, the circuit current for an RL circuit can be determined when the time constant is 1 t, 2 t, . . ., 5 t, according to the equations of i L ¼ ðE=RÞð1 À e Àt= Þ and i L ¼ I 0 e Àt= , respectively. These results are summar- ized in Table 7.2 and Figure 7.18. Example 7.5: In the RL circuit of Figure 7.14, the resistance R is 100 O and the transient state has lasted 25 ms. Determine the inductance L. Solution: The time of a transient state usually lasts 5, and this transient state is 5 ¼ 25 ms; ; ¼ ð25 ms=5Þ ¼ 5 ms. The time constant ¼ L=R; ; L ¼ R ¼ ð100 OÞð5 msÞ ¼ 500 mH. Table 7.2 Relationship between the time constant and the inductor current RL energy storing/releasing time Increasing the inductor current (storing): i L ¼ E R ð1 À e Àt= Þ Decreasing the inductor current (releasing): i L ¼ I 0 e Àt= 1 t 63.2% of E/R 36.8% of I 0 2 t 86.5% of E/R 13.5% of I 0 3 t 95.0% of E/R 5% of I 0 4 t 98.2% of E/R 1.8% of I 0 5 t 99.3% of E/R 0.67% of I 0 i L t 0 t i L 0 I 0 E ) ( ↑ τ ) ( ↑ τ R Figure 7.17 Effect of the time constant t on i L (L" or R#) Transient analysis of circuits 219 Example 7.6: In the circuit of Figure 7.14(b), R ¼ 2 kO, L ¼ 40 H, E ¼ 1 V and t ¼ 0.2 ms. Determine the circuit current i L in this energy releasing circuit. Solution: ¼ L R ¼ 40 H 2 kO ¼ 20 ms i L ¼ I 0 e Àt= ¼ E R e Àt= ¼ 1 V 2 kO e ðÀ0:2=20Þms % 0:5 mA Summary ● First-order circuit: ● The circuit that contains resistor(s), and a single energy storage ele- ment (L or C). ● RL or RC circuits that are described by the first-order differential equations. ● Transient state: The dynamic state that occurs when the physical quantities have been changed suddenly. ● Steady state: An equilibrium condition that occurs when all physical quantities have stopped changing and all transients have finished. ● Step response: The circuit response when the initial condition of the L or C is zero, and the input is not zero in a very short time, i.e. the charging/ storing process of the C or L. ● Source-free response: The circuit response when the input is zero, and the initial condition of L or C is not zero, i.e. the discharging/releasing process of the C or L. t i L 0 t 0 63.2% 86.5% 95% 98.2% 99.3% I 0 36.8% 13.5% 5% 1.8% 0.67% E R i L 1 τ 2 τ 3 τ 4 τ 5 τ 1 τ 2 τ 3 τ 4 τ 5 τ Figure 7.18 Relationship of inductor current and time constant 220 Understandable electric circuits ● The initial condition: ● t ¼ 07: the instant time before switching ● t ¼ 0þ: the instant time after switching ● v C ð0 þ Þ ¼ v C ð0 À Þ; i L ð0þÞ ¼ i L ð0ÀÞ ● Immediately before/after the switch is closed, v C and i L do not change instantly. ● The relationship between the time constants of RC/RL circuits and char- ging/storing or discharging/releasing: ● Summary of the first-order circuits (see p. 222). Experiment 7: The first-order circuit (RC circuit) Objectives ● Understand the capacitor charging/discharging characteristics in the RC circuit (the first-order circuit) by experiment. ● Construct an RC circuit, collect and evaluate experimental data to verify the capacitor charging/discharging characteristics in an RC circuit. ● Analyse and verify the capacitor’s charging/discharging time by experiment. ● Analyse the experimental data, circuit behaviour and performance, and compare them to the theoretical equivalents. Background information ● RC charging (the step response): v C ¼ Eð1 À e Àt= Þ; v R ¼ Ee Àt= ● RC discharging (the source-free response): v C ¼ V 0 e Àt= ; v R ¼ V 0 e Àt= ● RC time constant t. ¼ RC Time v C and i L increasing (charging/storing): v C ¼ Eð1 À e Àt=RC Þ; i L ¼ E R ð1 À e Àt= Þ v C and i L decaying (discharging/releasing): v C ¼ V 0 e Àt=RC ; i L ¼ I 0 e Àt= 1 t 63.2% 36.8% 2 t 86.5% 13.5% 3 t 95.0% 5% 4 t 98.2% 1.8% 5 t 99.3% 0.67% Transient analysis of circuits 221 S u m m a r y o f t h e f i r s t - o r d e r c i r c u i t s C i r c u i t s E q u a t i o n s W a v e f o r m s T i m e c o n s t a n t R C c h a r g i n g ( s t e p r e s p o n s e ) v C ¼ E ð 1 À e À t = Þ ; v R ¼ E e À t = ; i ¼ E R e À t = v C E t 0 E R E i v R t t 0 0 t ¼ R C R C d i s c h a r g i n g ( s o u r c e - f r e e r e s p o n s e ) v C ¼ V 0 e À t = ; v R ¼ V 0 e À t = ; i ¼ V 0 R e À t = t v C , v R V 0 0 t 0 i R V 0 t ¼ R C R L s t o r i n g ( s t e p r e s p o n s e ) i L ¼ E R ð 1 À e À t = Þ v R ¼ E ð 1 À e À t = Þ v L ¼ E e À t = i L t 0 v R t t 0 0 v L E E R E t ¼ L / R t ¼ L / R R L r e l e a s i n g ( s o u r c e - f r e e r e s p o n s e ) i L ¼ I 0 e À t = v R ¼ I 0 R e À t = v L ¼ À I 0 R e À t = v R t 0 t i L 0 I 0 I 0 R 0 v L t − I 0 R 222 Understandable electric circuits Equipment and components ● Multimeters (two) ● Breadboard ● DC power supply ● Stopwatch ● Z meter or LCZ meter ● Switch ● Resistors: 1 and 100 kO ● Capacitor: 100 mF electrolytic capacitor Procedure Part I: Charging/discharging process in an RC circuit 1. Use a jump wire to short circuit the 100 mF capacitor terminals to dis- charge it, then measure the value of capacitor using a Z meter or LCZ meter and record in Table L7.1. 2. Construct an RC circuit as shown in Figure L7.1 on the breadboard. 3. Turn on the switch to position 1 in the circuit of Figure L7.1, and observe the needles of the two multimeters (voltmeter function). Wait until voltage across the capacitor reaches to steady state (does not change any more), and record the observation of capacitor voltage v C and resistor voltage v R in Table L7.2 (such as 0–10 V, etc.). Table L7.1 Capacitor C Nominal value 100 mF Measured value 1 2 =1 kΩ 100 μF = E = 10 V V V R C Figure L7.1 An RC circuit Transient analysis of circuits 223 4. Turn on the switch to position 2 in the circuit of Figure L7.1, observe the needles of two multimeters (voltmeter function). Wait until voltage across the capacitor v C decreased to 0 V, record the observations of capacitor voltage v C and resistor voltage v R in Table L7.1 (such as 10–0 V, etc.). Part II: Capacitor’s characteristics in DC circuit 1. Measure the resistors listed in Table L7.3 using the multimeter (ohmmeter function), and record the measured values in Table L7.3. 2. Use a jump wire to short circuit the 100 mF capacitor terminals to dis- charge it, then measure the value of capacitor using a Z meter or LCZ meter and record in Table L7.3. 3. Construct a circuit as shown in Figure L7.2 on the breadboard. 4. Calculate the time constant t for the circuit in Figure L7.2 (use measured R and C values). Record the value in Table L7.4. 5. Calculate the capacitor charging/discharging voltage v C and v R when t ¼ t. Record the values in Table L7.4. Table L7.2 Switching position v C v R Turn on the switch to position 1 Turn on the switch to position 2 Table L7.3 Component R C Nominal value 100 kO 100 mF Measured value 1 2 =100 kΩ E = 10 V V V R 10 μF = C Figure L7.2 The RC circuit for Part II 224 Understandable electric circuits 6. Turn on the switch to position 1, observe time required for capacitor vol- tage v C charging to 6.3 V using both multimeter (voltmeter function) and stopwatch, this is the charging time constant t. Also measure v R at this time, and record t, v C and v R in Table L7.4. 7. Keep the switch at position 1 and make sure it does not change, and observe capacitor voltage v C using the multimeter (voltmeter function) until the capacitor voltage reaches and stays at 10 V (v C ¼ 10 V). Then turn on the switch to position 2, and observe the time required for the capacitor voltage to decrease to 3.6 V using both the multimeter (voltmeter function) and stopwatch (this is the discharging time constant t). Also measure v C and v R at this time and record the values in Table L7.4. Note: Since electrolyte capacitors may conduct leakage current, the measure- ment and calculation may be a little different, but it still can approximately verify the theory. Conclusion Write your conclusions below: Table L7.4 t v C v R Charging formula Discharging formula Calculated value for charging Calculated value for discharging Measured value for charging Measured value for discharging Transient analysis of circuits 225 Chapter 8 Fundamentals of AC circuits Objectives After completing this chapter, you should be able to: ● understand the difference between DC and AC ● understand the definitions of AC phase shift, period, frequency, peak to peak, peak, RMS values, phasor, etc. ● understand the relationship of period and frequency ● understand and define three important components of sinusoidal waveform ● define the phase difference between sinusoidal voltage and current ● convert sinusoidal time-domain quantities to phasor-domain forms, and vice versa ● analyse the sinusoidal AC circuits using phasors ● study the effect of resistive, inductive, and capacitive elements in ACcircuits 8.1 Introduction to alternating current (AC) 8.1.1 The difference between DC and AC Previous chapters have studied DC (direct current) circuits. The DC power supply provides a constant voltage and current; hence, all resulting voltages and currents in DC circuit are constant and do not change with time. That is, the polarity of DC voltage and direction of DC current do not change, only their magnitude changes. This chapter will discuss the alternating current (AC) circuits, in which the voltage alternates its polarity and the current alternates its direction periodi- cally. Since the AC power supply provides an alternating voltage and current, the resulting currents and voltages in AC circuit also periodically switch their polarities and directions. Similar to DC circuits, an alternating voltage is called AC voltage and alternating current is called AC current. Before the 19th century, the DC power supply was the main form of elec- trical energy to provide electricity. Since then, DC and AC have had constant competition; AC gradually showed its advantages and rapidly developed in the latter of the 19th century, and is still commonly used in current industries, businesses and homes throughout the world. This is because the AC power can be more cost-effective for long-distance transmission from power plants to industrial, commercial or residential areas. This is why power transmission for electricity today is nearly all AC. It is also easy to convert from AC to DC, allowing for a wide range of applications. 8.1.2 DC and AC waveforms The DC voltage and current do not change their polarity or direction over time, only their magnitude changes. A DC waveform (a graph of voltage and current versus time) is shown in Figure 8.1. There is also a type of DC waveform known as the pulsing DC, in which the amplitude of DC pulse changes periodically from zero to the maximum with time, but its polarity or direction does not change with time (always above zero), so it still belongs to the DC category. Figure 8.2 shows some pulsating DC waveforms. Direct current (DC) ● The polarity of DC voltage and direction of DC current do not change. ● The pulsing DC changes pulse amplitude periodically, but the polarity does not change. t V or I 0 Figure 8.1 DC waveform t V or I V or I V or I 0 t 0 t 0 Figure 8.2 Pulsing DC waveforms 228 Understandable electric circuits AC voltage and current periodically change polarity or direction with time. A few examples of AC waveforms are shown in Figure 8.3. The sinusoidal or sine AC wave is the most basic and widely used AC waveform, and is often referred to as AC, although other waveforms such as square wave, triangle wave, etc. also belong to AC. The sine AC wave energy is the type of power that is generated by the utility power industries around the world. Sine AC voltage and current vary with sine (or we could use cosine by adding 908 to the sine wave) function, the symbol of AC source is . AC quantities are represented by lowercase letters (e, v, i, etc.) and DC quantities use uppercase letters (E, V, I, etc.). Alternating current (AC) ● The polarity of voltage and direction of ACcurrent periodically change with time (such as sine wave, square wave, saw-tooth wave, etc.). ● Sine AC (or AC) varies over time according to sine (or cosine) function, and is the most widely used AC. A sine function can be described as a mathematical expression of f ðtÞ ¼ F m sinðot þcÞ. This is the expression of sine function in the time domain (the quantity versus time). Applying the expression of sine function to electrical quantities will obtain general expressions of AC voltage and current as follows: Sinusoidal voltage: vðtÞ ¼ V m sinðot þcÞ Sinusoidal current: iðtÞ ¼ I m sinðot þcÞ 8.1.3 Period and frequency The waveform of a sinusoidal function is shown in Figure 8.4. V or I V or I V or I t + _ t t 0 0 0 Figure 8.3 AC waveforms Fundamentals of AC circuits 229 ● Period T: It is the time to complete one full cycle of the waveform, or the positive and negative alternations of one revolution. T is measured in seconds (s). ● Frequency f: It is the number of cycles of waveforms within 1 s. The fre- quency is measured in hertz (Hz). For instance, in Figure 8.5, the number of complete cycles in 1 s is 2, so it has a frequency of 2 Hz. ● Relationship of T and f: The frequency f of the waveform is inversely proportional to period T of the waveform, i.e. f ¼ 1=T. Period and frequency ● Period T: Time to complete one full cycle. ● Frequency f: Number of cycles per second. ● f ¼ 1=T 8.1.4 Three important components of a sine function There are three important components in the expression of the sine function f ðtÞ ¼ F m sinðot þcÞ: peak value F m , angular velocity O and phase shift c. ● Peak value F m : In the expression f ðtÞ ¼ F m sinðot þcÞ, F m is the peak value or amplitude of the sine wave (I m for current or V m for voltage). It is the distance from zero of the horizontal axis to the maximum point wt f(t) 0 2p p F m T Figure 8.4 Sinusoidal waveform t(s) f(t) 0 t = 1s Figure 8.5 Frequency of sine waveform 230 Understandable electric circuits (positive or negative) that a waveform can reach during its entire cycle (Figure 8.6 (a)). It is measured in volts or amperes. ● Angular velocity o (the Greek letter omega): Angular velocity or angular frequency of a sine wave reflects the rate of change of the rotation of the wave. Angular velocity ¼ Rotating distance/Time (Same with the linear motion: Velocity ¼ Distance/Time) Since the time required for a sine wave to complete one cycle is period T, the distance of one cycle is 2p as shown in Figure 8.4, so the angular velocity can be determined by o ¼ 2p T The relationship between the angular velocity and frequency is o ¼ 2p T ¼ 2pf f ¼ 1 T Since the angular velocity is directly proportional to the frequency, it is also called the angular frequency. It is measured in radian per second (rad/s). ● Phase c (the Greek letter phi): The phase or phase shift of a sine wave is an angle that represents the position of the wave shifted from a reference point at the vertical axis (08). It is measured in degrees or radians. A sine wave may shift to the left or right of 08. The range of phase shift is between 7p and þp. ● If phase shift c ¼ 0, the waveform of sine function f ðtÞ ¼ F m sinot starts from t ¼ 0 as shown in Figure 8.6(a). ● If phase shift c has a negative value (c 5 0), the waveform of sine function f ðtÞ ¼ F m ðot ÀcÞ will shift to the right side of 08 as shown in Figure 8.6(b). ● If phase shift c has a positive value (c 4 0), the waveform of sine function f ðtÞ ¼ F m ðot þcÞ will shift to the left side of 08 as shown in Figure 8.6(c). 0 F m 0 0 f(t) wt wt wt f(t) f(t) 2p p ψ ψ (a) (b) (c) Figure 8.6 The peak value and phase of the sine wave Fundamentals of AC circuits 231 Three important components of sine function f ðtÞ ¼ F m sinðot þcÞ ● F m : Peak value (amplitude) ● o: Angular velocity or angular frequency ● o ¼ 2p=T ¼ 2pf (p ¼ 1808) ● c: Phase or phase shift ● c 4 0: Waveform shifted to the left side of 08 ● c 5 0: Waveform shifted to the right side of 08 Example 8.1: Given a sinusoidal voltage vðtÞ ¼ 6sinð25t À30 ÞV, determine its peak voltage, phase angle and frequency, and plot its waveform. Solution: Peak value: V m ¼ 6 V Phase: c ¼ 7308 (c 5 0, waveform shifted to the right side of 08) Frequency: f ¼ 1=T Since o ¼ 2p=T and o ¼ 25 rad=s T ¼ 2p o ¼ 2p rad 25 rad=s % 0:25s f ¼ 1 T ¼ 1 0:25 s ¼ 4 Hz The waveform is shown in Figure 8.7. 8.1.5 Phase difference of the sine function For two different sine waves with the same frequency, the angular displace- ment of their phases is called phase difference and is denoted by (lowercase Greek letter phi). It is a phase angle by which one wave leads or lags another. wt 2p p = 30° 0 V m = 6V T = 0.25 s ψ v(t) Figure 8.7 Waveform for Example 8.1 For instance, given the general expressions of sinusoidal voltage and cur- rent as vðtÞ ¼ V m sinðot þc v Þ and iðtÞ ¼ I m sinðot þc i Þ 232 Understandable electric circuits the phase difference between voltage and current is ¼ ðot þc v Þ Àðot þc i Þ ¼ c v Àc i ● If ¼ c v Àc i ¼ 0, the two waveforms are in phase as shown in Figure 8.8(a). ● If ¼ c v Àc i > 0, voltage leads current, or current lags voltage as shown in Figure 8.8(b). ● If ¼ c v Àc i < 0, current leads voltage, or voltage lags current, as shown in Figure 8.8(c). ● If ¼ c v Àc i ¼ Æp=2 (or +908), then voltage and current are orthogo- nal, or is a right angle (orthos means ‘straight’, and gonia means ‘angle’). It is shown in Figure 8.8(d). wt 0 v i Figure 8.8(a) Two waveforms are in phase wt 0 v i v i 0 y v y i f f wt Figure 8.8(b) Current lags voltage 0 v i v i 0 y v y i f f wt wt Figure 8.8(c) Current leads voltage Fundamentals of AC circuits 233 ● If ¼ c v Àc i ¼ Æp (or Æ180 ), voltage and current are out of phase as shown in Figure 8.8(e). Phase difference ¼ c v Àc i For two waves with the same frequency such as vðtÞ ¼ V m sinðot þc v Þ and iðtÞ ¼ I m sinðot þc i Þ: ● If ¼ 0: v and i are in phase ● If 40: v leads i ● If 50: v lags i ● If ¼ Æp=2: v and i are orthogonal ● If ¼ Æp: v and i are out of phase Example 8.2: Determine the phase difference of the following functions and plot their waveforms. (a) vðtÞ ¼ 20 sinðot þ30 ÞV; iðtÞ ¼ 12 sinðot þ60 ÞA (b) vðtÞ ¼ 5 sinðot þ60 ÞV; iðtÞ ¼ 2:5 sinðot þ20 ÞA Solution: (a) ¼ c v Àc i ¼ 30 À60 ¼ À30 < 0 So voltage lags current 308 as shown in Figure 8.9(a). (b) ¼ c v Àc i ¼ 60 À20 ¼ 40 > 0 So voltage leads current 408 as shown in Figure 8.9(b). wt 2p p 0 v i Figure 8.8(e) Voltage and current are out of phase 0 v i v i 0 2 f = 2 f = − p p wt wt Figure 8.8(d) Voltage and current are orthogonal 234 Understandable electric circuits 8.2 Sinusoidal AC quantity A sinusoidal AC quantity such as AC voltage or current can be described in a number of ways. They can be described by their peak value, peak–peak value, instantaneous value, average value or root mean square (RMS) value. The different expressions will provide different ways to analyse the sinusoidal AC quantity, and it is also because a sinusoidal wave always varies periodically and there is no one single value that can truly describe it. 8.2.1 Peak and peak–peak value As previously mentioned, the peak value of the sinusoidal waveform is one of three important components of the sine function, and is the amplitude or maximum value F m in sine function f ðtÞ ¼ F m sinðot þcÞ. The peak value is denoted by F pk as shown in Figure 8.10. wt 0 v i 20 30° 30° 60° Figure 8.9(a) Figure for Example 8.2(a) w t 0 v i 5 40° 20° 60° Figure 8.9(b) Figure for Example 8.2(b) wt 0 f(t) F p–p F m = F pk Figure 8.10 Peak and peak–peak value Fundamentals of AC circuits 235 The peak–peak value F p–p represents the distance from negative to positive peak, or minimum to maximum peak, or between peak and trough of the waveform, so F p–p ¼ 2F pk as shown in Figure 8.10. To determine the maximum values that electrical equipment can withstand, the peak values or peak–peak values of the AC quantities should be considered. 8.2.2 Instantaneous value The instantaneous value of the sinusoidal waveform f(t) varies with time, and it is the value at any instant time t (or ot) in any particular point of a waveform. Instantaneous values of the variables are denoted by lowercase letters, such as voltage v, current i, etc. Example 8.3: Given a sinusoidal AC voltage vðtÞ ¼ V m sin ot as shown in Figure 8.11, determine the instantaneous voltage v 1 (voltage at 308) and v 2 (voltage at 1358) when V m ¼ 5 V. Solution: v 1 ¼ V m sin ot ¼ 5 sin30 ¼ 2:5 V v 2 ¼ V m sin ot ¼ 5 sin135 % 3:54 V 8.2.3 Average value Because of the symmetry of the sinusoidal waveform, its average value in a com- plete full cycle is always zero. For a sinusoidal function f ðtÞ ¼ F m sinðot þcÞ, its average value is defined as the average of its half-cycle (0 to p), as shown in Figures 8.12 and 8.13. wt v 2 0 v(t) v 1 135° 30° V m = 5V 90° Figure 8.11 Figure for Example 8.3 wt 0 π F m f(t) Figure 8.12 Average value 236 Understandable electric circuits The average value of a half-cycle sinusoidal wave with a zero phase shift can be derived by using integration as follows: Note: If you haven’t learned calculus, then just keep in mind that F avg ¼ 0:637F m is the equation for the average value of a half-cycle sinusoidal wave, and skip the following mathematic derivation process. F avg ¼ Area p ¼ 1 p p 0 f ðtÞdt ¼ 1 p p 0 F m sinotdot ¼ F m p ½ Àcosot p 0 ¼ À F m p ½cosp Àcos0 ¼À F m p ðÀ1 À1Þ ¼ 2F m p % 0:637F m F avg % 0:637F m Therefore, the average value of a half-cycle sinusoidal wave is 0.637 times the peak value, as shown in Figure 8.13. Peak value, peak–peak value, instantaneous value and average value For a sinusoidal waveform: ● Peak value F pk ¼ F m : The amplitude or maximum value ● Peak–peak value F p–p : F p–p ¼ 2F pk ● Instantaneous value f(t): The value at any time at any particular point of the waveform ● Average value F avg : F avg ¼ 0:637F m 8.2.4 Root mean square (RMS) value 1. Applications of RMS value: RMS value (also referred to as the effective value) of the sinusoidal waveform is widely used in practice. For example, the values measured and displayed on instruments and the nominal rat- ings of the electrical equipment are RMS values. In North America, the single-phase AC voltage 110 V from the wall outlet is an RMS value. 0 F m π f(t) 0.637 wt Figure 8.13 Average value Fundamentals of AC circuits 237 2. The physical meaning of RMS value: For a sinusoidal waveform, the phy- sical meaning of the AC RMS value is the heating effect of the sine wave. That is, an AC source RMS value will deliver the equivalent amount of average power to a load as a DC source. For instance, whether turning on the switch 1 (connect to DC) or switch 2 (connect to AC) in Figure 8.14, 20-V DC or 20-V AC RMS value will deliver the same amount of power (40 W) to the resistor (lamp). If the lamp is replaced by an electric heater, then the heating effect delivered by 20-VDCand 20-VAC RMS will be the same. 1 2 20-V DC 20-V RMS P = 40W R = 10Ω I = 2A Figure 8.14 RMS value 3. Quantitative analysis of RMS value: The average power generated by an AC power supply is p AC ¼ i 2 AC R ¼ ðI m sinotÞ 2 R ¼ ðI 2 m sin 2 otÞR p AC ¼ I 2 m 1 2 ð1 Àcos2otÞ ¸ R ¼ I 2 m R 2 À I 2 m R 2 cos2ot as sin 2 ot ¼ 1=2ð1 Àcos2otÞ. Only the first part in the above power expression represents the average power of AC, since the average value of the second part in the power expression (a cosine function) is zero, i.e. P AC ¼ ðI 2 m RÞ=2. ● The average power generated by DC voltage is P avg ¼ I 2 R. ● RMS value of AC current: According to the physical meaning of RMS, the average AC power is equivalent to the average DC power when the AC source is an RMS value, so ðI 2 m RÞ=2 ¼ I 2 R or I 2 ¼ I 2 m =2. Taking the square root of both sides of the equation gives I ¼ ffiffiffiffiffi I 2 m 2 ¼ I m ffiffiffi 2 p % 0:707I m or I m ¼ ffiffiffi 2 p I % 1:414I ð8:1Þ 238 Understandable electric circuits The current I in (8.1) is the RMS value of the AC current, and I m is the peak value or amplitude of the AC current. ● RMS value of AC voltage: It can be obtained by the same approach by determining the RMS value of the AC current, i.e. V ¼ V m ffiffiffi 2 p ¼ 0:707 V m or V m ¼ ffiffiffi 2 p V ¼ 1:414 V ð8:2Þ The voltage V in (8.2) is the RMS value of AC voltage, and V m is the peak value or amplitude of the AC voltage. ● RMS value of a periodical function f(t): Equations (8.1) and (8.2) indicate the relationship between the RMS value and the peak value, which is related by ffiffiffi 2 p . However, this relation only applies to the sine wave. For a non sine wave function f(t), the following general equation can be used to determine its RMS value. F ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 T T 0 f 2 ðtÞdt ðT is the period of the functionÞ The name root mean square (RMS) is obtained from the above equation, in which the term 1/T denotes the average (mean), f 2 (t) denotes the square (square) and ffi p denotes the square root (root) value. RMS value of AC function ● RMS value or effective value of AC: An AC source with RMS value will deliver the equivalent amount of power to a load as a DC source. ● V ¼ 0:707 V m , I ¼ 0:707I m or V m ¼ ffiffiffi 2 p V, I m ¼ ffiffiffi 2 p I ● The general equation to calculate RMS value: F ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 T T 0 f 2 ðtÞdt 8.3 Phasors 8.3.1 Introduction to phasor notation Charles Proteus Steinmetz, a German-American mathematician and electrical engineer, developed the phasor notation in 1893. A phasor is a vector that contains both magnitude and direction or amplitude and phase information. It can be used to represent AC quantities. Since phasors have magnitudes and directions, they can be represented as complex numbers. A phasor notation or phasor domain is a method that uses complex numbers to represent the sinusoidal quantities for analysing AC circuits. It can Fundamentals of AC circuits 239 represent sine waves in terms of their peak value (magnitude) and phase angle (direction). The peak value can be easily converted to the RMS value. The phasor notation can simplify the calculations for AC sinusoidal cir- cuits; therefore, it is widely used in circuit analysis and calculations. Note that the phasor notation can be used for sinusoidal quantities only when all wave- forms have the same frequency. We have learned that a sinusoidal wave can be represented by its three important components: the peak value (or RMS value), the phase angle and the angular frequency. In an AC circuit, the AC source voltage and the current are the sinusoidal values with the same frequency, so the resulting voltages and currents in the circuit should also be sinusoidal values with the same frequency or angular frequency. Therefore, voltages and currents in an AC circuit can be analysed by using the phasor notation, i.e. they can be determined by the peak value or RMS value and the phase shift of the phasor notation. Phasor ● A phasor is a vector that contains both amplitude and angle infor- mation, and it can be represented as complex number. ● Phasor notation is a method that uses complex numbers to represent the sinusoidal quantities for analysing AC circuits when all quantities have the same frequency. The key for understanding the phasor notation is to know how to use complex numbers. Therefore, we will review some important formulas of complex numbers that you may have learned in previous mathematics courses. 8.3.2 Complex numbers review The complex number has two main forms, the rectangular form and the polar form. ● Rectangular form: A ¼ x þjy ð j ¼ ffiffiffiffiffiffiffi À1 p Þ where x is the real part and y is the imagery part of the complex number A; j is called the imagery unit. Note: The symbol i is used to represent imagery unit in mathematics. Since i has been used to represent AC current in the circuit analysis, j is used to denote the imagery unit rather than i to avoid confusion. ● Polar form: A ¼ affc This is the abbreviated form of the exponential form A ¼ ae jc , in which a is called modulus of the complex number, and the angle c is called argu- ment of the complex number. 240 Understandable electric circuits ● Convert rectangular form to polar form (refer to Figure 8.15). Let A ¼ x þ jy ¼ affc Applying a ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffi x 2 þy 2 (Pythagorus theory) gives A ¼ x þjy ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffi x 2 þy 2 tan À1 y=x ¼ affc ● Convert polar form to rectangular or triangular form. x ¼ acosc and y ¼ asinc can be obtained from Figure 8.15. So A ¼ affc ¼ x þ jy ¼ a(cosc þ jsinc). Euler’s formula can also be used for the conversion of triangular form to exponential form e jc ¼ cosc þjsinc or ae jc ¼ aðcosc þjsincÞ ● Operations on complex numbers: Given two complex numbers A 1 ¼ x 1 þjy 1 ¼ a 1 ffc 1 and A 2 ¼ x 2 þjy 2 ¼ a 2 ffc 2 The basic algebraic operations of these two complex numbers are given as follows: Addition: A 1 þA 2 ¼ ðx 1 þx 2 Þ þjðy 1 þy 2 Þ Subtraction: A 1 ÀA 2 ¼ ðx 1 Àx 2 Þ þjðy 1 Ày 2 Þ Multiplication: ● Polar form: A 1 Á A 2 ¼ a 1 Á a 2 ffðc 1 þc 2 Þ ● Rectangular form: A 1 Á A 2 ¼ ðx 1 þjy 1 Þðx 2 þjy 2 Þ ¼ ðx 1 x 2 Ày 1 y 2 Þ þ jðx 2 y 1 þx 1 y 2 Þ Here j 2 ¼ ffiffiffiffiffiffiffi À1 p ffiffiffiffiffiffiffi À1 p ¼ ð ffiffiffiffiffiffiffi À1 p Þ 2 ¼ À1 is used. Division: ● Polar form: A 1 A 2 ¼ a 1 a 2 ffðc 1 Àc 2 Þ ● Rectangular form: A 1 A 2 ¼ x 1 þjy 1 x 2 þjy 2 ¼ ðx 1 þjy 1 Þðx 2 Àjy 2 Þ ðx 2 þjy 2 Þðx 2 Àjy 2 Þ ¼ x 1 x 2 þy 1 y 2 x 2 2 þy 2 2 þj x 2 y 1 Àx 1 y 2 x 2 2 þy 2 2 x y 0 a + +j y Figure 8.15 Complex number Fundamentals of AC circuits 241 It will be much simpler to use the polar formon operations of multiplication and division. Complex numbers ● Rectangular form: A ¼ x þ jy ● Polar form: A ¼ affc ● Conversion between rectangular and polar forms: A ¼ x þjy ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffi x 2 þy 2 tan À1 y x ¼ affc A ¼ affc ¼ x þjy ¼ aðcosc þjsincÞ ● Addition and subtraction: A 1 ÆA 2 ¼ ðx 1 Æx 2 Þ þjðy 1 Æy 2 Þ ● Multiplication: A 1 Á A 2 ¼ a 1 Á a 2 ffðc 1 þc 2 Þ ¼ ðx 1 þjy 1 Þðx 2 þjy 2 Þ ● Division: A 1 A 2 ¼ a 1 a 2 ffðc 1 Àc 2 Þ ¼ x 1 þjy 1 x 2 þjy 2 8.3.3 Phasor Using the phase notation to represent the sinusoidal function is based on Euler’s formula e j ¼ cos þjsin. For a sinusoidal function f ðtÞ ¼ F m sinðot þcÞ, replacing with (ot þ c) in Euler’s formula gives e jðotþcÞ ¼ cosðot þcÞ þjsinðot þcÞ where cosðot þcÞ ¼ R e ½e jðotþcÞ and sinðot þcÞ ¼ J m ½e jðotþcÞ . ‘R e [ ]’ and ‘J m [ ]’ stand for ‘real part’ and ‘imaginary part’ of complex numbers, respectively. Therefore, sine function f ðtÞ ¼ F m sinðot þcÞ ¼ J m ½F m e jðotþcÞ ¼ J m ½F m e jc e jot . That is, a sinusoidal function is actually taking the imaginary part of the complex number f ðtÞ ¼ J m ½F m e jc e jot ð8:3Þ There are two terms in (8.3), F m e jc and e jot . The second term e jot is called the rotating factor that varies with time t, which will be discussed later. The first term is the phasor of the sinusoidal function F m e jc ¼ F m ffc ¼ F 242 Understandable electric circuits So (8.3) of sine function can be written as f ðtÞ ¼ F m sinðot þcÞ ¼ J m ½Fe jot Therefore, the first term in (8.3) is F ¼ F m ffc, where boldface letter F represents a phasor (vector) quantity, similar to the boldface that indicates a vector quantity in maths and physics. A phasor quantity can also be repre- sented by a little dot on the top of the letter, such as _ F ¼ F m ffc. There is no difference between operations on phasors and complex numbers, since both of them are vectors. If the sinusoidal currents and voltages in an AC circuit are represented by vectors with the complex numbers, this is known as phasors. The sinusoidal voltage vðtÞ ¼ V m sinðot þcÞ and current iðtÞ ¼ I m sinðot þcÞ in an AC circuit can be expressed in the phasor domain as: ● Peak value: _ V ¼ V m ffc v or V ¼ V m ffc v _ I ¼ I m ffc i or I ¼ I m ffc i ● RMS value: _ V ¼ Vffc v or V ¼ Vffc v _ I ¼ Iffc i or I ¼ Iffc i 8.3.4 Phasor diagram Since a phasor is a vector that can be represented by a complex number, it can be presented with a rotating line in the complex plane as shown in Figure 8.16. The length of the phasor is the peak value F m (or RMS value F). The angle between the rotating line and the positive horizontal axis is the phase angle c of the sinusoidal function. This diagram is called the phasor diagram. y y 0 F m +j F = F m + Figure 8.16 Phasor diagram Fundamentals of AC circuits 243 Example 8.4: Use the phasor notation to express the following voltage and current in which 710 and 12 are the peak values. (a) v ¼ À10sinð60t þ25 ÞV (b) i ¼ 12sinð25t À20 ÞA Solution: (a) _ V ¼ À10ff25 V (b) _ I ¼ 12ff À20 A Example 8.5: Use the instantaneous value to express the following voltage and current in which 120 and 12 are RMS values. (a) _ V ¼ 120ff30 V (b) _ I ¼ 12ff0 A Solution: (a) v ¼ 120 ffiffiffi 2 p sinðot þ30 Þ V (b) i ¼ 12 ffiffiffi 2 p sinot A 8.3.5 Rotating factor In the sinusoidal expression of f ðtÞ ¼ F m sinðot þcÞ ¼ J m ½F m e jc e jot , the term e jot varies with time t, known as the rotating factor or time factor. As time changes, it rotates counterclockwise at angular frequency o in a radius F m of the circle, as shown in Figure 8.17. The rotating factor e jot can be represented by Euler’s formula e jot ¼ cosot þjsinot when ot ¼ Æ90 : e Æj90 ¼ cosð Æ90 Þ þjsinð Æ90 Þ ¼ Æj. Therefore, +908 is also the rotating factor (Æj ¼ Æ90 ). Phasor Time domain Phasor domain f ðtÞ ¼ F m sinðot þcÞ F m ¼ F m ffc or _ F m ¼ F m ffc ðPeak valueÞ F ¼ Fffc or _ F ¼ Fffc ðRMS valueÞ vðtÞ ¼ V m sinðot þcÞ _ V m ¼ V m ffc v , _ V ¼ Vffc v iðtÞ ¼ I m sinðot þcÞ _ I m ¼ I m ffc i , _ I ¼ Iffc i 244 Understandable electric circuits Rotating factor e jot or Æj ¼ Æ90 Asinusoidal function can be represented by a rotating phasor that rotates 3608 in a complex plane as shown in Figure 8.18. The instantaneous value of the sinu- soidal wave at any time is equal to the projection of its relative rotating phasor on the vertical axis ( j ) at that time. The geometric meaning of the sinusoidal function f ðtÞ ¼ F m sinðot þcÞ ¼ I m ½F m e jc e jot represented by the rotational phasor motion can be seen from the following example. Example 8.6: In Figure 8.18, When t ¼ t 0 ¼ 0, the phasor is F ¼ F m ffc. When t ¼ t 1 , the phasor is F ¼ F m ff90 . And it goes from c to 3608. 0 +1 +j F m – j w ψ Figure 8.17 Rotating factor F m F m –j w wt ψ ψ 0 +j 90° 0 t = t 0 t = t 1 t = t 0 t = t 1 Figure 8.18 Sine wave and phasor motion Fundamentals of AC circuits 245 8.3.6 Differentiation and integration of the phasor Note: Skip the following part and start from Example 8.8 if you haven’t learned calculus. For a sinusoidal function f ðtÞ ¼ F m sinðot þcÞ, the derivative of the sinu- soidal function with respect to time can be obtained by its phasor F multiplying with jo, i.e. df ðtÞ dt , joF This is equivalent to a phasor that rotates counterclockwise by 908 on the complex plane since þj ¼ þ90 . (Appendix B provides the details for how to derive the above differentiation of the sinusoidal function in phasor notation.) The integral of the sinusoidal function with respect to time can be obtained from its phasor divided by jo, i.e. f ðtÞdt ¼ _ F jo This is equivalent to a phasor that rotates clockwise on the complex plane by 908 (since 1=j ¼ Àj ¼ À90 ). Differentiation and integration of the sinusoidal function in phasor notation Differentiation: df ðtÞ=dt , joF or jo _ F ðþj ¼ þ90 Þ Integration: f ðtÞdt , F=jo or ð1=joÞ _ F (1=j ¼ Àj ¼ À90 ) Example 8.7: Convert the following sinusoidal time-domain expression to its equivalent phasor domain, and determine voltage _ V (or V). 2v À6 dv dt þ4 vdt ¼ 20sinð4t þ30 Þ Solution: 2 _ V À6jo _ V þ4 _ V jo ¼ 20ff30 Since o ¼ 4 in the original expression, so 2 _ V À6 j4 _ V þ4 _ V j4 ¼ 20ff30 _ Vð2 À24j ÀjÞ ¼ 20ff30 _ V ¼ 20ff30 2 Àj25 % 20ff30 25ff À85:43 ¼ 0:8ff115:43 246 Understandable electric circuits Note: The complex number of the denominate is Z ¼ x þjy ¼ 2 Àj25 ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffi x 2 þy 2 tan À1 y x (Since x is positive and y is negative in 2 7j25, the angle should be in the fourth quadrant, i.e. 785.438.) Example 8.8: Convert the phasor-domain voltage and current to their equiva- lent sinusoidal forms (time domain). (a) _ I ¼ j5e Àj30 mA (b) _ V ¼ À6 þj8 V Solution: (a) _ I ¼ j5ff À30 mA ¼5ff90 ff À30 mA ð j ¼ 90 Þ ¼5ffð90 À30 Þ mA ¼ 5ff60 mA iðtÞ ¼ 5sinðot þ60 Þ mA (b) _ V ¼ À6 þj8 V ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðÀ6Þ 2 þ8 2 tan À1 ff 8 À6 V % 10ff126:87 V (Since y is positive and x is negative, it should be in the second quadrant.) vðtÞ ¼ 10sinðot þ126:87 Þ V If the phasors are used to express sinusoidal functions, the algebraic operations of sinusoidal functions of the same frequency can be replaced by algebraic operations of the equivalent phasors, which is shown in Example 8.9. Example 8.9: Calculate the sum of the following two voltages v 1 ðtÞ ¼ 2 sin ðot þ60 Þ V and v 2 ðtÞ ¼ 10 sin ðot À40 Þ V Solution: Convert the sinusoidal time-domain voltages to their equivalent phasor forms _ V 1 ¼ 2ff60 V and _ V 2 ¼ 10ff À40 V So _ V 1 þ _ V 2 ¼ 2ff60 þ10ff À40 ¼ 2 cos 60 þ j2 sin 60 þ10 cos ðÀ40 Þ þ j10 sin ðÀ40 Þ % 1 þj 1:732 þ7:66 À j6:43 ¼ 8:66 À j4:698 ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 8:66 2 þðÀ4:698Þ 2 tan À1 À4:698 8:66 % 9:85ff À28:48 V (Since y is negative and x is positive, it should be in the fourth quadrant.) ; vðtÞ ¼ 9:85 sin ðot À28:48 Þ V Fundamentals of AC circuits 247 8.4 Resistors, inductors and capacitors in sinusoidal AC circuits Any AC circuit may contain a combination of three basic circuit elements: resistor, inductors and capacitors. When these elements are connected to a sinusoidal AC voltage source, all resulting voltages and currents in the circuit are also sinusoidal and have the same frequency as AC voltage source. Therefore, they can all be converted from the sinusoidal time-domain form f ðtÞ ¼ F m sinðot þcÞ to the phasor-domain form F ¼ F m ffc. 8.4.1 Resistor’s AC response Aresistor is connected to a sinusoidal voltage source as shown in Figure 8.19(a). Where the source voltage is e ¼ V m sinðot þcÞ The sinusoidal current in the circuit can be obtained by applying Ohm’s law for AC circuits (v ¼ Ri), i.e. i R ¼ e R ¼ V Rm R sinðot þcÞ ¼ I Rm sinðot þcÞ where I Rm ¼ V Rm =R (peak values) or I ¼ V R =R (RMS value), and voltage across the resistor is the same as the source voltage, i.e. e ¼ v R or v R ¼ V m sinðot þcÞ. The above sinusoidal expressions of resistor voltage v R and current i R indicate that voltage and current in the circuit have the same frequency f (since o ¼ 2pf ) and the same phase angle c (or v R and i R are in phase). This is also illustrated in Figure 8.19(b). Assuming the initial phase angle is zero, i.e. c ¼ 08, then i R ¼ v R R ¼ I Rm sinot v R ¼ Ri R ¼ V Rm sinot This is illustrated in Figure 8.19(c). w t φ 0 e v R v R i R R (a) (b) i R + – Figure 8.19 Resistor’s AC response 248 Understandable electric circuits Relationship of voltage and current of a resistor in an AC circuit ● Instantaneous values (time domain): v R ¼ V Rm sinðot þcÞ i R ¼ I Rm sinðot þcÞ ● Ohm’s law: V Rm ¼ I Rm R ðpeak valueÞ V R ¼ I R R ðRMS valueÞ The sinusoidal expressions of resistor voltage (v R ) and current (i R ) are in the time domain. The peak and RMS values of the resistor voltage and the current in phasor domain also obey the Ohm’s law as follows: Peak value: _ I Rm ¼ _ V Rm =R or V Rm ¼ I Rm R RMS value: _ I R ¼ _ V R =R or V R ¼ I R R If it is expressed in terms of conductance, it will give _ I R ¼ G _ V R G ¼ 1=R ð Þ The relationship of the resistor voltage and current in an AC circuit can be presented by a phasor diagram illustrated in Figure 8.20(b). w t v R i R 0 Figure 8.19(c) When c ¼ 08 (a) (b) v R R + – I R E • • • v R 0 +j + I R y • • Figure 8.20 The phasor diagram of the AC resistive circuit Fundamentals of AC circuits 249 Example 8.10: If R ¼ 10 O, i R ¼ 6 ffiffiffi 2 p sinðot À30 ÞA in Figure 8.20(a), deter- mine the voltage across resistor in phasor domain. Solution: v R ¼ Ri R ¼ 10 Â6 ffiffiffi 2 p sinðot À30 Þ ¼ 60 ffiffiffi 2 p ðsinot À30 Þ So _ V Rm ¼ 60 ffiffiffi 2 p ff À30 V. Resistor’s AC response in phasor domain ● Ohm’s law: Peak value: _ V Rm ¼ _ I Rm R or V Rm ¼ I Rm R RMS value: _ V R ¼ _ I R R or V R ¼ I R R Using conductance: _ I R ¼ G _ V R ðG ¼ 1=RÞ ● Phasor diagram: À! _ I R À! _ V R (AC resistor voltage and current are in phase) Note that we can use Ohm’s law in AC circuits as long as the circuit quantities are consistently expressed, i.e. both the voltage and current are peak values, RMS values, instantaneous values, etc. 8.4.2 Inductor’s AC response If an AC voltage source is applied to an inductor as shown in Figure 8.21(a), the current flowing through the inductor will be i L ¼ I Lm sinðot þcÞ ð8:4Þ We have learned from chapter 6 that the relationship between the voltage across the inductor and the current that flows through it is v L ¼ L di dt ð8:5Þ (a) (b) v L + – e L i L 0 v L i L 90° wt Figure 8.21 Inductor’s AC response 250 Understandable electric circuits Note: If you haven’t learned calculus, then just keep in mind that v L ¼ oLI Lm sinðot þc þ90 Þ is the sinusoidal expression of the inductor vol- tage, and skip the following mathematic derivation process. Substituting (8.4) into (8.5) and applying differentiation gives v L ¼ L di L dt ¼ L d½I Lm sinðot þcÞ dt ¼ oLI Lm cosðot þcÞ ¼ oLI Lm sinðot þc þ90 Þ Therefore v L ¼ oLI Lm sinðot þc þ90 Þ ð8:6Þ Note: cos ¼ sinðot þ90 Þ The sinusoidal expressions of the inductor voltage v L and current i L indi- cate that in an AC inductive circuit, the voltage and current have the same angular frequency (o) and a phase difference. The inductor voltage v L leads the current i L by 908 as illustrated in Figure 8.21(b) if we assume that initial phase angle c ¼ 08. The relationship between the voltage and current in an inductive sinusoidal AC circuit can be obtained from (8.6), which is given by V Lm ¼ oLI Lm ðpeak valueÞ or V L ¼ oLI L ðRMS valueÞ This is also known as Ohm’s law for an inductive circuit, where oL is called inductive reactance and is denoted by X L , i.e. X L ¼ oL ¼ 2pfL ðo ¼ 2pf Þ So V Lm ¼ X L I Lm ðpeak valueÞ and V L ¼ X L I L ðRMS valueÞ or X L ¼ V Lm =I Lm ; X L ¼ V L =I L where X L is measured in ohms (O) and is the same as resistance R. Recall that conductance G is the reciprocal of resistance R, and in an inductive circuit, the reciprocal of reactance is called inductive susceptance and is denoted by B L , i.e. B L ¼ 1=X L , and is measured in siemens (S) or mho ( O ). Relationship of voltage and current of inductor in an AC circuit ● Instantaneous values (time domain) i L ¼ I Lm sinðot þcÞ v L ¼ X L I Lm sinðot þc þ90 Þ Fundamentals of AC circuits 251 ● Ohm’s law: V Lm ¼ X L I Lm (peak value) V L ¼ X L I L (RMS value) ● Inductive reactance: X L ¼ oL ¼ 2pfL ● Inductive susceptance: B L ¼ 1=X L In an AC inductive circuit, the relationship between the voltage and cur- rent is not only determined by the value of inductance L in the circuit, but also related to the angular frequency o. If an inductor has a fixed value in the circuit of Figure 8.21(a), inductance L in the circuit is a constant, and the higher the angular frequency o, the greater the voltage across the inductor V L "¼ X L I L ¼ ðo " LÞI L When o ! 1, V L ! 1, i.e. when the angular frequency approaches to infi- nite, the inductor behaves as an open circuit in which the current is reduced to zero. The lower the angular frequency o, the lower the voltage across the inductor V L #¼ X L I L ¼ ðo # LÞI L When o ¼ 0, V L ¼ 0, i.e. the AC voltage across the inductor now is equivalent to a DC voltage since the frequency (o ¼ 2pf Þ does not change any more. Recall that the inductor is equivalent to a short circuit at DC. In this case, the inductor is shortened because of zero voltage across the inductor. This indicates that an inductor can pass the high-frequency signals (pass AC) and block the low-frequency signals (block DC). Characteristics of an inductor ● An inductor can pass AC (open-circuit equivalent). ● An inductor can block DC (short-circuit equivalent). The sinusoidal expressions of the inductor voltage v L and current i L are in the time domain. The peak and RMS values of the inductor voltage and the current in phasor domain also obey Ohm’s law as follows: Peak value: _ V Lm ¼ jX L _ I Lm or V Lm ¼ jX L I Lm RMS value: _ V L ¼ jX L _ I L or V L ¼ jX L I L 252 Understandable electric circuits This is because v L ¼ L di L dt , LjoI L (differentiating: multiply by jo) So _ V L ¼ ð joLÞ _ I L or _ V L ¼ jX L _ I L ðX L ¼ oLÞ. The relationship of the inductor voltage and current in an AC circuit can be presented by a phasor diagram illustrated in Figure 8.22(b and c). Figure 8.22(b) is when the initial phase angle is zero, i.e. c ¼ 08, and Figure 8.22(c) is when c6¼08 (the inductor current lags voltage by 908). Inductor’s AC response in phasor domain ● Ohm’s law: Peak value: _ V Lm ¼ jX L _ I Lm or V Lm ¼ jX L I Lm RMS value: _ V L ¼ jX L _ I L or V L ¼ jX L I L ● Phasor diagram: 0 90º V L I L ● Inductor voltage leads the current by 908. Example 8.11: In an AC inductive circuit, given v L ¼ 6 ffiffiffi 2 p sinð60t þ35 ÞV and L is 0.2 H, determine the current through the inductor in time domain. Solution: Inductive reactance X L ¼ oL ¼ ð60 rad=sÞð0:2 HÞ ¼ 12 O _ I Lm ¼ _ V Lm jX L ¼ 6 ffiffiffi 2 p ff35 V j12 O ¼ 6 ffiffiffi 2 p ff35 V 12ff90 O ¼ 0:5 ffiffiffi 2 p ff À55 A v L + – v L i L 90° 90° 0 e L +j + + +j 0 y v L • I L • I L • • (a) (b) (c) Figure 8.22 The phasor diagram of the AC inductive circuit Convert the inductor current from the phasor domain to the time domain i L ¼ 0:5 ffiffiffi 2 p sinð60t À55 Þ A Fundamentals of AC circuits 253 8.4.3 Capacitor’s AC response If an AC voltage source is applied to a capacitor as shown in Figure 8.23(a), the voltage across the capacitor will be v C ¼ V Cm sinðot þcÞV As we have learned from chapter 6, the relationship between the voltage across the capacitor and the current through it is i C ¼ C dv C dt Substituting v C into the above expression and applying differentiation gives i C ¼ C d½V Cm sinðot þcÞ dt ¼ oCV Cm sinðot þc þ90 Þ That is i C ¼ oCV Cm sinðot þc þ90 Þ ð8:7Þ The sinusoidal expressions of the capacitor voltage v C and current i C indicated that in an AC capacitive circuit, the voltage and current have the same angular frequency (o) and a phase difference. The capacitor current leads the voltage by 908 as illustrated in Figure 8.23(b), if we assume that the initial phase angle c¼08. The relationship between voltage and current in an inductive sinusoidal AC circuit can be obtained from (8.7), which is given by I Cm ¼ ðoCÞV Cm ðpeak valueÞ or I C ¼ ðoCÞV C ðRMS valueÞ This is also known as Ohm’s law for a capacitive circuit, where oC is called capacitive reactance that is denoted by the reciprocal of X C , i.e. (a) (b) – e C I c + v c • 90° 0 v c I c w t • Figure 8.23 Capacitor’s AC response 254 Understandable electric circuits X C ¼ 1 oC ¼ 1 2pfC ðo ¼ 2pf Þ So X C ¼ V Cm I Cm ðpeak valueÞ or X C ¼ V C I C ðRMS valueÞ X C is measured in ohms (O) and that is the same as resistance R and inductive reactance X L . Recall that the inductive susceptance B L is the reciprocal of the inductive reactance X L . The reciprocal of capacitive reactance is called capacitive sus- ceptance and is denoted by B C , i.e. B C ¼ 1=X C , and it is also measured in sie- mens or mho (same as B L ). The relationship of voltage and current of capacitor in an AC circuit ● Instantaneous values (time domain): v C ¼ V Cm sinðot þcÞ i C ¼ ocV Cm sinðot þc þ90 Þ ● Ohm’s law: V Cm ¼ X C I Cm (peak value) V C ¼ X C I C (RMS value) ● Capacitive reactance: X C ¼ 1=oC ¼ 1=2pfC ● Capacitive susceptance: B C ¼ 1=X C Similar to an inductor, in an AC capacitive circuit not only is the rela- tionship between voltage and current determined by the value of capacitive C in the circuit but it is also related to angular frequency o. If there is a fixed capacitor in Figure 8.23(a), the conductance C in the circuit is a constant, and the higher the angular frequency o, the lower the voltage across the capacitor. V C # ¼ X C I C ¼ I C o " C When o ! 1, V C ! 0, i.e. when the angular frequency approaches infinite, the capacitor behaves as a short circuit in which the voltage across it will be reduced to zero. Fundamentals of AC circuits 255 The lower the angular frequency o, the higher the voltage across the capacitor. V C "¼ I C o # C When o ! 0, V C ! 1, i.e. the ACvoltage across the capacitor nowis equivalent to a DCvoltage since the frequency ðo ¼ 2pf Þ does not change any more. Recall that a capacitor is equivalent to an open circuit at DC. In this case, the capa- citor is open because there will be no current flowing through the capacitor. This indicates that a capacitor can block the high-frequency signal (block AC) and pass the low-frequency signal (pass DC). The characteristics of a capacitor are opposite to those of an inductor. Characteristics of a capacitor ● A capacitor can pass DC (short-circuit equivalent). ● A capacitor can block AC (open-circuit equivalent). The sinusoidal expressions of the capacitor voltage v C and current i C are in the time domain. The peak and RMS values of the capacitor voltage and the current in phasor domain also obey Ohm’s law as follows: Peak value: _ V Cm ¼ ÀjX C _ I Cm or V Cm ¼ ÀjX C I Cm RMS value: _ V C ¼ ÀjX C _ I C or V C ¼ ÀjX C I C This is because i C ¼ C dv C dt , CjoV C (differentiating: multiply by jo). So _ I C ¼ joC _ V C ¼ jð1=X C Þ _ V C X C ¼ 1=oC ð Þ or _ V C ¼ ÀjX C _ I C ð1=j ¼ÀjÞ: The relationship of the capacitor voltage and current in an AC circuit can be presented by a phasor diagram and is illustrated in Figure 8.24(b and c). Figure 8.24(b) is when the initial phase angle is zero, i.e. c ¼ 08 (capacitor voltage lags current by 908), and Figure 8.24(c) is when c 6¼ 08 (a) (b) (c) 90° 0 +j + V c • I c • – C + v c e I c • 90° v c +j + 0 • I c • Figure 8.24 The phasor diagram of an AC capacitive circuit 256 Understandable electric circuits Capacitor’s AC response in phasor domain ● Ohm’s law: Peak value: _ V Cm ¼ ÀjX C _ I Cm or V Cm ¼ ÀjX C I Cm RMS value: _ V C ¼ Àj _ X C I C or V C ¼ ÀjX C I C ● Phasor diagram: 0 90º I C • V C • Capacitor current leads voltage by 908. Example 8.12: Given a capacitive circuit in which v C ¼ 50 ffiffiffi 2 p sinðot À20 ÞV, capacitance is 5 mF and frequency is 500 Hz, determine the capacitor current in the time domain. Solution: o ¼ 2pf ¼ 2pð500 HzÞ % 3142 rad=s X C ¼ 1 oC ¼ 1 ð3 142 rad=sÞð5 Â10 À6 FÞ % 63:65 O ; I Cm ¼ V Cm X C ¼ 50 ffiffiffi 2 p V 63:65 O % 786 ffiffiffi 2 p mA i C ¼ 786 ffiffiffi 2 p sinðot À20 þ90 Þ ¼ 786 ffiffiffi 2 p sinðot þ70 ÞmA Summary ● Direct current (DC) ● The polarity of DC voltage and direction of DC current do not change. ● The pulsing DC changes the amplitude of the pulse, but does not change the polarity. ● Alternating current (AC) ● The voltage and current periodically change polarity with time (such as sine wave, square wave, saw-tooth wave, etc.). ● Sine AC varies over time according to the sine function, and is the most widely used AC. ● Period and frequency ● Period T is the time to complete one full cycle of the waveform. ● Frequency f is the number of cycles of waveforms within 1 s: f ¼ 1=T. ● Three important components of the sinusoidal function f ðtÞ ¼ F m sinðot þcÞ ● F m : Peak value (amplitude) ● o: Angular velocity (or angular frequency) o ¼ 2p=T ¼ 2pf ● c: Phase or phase shift Fundamentals of AC circuits 257 ● c 40: Waveform shifted to the left side of 08 ● c 50: Waveform shifted to the right side of 08 ● Phase difference : For two waves with the same frequency such as vðtÞ ¼ V m sin ðot þc v Þ; iðtÞ ¼ I m sinðot þc i Þ ¼ c v Àc i ● If ¼ 0: v and i in phase ● If 40: v leads i ● If 50: v lags i ● If ¼ Æp=2: v and i are orthogonal ● If ¼ +p: v and i are out of phase ● Peak value, peak–peak value, instantaneous value and average value of sine waveform. ● Peak value F pk ¼ F m : the amplitude ● Peak–peak value F p–p : F p–p ¼ 2F pk ● Instantaneous value f(t): value at any time at any particular point of the waveform ● Average value: average value of a half-cycle of the sine waveform F avg ¼ 0:637F m ● RMS value (or effective value) of AC sinusoidal function ● If an AC source delivers the equivalent amount of power to a resistor as a DC source, which is the effective or RMS value of AC. V ¼ V m ffiffiffi 2 p ¼ 0:707 V m ; I ¼ I m ffiffiffi 2 p ¼ 0:707 I m ● The general formula to calculate RMS value is F ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 T T 0 f 2 ðtÞdt ● Complex numbers ● Rectangular form: A ¼ x þ jy ● Polar form: A ¼ affc ● Conversion between rectangular and polar forms: A ¼ x þjy ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffi x 2 þy 2 tan À1 y x ¼ affc A ¼ affc ¼ x þjy ¼ aðcosc þjsincÞ ● Addition and subtraction: A 1 ÆA 2 ¼ ðx 1 Æx 2 Þ þjðy 1 Æy 2 Þ ● Multiplication: A 1 Á A 2 ¼ a 1 Á a 2 ffðc 1 þc 2 Þ ¼ ðx 1 þjy 1 Þðx 2 þjy 2 Þ 258 Understandable electric circuits ● Division: A 1 A 2 ¼ a 1 a 2 ffðc 1 Àc 2 Þ ¼ x 1 þjy 1 x 2 þjy 2 ● Phasor ● A phasor is a vector that contains both amplitude and angle infor- mation, and can be represented as a complex number. ● The phasor notation is a method that uses complex numbers to represent the sinusoidal quantities for analysing AC circuits when all quantities have the same frequency. ● Rotation factor: e jot or Æj ¼ Æ90 ● Differentiation and integration of the sinusoidal function in phasor notation: ● Differentiation: df ðtÞ=dt ¼ joF or jo _ F (þj ¼ þ90 ) ● Integration: f ðtÞdt ¼ F=jo or ð1=joÞ _ F ð1=j ¼ Àj ¼ À90 Þ ● Characteristics of the inductor and capacitor: ● Three basic elements in an AC circuit Time domain Phasor domain f ðtÞ ¼ F m sinðot þcÞ F m ¼ F m ffc or _ F m ¼ F m ffc ðpeak valueÞ F ¼ Fffc or _ F ¼ Fffc ðRMS valueÞ vðtÞ ¼ V m sin ðot þcÞ _ V m ¼ V m ffc v , _ V ¼ Vffc v iðtÞ ¼ I m sin ðot þcÞ _ I m ¼ I m ffc i , _ I ¼ Iffc i Element DC (v ¼ 0) AC (v ! `) Characteristics Inductor Short circuit Open circuit Pass DC and block AC Capacitor Open circuit Short circuit Pass AC and block DC Element Time domain Phasor domain Resistance and reactance Conductance and susceptance Phasor diagram Resistor v R ¼ Ri R _ V R ¼ _ I R R R G ¼ 1=R • I R V R Inductor v L ¼ Lðdi=dtÞ _ V L ¼ jX L _ I L X L ¼ oL B L ¼ 1=X L 0 90º v L I L Capacitor i C ¼ Cðdv C =dtÞ _ V C ¼ ÀjX C _ I C X C ¼ 1=oC B C ¼ 1=X C 0 90º • • I C v C Fundamentals of AC circuits 259 Experiment 8: Measuring DC and AC voltages using the oscilloscope Objectives ● Become familiar with the operations of a function generator. ● Become familiar with the settings and correction of an oscilloscope. ● Become familiar with the operations of an oscilloscope. ● Become familiar with the method to measure DC and AC voltages with an oscilloscope. Background information 1. Function generator: The function generator is an electronic equipment that can generate various types of waveforms that can have different fre- quencies and amplitudes. A function generator can be used as an AC vol- tage source to provide time-varying signals such as sine waves, square waves, triangle waves, etc. 2. Oscilloscope: The oscilloscope is one of the most important experimental and measurement instruments available for testing electric and electronic circuits. Its main function is to display waveforms to observe and analyse voltage, frequency, period and phase difference of DC or AC signals. The oscilloscope is a complex testing equipment and it is important to be familiar with its operations. There are various types of oscilloscopes that may look different, but most of their controls (knobs and buttons) in Table L8.1 have similar functions. Figure L8.1 shows the front panel of an oscilloscope. We will use this oscilloscope as an example for a brief description of the operations of the oscilloscope. ● Intensity control (INTENSITY): It can adjust the brightness of the display. ● Focus control (FOCUS): It can adjust the sharpness and clarity of the display. Table L8.1 The main controls of an oscilloscope Display Horizontal control Vertical control Selecting switch Probe INTENSITY Time base setting (TIME/DIV) Volts per division (VOLTS/DIV) Channel coupling (CH I–DUAL– CH II) 61 FOCUS Horizontal position control (X-POS $) Vertical position control (Y-POS) Input coupling (DC–GND–AC) 610 260 Understandable electric circuits ● Time base control (TIME/DIV – seconds per division): It can set up the length of time displayed per horizontal square (division) on the screen. ● Volts per division selector (VOLTS/DIV – volts per division): It can set up the waveform amplitude value per vertical square (division) on the screen. Measured amplitude ¼ ðNumber of vertical divisionsÞ ÂðVOLTS=DIVÞ Note: There is a small calibration (CAL) knob in the centre of both the VOLTS/DIV and TIME/DIV knobs. It should be in the fully clockwise posi- tion for the accuracy of the measurement. ● Horizontal position control (X-POS$): It can adjust the horizontal posi- tion of the waveform. ● Vertical position control (Y-POSl): It can adjust the vertical position of the waveform. ● Channel coupling (CH I–DUAL–CH II): CH I: Displays the input signal from channel I. CH II: Displays the input signal from channel II. DUAL: Displays the input signals from both channels I and II. ● Input coupling (DC–GND–AC): The connection from the test circuit to the oscilloscope. Figure L8.1 An oscilloscope Fundamentals of AC circuits 261 DC: The DC position can display both DC and AC waveforms (the AC signal is superimposed on the DC waveform). AC: The AC position blocks the DC waveform and only displays AC waveform. GND: The GND position has a horizontal line on the screen that repre- sents zero reference. ● 61 Probe: Can measure and read the signal directly but may load the circuit under test and distort the waveform. ● 610 Probe: Needs to multiply by 10 for each measured reading (more accurate). Equipment and components ● Digital multimeter ● Breadboard ● DC power supply ● Oscilloscope ● Function generator Procedure Part I: Measure DC voltage using an oscilloscope 1. Set up the oscilloscope controls to the following positions: ● Channel coupling: CH I or CH II ● Input coupling: Set up to GND and adjust the trace to the central reference line (0 V) first, then switch to DC ● TIME/DIV: 1 ms/DIV ● Trigger: Auto (The trigger can stabilize repeating waveforms and capture single-shot waveforms.) 2. Connect a circuit as shown in Figure L8.2. The negative terminal of DC power, ground of the oscilloscope probe, and negative terminal of the multimeter (voltmeter function) should be connected together. 3. Set up the oscilloscope probe to 61, adjust VOLTS/DIV of the oscillo- scope to 1 V/DIV, and adjust DC power supply to 3 V. The voltmeter reading should be 3 V now. The DC wave on the oscilloscope screen E Oscilloscope V + – Figure L8.2 Measuring DC voltage using an oscilloscope 262 Understandable electric circuits occupies three vertical grids (squares) at this time, so the voltage measured by the oscilloscope is also 3 V. ð3 vertical divisionsÞ Âð1 V=DIVÞ ¼ 3 V 4. Keep the oscilloscope probe at 61, adjust VOLTS/DIV of the oscilloscope to 0.5 V/DIV, and adjust DC power supply to 4 V. The DC wave on the oscilloscope screen occupies eight vertical divisions at this time. ð8 vertical divisionsÞ Âð0:5 V=DIVÞ ¼ 4 V Read the value on the voltmeter, and record it in Table L8.2. 5. Keep the oscilloscope probe at 61, adjust VOLTS/ DIV of the oscillo- scope to 2 V/DIV, and adjust DC power supply to 5 V. Read the voltage value on the voltmeter and oscilloscope, and record them in Table L8.2. 6. Set up the oscilloscope probe to 610, adjust DC power supply to 8, 12 and 16 V, respectively, and adjust VOLTS/DIV to suitable positions. Read the vol- tage values on the voltmeter and oscilloscope, and record them in Table L8.2. Part II: AC measurements using an oscilloscope 1. Replace the DC power supply by a function generator in Figure L8.2. The ground of the function generator, ground of the oscilloscope probe and nega- tive terminal of multimeter (voltmeter function) should be connected together. ● Set up the function generator: Waveform: sine Frequency: 1.5 kHz DC offset: 0 V Amplitude knob: minimum (Fully counter clockwise) ● Set up the oscilloscope: VOLTS/DIV: 0.5 V/DIV Channel coupling: CH I TIME/DIV: 0.2 ms/DIV Table L8.2 Probe DC power supply (V) Vertical division (DIV) VOLTS/DIV (V/DIV) Voltmeter (V) Oscilloscope (V) 61 Example: 3 3 1 3 3 4 8 0.5 4 5 2 610 8 12 16 Fundamentals of AC circuits 263 Input coupling: Set up to GND and adjust the trace to the central reference line (0 V) first, then switch to AC 2. Adjust the amplitude knob of the function generator until that sine wave on the vertical division of the oscilloscope screen occupies six divisions (squares). The voltage amplitude at this time is V PÀP ¼ ð6 DIVÞ Âð0:5 V=DIVÞ ¼ 3 V Note that the reading of the multimeter is RMS value, and it can be converted to the peak value comparing with the waveform obtained from the oscilloscope. 3. Adjust the horizontal position control of the oscilloscope (X-POS) until the sine wave on the oscilloscope screen occupies four horizontal divisions. ● Determine the period of the sine wave T: Period ðTÞ ¼ ðNumber of horizontal divisionsÞ ÂðTIME=DIVÞ T ¼ ð4 divisionsÞ Âð0:2 ms=DIVÞ ¼ 0:8 ms ● Determine the frequency f: f ¼ 1 T ¼ 1 0:8 ms ¼ 1:25 kHz 4. Adjust the horizontal position control of the oscilloscope (X-POS) until the sine wave on the oscilloscope screen occupies six horizontal divisions (adjust the frequency knob on the function generator if necessary). Deter- mine the period T and frequency f of the sine wave, and record the values in Table L8.3 (keep TIME/DIV ¼ 0.2 ms/DIV). 5. Adjust TIME/DIV of the oscilloscope to 0.5 ms/DIV, and adjust the hor- izontal position control of the oscilloscope (X-POS) until the sine wave on the oscilloscope screen occupies five horizontal divisions. Determine the period T and frequency f of the sine wave, and record the values in Table L8.3. Conclusion Write your conclusions below: Table L8.3 Period T Frequency f Step 4 Step 5 264 Understandable electric circuits Chapter 9 Methods of AC circuit analysis Objectives After completing this chapter, you will be able to: ● understand concepts and characteristics of the impedance and admittance of AC circuits ● define the impedance and admittance of resistor R, inductor L and capacitor C ● determine the impedance and admittance of series and parallel AC circuits ● apply the voltage divider and current divider rules to AC circuits ● apply KCL and KVL to AC circuits ● understand the concepts of instantaneous power, active power, reactive power, apparent power, power triangle and power factor ● apply the mesh analysis, node voltage analysis, superposition theorem and Thevinin’s and Norton’s theorems, etc. to analyse AC circuits 9.1 Impedance and admittance 9.1.1 Impedance In the previous chapter, we had learned that the phasor forms of relationship between voltage and current for resistor, inductor and capacitor in an AC circuit are as follows: _ V R ¼ _ I R R; _ V L ¼ j _ I L X L ; _ V C ¼ Àj _ I C X C The above equations can be changed to a ratio of voltage and current _ V R _ I R ¼ R; _ V L _ I L ¼ jX L ; _ V C _ I C ¼ ÀjX C The ratio of voltage and current is the impedance of an AC circuit, and it can be generally expressed as Z ¼ _ V= _ I. This equation is also known as Ohm’s law of AC circuits. The physical meaning of the impedance is that it is a measure of the opposition to AC current in an AC circuit. It is similar to the concept of resistance in DC circuits, so the impedance is also measured in ohms. The impedance can be extended to the inductor and capacitor in an AC circuit. It is a complex number that describes both the amplitude and phase characteristics. The impedances of resistor, inductor and capacitor are as follows: Z R ¼ R ¼ _ V R _ I R ; Z L ¼ jX L ¼ _ V L _ I L ; Z C ¼ ÀjX C ¼ _ V C _ I C Impedance Z ● Z is a measure of the opposition to AC current in an AC circuit. ● Ohm’s law in AC circuits: Z ¼ _ V= _ I: 9.1.2 Admittance Recall that the conductance G is the inverse of resistance R, and it is a measure of how easily current flows in a DC circuit. It is more convenient to use the con- ductance in a parallel DC circuit. Similarly, the admittance is the inverse of impedance Z, it is denoted by Y, Y = 1/Z, and is measured in siemens (S). The admittance is a measure of how easily a current can flow in an AC circuit. It can be expressed as the ratio of current and voltage of an AC circuit, i.e. Y ¼ _ I= _ V. It is more convenient to use the admittance in a parallel AC circuit. Admittance Y ● Y is the measure of how easily current can flow in an AC circuit. ● Y is the inverse of impedance: Y ¼ 1/Z. ● Ohm’s law in AC circuits: _ I ¼ _ VY: The admittance of resistor, inductor and capacitor are as follows: Y R ¼ 1 R ; Y L ¼ 1 jX L ¼ Àj 1 X L ; Y C ¼ 1 ÀjX C ¼ j 1 X C j ¼ 1 Àj Quantity Quantity Symbol Unit Unit symbol Admittance Y Siemens S Quantity Quantity Symbol Unit Unit symbol Impedance Z Ohm O 266 Understandable electric circuits 9.1.3 Characteristics of the impedance Since the impedance is a vector quantity, it can be expressed in both polar form and rectangular form (complex number) as follows: Z ¼ zfff ¼ R þ j X ¼ zðcos f þ j sin fÞ ð9:1Þ The rectangular form is the sum of the real part and the imaginary part, where the real part of the complex is the resistance R, and the imaginary part is the reactance X. The reactance is the difference of inductive reactive and capacitive reactance, i.e. X ¼ X L À X C The lower case letter z in (9.1) is the magnitude of the impedance, which is z ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi R 2 þ X 2 p The corresponding angle f between the resistance R and reactance X is called the impedance angle and can be expressed as follows: f ¼ tan À1 X R The relationship between R, X and Z in the expression of the impedance is a right triangle, and can be described using the Pythagoras’ theorem. This can be illu- strated in Figure 9.1(a). Figure 9.1(a) is an impedance triangle. If we multiply each side of the quantity in the impedance triangle by current _ I the following expressions will be obtained: _ V z ¼ Z _ I z ; _ V X ¼ _ I X X; _ V R ¼ _ I X R These can form another triangle that is called the voltage triangle, which is illustrated in Figure 9.1(b). If we divide each side of the value by voltage _ V in the impedance triangle, the following expressions will be obtained: _ I z ¼ _ V z Z ; _ I X ¼ _ V X X ; _ I R ¼ _ V R R R f f f X z Vz V X V R I R I X Iz (a) (b) (c) Figure 9.1 Impedance, voltage and current triangles Methods of AC circuit analysis 267 The above expression can form another triangle that is called the current triangle, and it is illustrated in Figure 9.1(c). The characteristics of the impedance triangle in Figure 9.1(a) can be summarized as follows: ● If X 4 0 or X ¼ X L À X C > 0; X L > X C : The reactance X is above the hor- izontal axis, and the impedance angle f 40. The circuit is more inductive as shown in Figure 9.2(a). ● If X 5 0 or X ¼ X L À X C < 0; X C > X L : The reactance X is below the hor- izontal axis, and the impedance angle f < 0. The circuit is more capacitive as shown in Figure 9.2(b). ● If X = 0 or X ¼ X L À X C ¼ 0; X C ¼ X L : The impedance angle f ¼ 0, the circuit will look like a purely resistive circuit (z = R) as shown in Figure 9.2(c). Example 9.1: Determine the impedance Z in the circuit of Figure 9.3 and plot the phasor diagram of the impedance. Solution: The impedances in series in an AC circuit behave like resistors in series. 0 X L X L X L + +j R X L – X C (X L > X C ) (X L = X C ) (X C > X L ) X L – X C X C X C X C z +j + f > 0 f < 0 f = 0 R z 0 + 0 +j R R = z (a) (b) (c) f f Figure 9.2 The phasor diagrams of the impedance Z R Z L Z C R = 1.5 Ω X L = 2.5 Ω X C = 3 Ω Z Figure 9.3 Circuit for Example 9.1 268 Understandable electric circuits So, Z ¼ Z R þ Z L þ Z C ¼ R þ jX ¼ R þ jðX L À X C Þ ¼ 1:5 O þ jð2:5 À 3ÞO ¼ 1:5 O À j0:5 O ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1:5 2 þ ðÀ0:5Þ 2 q tan À1 À0:5 1:5 % 1:58ff À 18:44 O Note: Since the imaginary term is 70.5 on y-coordinate, and the real term is þ1.5 on the x-coordinate, the impedance angle for this circuit is located in the 4th quadrant. The circuit for Example 9.1 is more capacitive since X C > X L , and f < 0 as shown in Figure 9.4. 9.1.4 Characteristics of the admittance The admittance is also a complex number; it can be expressed in both polar and rectangular forms as follows: Y ¼ yfff y ¼ G þ jB ¼ yðcos f þ j sin fÞ ð9:2Þ The real part of the complex is the conductance G, and the imaginary part is called the susceptance B. The susceptance is measured in the same way as the admittance, i.e. siemens (S). The susceptance is the difference of the capacitive susceptance and inductive susceptance, i.e. B ¼ B C À B L : The lower case letter y in (9.2) is the magnitude of the admittance, i.e. y ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi G 2 þ B 2 p +j + X L = 2.5 Ω X C = 3 Ω X L – X C = –0.5 Ω R = 1.5 Ω f z 0 Figure 9.4 Impedance angle for Example 9.1 Methods of AC circuit analysis 269 The corresponding angle f between the conductance G and susceptance B is called the admittance angle and can be expressed as f y ¼ tan À1 B G The admittance of resistor, inductor and capacitor are as follows: Y R ¼ 1 R ¼ G; Y L ¼ 1 jX L ¼ Àj 1 X L ; Y C ¼ 1 ÀjX C ¼ j 1 X C The impedance, admittance, susceptance and their relationship can be sum- marized as given in Table 9.1. The characteristics of the admittance triangle in Figure 9.5 can be summarized as follows: X L ¼ oL; X C ¼ 1 oC ; j ¼ 1 Àj Table 9.1 Impedance and admittance Component Impedance Z ¼ _ V= _ I Admittance Y ¼ 1/Z Conductance and susceptance Resistor (R) Z R ¼ R Y R ¼ G Conductance: G ¼ 1/R Inductor (L) Z L ¼ jX L Y L ¼ 7jB L Inductive susceptance: B L ¼ 1/X L Capacitor (C) Z C ¼ 7jX C Y C ¼ jB C Capacitive susceptance: B C ¼ 1/X C Z ¼ zfff ¼ R þ jX Y ¼ yfff y ¼ G þ jB Reactance: X ¼ X L À X C z ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi R 2 þ X 2 p y ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi G 2 þ B 2 p Susceptance: B ¼ B C À B L f ¼ tan À1 X R f y ¼ tan À1 B G 0 + G G = Y Y + f y f y G Y 0 + 0 G +j +j +j (B L = B C ) (B L > B C ) (B C > B L ) B C – B L B C – B L B C B C B C B L B L B L f y = 0 f y < 0 f y > 0 (a) (b) (c) Figure 9.5 The phasor diagrams of the admittance 270 Understandable electric circuits ● If B 4 0 or B ¼ ðB C À B L Þ > 0, B C > B L : The susceptance B is above the horizontal axis, the admittance angle f y 40, and the circuit is more capacitive as shown in Figure 9.5(a). ● If B 5 0 or B ¼ ðB C À B L Þ < 0, B L > B C : The susceptance B is below the horizontal axis, the admittance angle f y < 0, and the circuit is more inductive as shown in Figure 9.5(b). ● If B ¼ 0 or B ¼ ðB C À B L Þ ¼ 0, B L ¼ B C : The admittance angle f y ¼ 0, the circuit will look like a purely resistive circuit (Y ¼ G) as shown in Figure 9.5(c). Characteristics of impedance and admittance ● If X 4 0, f 4 0, B 50, f y 5 0: The circuit is more inductive. ● If X 5 0, f 5 0, B 40, f y 4 0: The circuit is more capacitive. ● If X ¼ 0, f ¼ 0, B ¼ 0, f y ¼ 0: The circuit is purely resistive. Example 9.2: Determine the admittance in the circuit of Figure 9.6 and plot the phasor diagrams of the admittance. Solution: The admittances in parallel in AC circuits behave like the conductances in parallel in DC circuits. So, Y ¼Y R þ Y L þ Y C ¼ G þ jB ¼ G þ jðB C À B L Þ ¼ 1 13:3 O þ j 1 13:3 O À 1 5:72 O % 0:075 S þ jð0:075 À 0:175ÞS ¼0:075 S À j0:1 S ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0:075 2 þ ðÀ0:1Þ 2 q tan À1 À0:1 0:075 ¼ 0:125ff À 53:13 S Note: The admittance angle for this circuit is located in the fourth quadrant since the imaginary term is 70.1 and the real term is þ0.075. Y X L X C 13.3 Ω 5.72 Ω 13.3 Ω R Figure 9.6 Circuit for Example 9.2 Methods of AC circuit analysis 271 Since B L 4B C (B L ¼ 0.175, B C ¼ 0.075) and f y 50, the circuit is more inductive as shown in Figure 9.7. 9.2 Impedance in series and parallel 9.2.1 Impedance of series and parallel circuits The impedances in series and parallel AC circuits behave like resistors in series and parallel DC circuits, except the phasor form (complex number) is used. The equivalent impedance (or total impedance) for a series circuit in Figure 9.8 is given as Z eq ¼ Z 1 þ Z 2 þ Á Á Á þ Z n The equivalent impedance (or total impedance) for a parallel circuit in Figure 9.9 is given as Z eq ¼ 1 ð1=Z 1 Þ þ ð1=Z 2 Þ þ Á Á Á þ ð1=Z n Þ ¼ Z 1 ==Z 2 == . . . ==Z n Y eq ¼ Y 1 þ Y 2 þ Á Á Á þ Y n + j + B C = 0.075 S B L = 0.175 S B C – B L Y = 0.125 S 0 G = 0.075 S f y = –53.1° Figure 9.7 Admittance angle for Example 9.2 Z 1 Z 2 Z eq Z n … Figure 9.8 Impedance of a series circuit … Z 1 Z 2 Z eq Z n Figure 9.9 Impedance of a parallel circuit 272 Understandable electric circuits The equivalent impedance is the reciprocal of equivalent admittance, Z eq ¼ 1=Y eq . If only have two impedances in parallel, the equivalent impedance is given as Z eq ¼ Z 1 Z 2 Z 1 þ Z 2 ¼ Z 1 ==Z 2 Impedances in series and parallel ● Impedances in series: Z eq ¼ Z 1 þ Z 2 þ Á Á Á þ Z n ● Impedances in parallel: Z eq ¼ Z 1 ==Z 2 == Á Á Á ==Z n ; Y eq ¼ Y 1 þ Y 2 þ Á Á Á þ Y n ● Two impedances in parallel: Z eq ¼ Z 1 Z 2 Z 1 þ Z 2 To determine the equivalent impedance in series and parallel AC circuits, use the same method that determines the equivalent resistance in series and parallel DC circuits. 9.2.2 Voltage divider and current divider rules The voltage divider and current divider rules in phasor form in AC circuits are very similar to the DC circuits as follows: _ V 1 ¼ Z 1 Z 1 þ Z 2 _ E; _ V 2 ¼ Z 2 Z 1 þ Z 2 _ E _ I 1 ¼ Z 2 Z 1 þ Z 2 _ I T ; _ I 2 ¼ Z 1 Z 1 þ Z 2 _ I T (a) (b) Z 1 Z 2 1 V 2 V E • • • Z 1 Z 2 E 1 I 2 I T I • • • • Figure 9.10 Voltage and current dividers Methods of AC circuit analysis 273 9.2.3 The phasor forms of KVL and KCL The phasor forms of Kirchhoff’s voltage law (KVL) and Kirchhoff’s current law (KCL) also hold true in AC circuits. Phasor forms of KVL and KCL ● KVL: S _ V ¼ 0 or _ V 1 þ _ V 2 þ Á Á Á þ _ V n ¼ _ E ● KCL: S _ I ¼ 0 or _ I in ¼ _ I out The following examples show how to use the above equations in series-parallel AC circuits. Example 9.3: Determine the following values for the circuit in Figure 9.11. ● the input equivalent impedance Z eq and ● the current _ I 3 in the branch of R L and X L . Solution: (a) Z eq ¼ Z 1 þ Z 2 ==Z 3 Z 1 ¼ R 1 ¼ 4 kO Z 2 ¼ ÀjX C ¼ Àj8 kO Z 3 ¼ R L þ jX L ¼ 4 kO þ j8 kO % 8:94ff63:44 kO Z 2 ==Z 3 ¼ Z 2 Z 3 Z 2 þ Z 3 ¼ ðÀj8Þð4 þ j8Þ Àj8 þ 4 þ j8 kO ¼ 64 À j32 4 kO % 71:55ff À 26:57 4ff0 kO % ð16 À j8ÞkO ¼ 17:9ff À 26:57 kO ¼ 17:9½cosðÀ26:57 Þ þ j sinðÀ26:57 ÞkO Z eq ¼ Z 1 þ Z 2 ==Z 3 ¼ ½4 þ ð16 À j8ÞkO ¼ ð20 À j8ÞkO % 21:54ff À 21:8 kO Z 1 Z 3 E = 100V∠0° Z 2 I 3 I 1 I 2 R 1 = 4kΩ R L = 4kΩ X L = 8kΩ X C = 8kΩ Figure 9.11 Circuit for Example 9.3 274 Understandable electric circuits (b) _ I 3 ¼ Z 2 Z 2 þ Z 3 _ I 1 Here _ I 1 ¼ _ E Z eq ¼ 100ff0 V 21:54ff À 21:8 O % 4:64ff 21:8 mA ; _ I 3 ¼ Z 2 Z 2 þ Z 3 _ I 1 ¼ð4:64ff21:8 ÞmA 8ff À 90 kO ðÀj8 þ 4 þ j8ÞkO ¼ 37:12ff À 68:2 4ff0 mA ¼9:28ff À 68:2 mA Example 9.4: Determine the voltage across the inductor L for the circuit in Figure 9.12. Z 1 = 120Ω (a) (b) V L V L E = 40 sin(2t-30°) V R = 120Ω L = 10H C = 20mF + – E = 40V ∠–30° Z 3 = –j25Ω Z 2 = j20Ω + – Figure 9.12 Circuits for Example 9.4 Solution: ● Convert the time domain to the phasor domain as shown in Figure 9.12(b) first. Z 1 ¼ R ¼ 120 O Z 2 ¼ jX L ¼ jðoLÞ ¼ jð2  10 HÞ ¼ j20 O Z 3 ¼ ÀjX C ¼ Àj 1 oC ¼ Àj 1 2  20 mF ¼ Àj25 O e ¼ 40 sinð2t À 30 ÞV ) _ E ¼ 40ff À 30 V ● _ V L ¼ _ E Z 2 ==Z 3 Z 1 þ Z 2 ==Z 3 ðZ 2 ==Z 3 ¼ ?Þ Z 2 ==Z 3 ¼ Z 2 Z 3 Z 2 þ Z 3 ¼ j20ðÀj25Þ j20 À j25 O ¼ 500 Àj5 O ¼ j100 O Methods of AC circuit analysis 275 ; _ V L ¼ _ E ¼ Z 2 ==Z 3 Z 1 þ Z 2 ==Z 3 ¼ ð40ffÀ 30 ÞV j100 O ð120 þ j100ÞO % 4000ff 60 156:2ff 39:8 V %25:61ff 20:2 V After converting the phasor form to the time form gives v L ¼ 25:61 sinð2t þ 20:2 ÞV 9.3. Power in AC circuits There are different types of power in AC circuits, such as instantaneous power, active power, reactive power and apparent power. 9.3.1 Instantaneous power p The instantaneous power p is the power dissipated in a component of an AC circuit at any instant time. It is the product of instantaneous voltage v and current i at that particular moment (Figure 9.13), i.e. instantaneous power can be expressed as p ¼ vi If v ¼ V m sinðot þfÞ and i ¼ I m sin ot Then, p ¼ vi ¼ V m I m sin ot sinðot þfÞ , À sin x sin y ¼ 1 2 ½cosðx þ yÞ À cosðx À yÞ ; p ¼ À 1 2 V m I m ½cosð2ot þfÞ À cos f ¼ VI cos f À VI cosð2ot þfÞ ðV m ¼ ffiffiffi 2 p V; I m ¼ ffiffiffi 2 p IÞ ¼ VI cos f À VIðcos 2ot cos f À sin 2ot sin fÞ ½, cosðx þ yÞ ¼ cos x cos y À sin x sin y Load v + – i Figure 9.13 Instantaneous power 276 Understandable electric circuits Therefore, instantaneous power is given as p ¼ VI cos fð1 À cos 2otÞ þ VI sin fsin 2ot The waveform of the instantaneous power can be obtained from the product of instantaneous voltage and current at each point on their waveforms as shown in Figure 9.14. Such as: ● at time t ¼ 0 : i ¼ 0, p ¼ vi ¼ 0 ● at time t ¼ t 1 : v ¼ 0, p ¼ vi ¼ 0 ● between time 0 * t 1 : v 40 and i 40, \ p ¼ vi 40 ● between time t 1 *t 2 : v 5 0 and i 4 0, \ p ¼ vi 5 0 ● between time t 2 *t 3 : v 5 0 and i 5 0, \ p ¼ vi 4 0 When instantaneous power p is 40 (p is positive), the component stores energy provided by the source. When instantaneous power p is 50 (p is negative), the component returns the stored energy to the source. Instantaneous power p p is the product of instantaneous voltage and current at any instant time: p ¼ vi ¼ VI cosfð1 À cos 2otÞ þ VIðsinfsin 2otÞ p 40: The component absorbs (stores) energy. p 50: The component returns (releases) energy. ● Instantaneous power for a resistive component p R : Since voltage and current in a purely resistive circuit is in phase, i.e. f ¼ 0, substituting this into the equation of the instantaneous power gives 0 v i t p t 1 t 3 t 2 Figure 9.14 The waveform of instantaneous power Methods of AC circuit analysis 277 p R ¼ vi ¼ VI cos0 ð1 À cos 2otÞ þ VI sin0 sin 2ot ¼ VI À VI cos 2ot ¼ VIð1 À cos 2otÞ ð9:3Þ The first part VI in (9.3) is average power dissipated in the resistive load (p 40, the load absorbs power). The second part in (9.3) is a sinusoidal quantity with a double frequency 2o, this indicates that when voltage and current waveforms oscillate one full cycle in one period of time, power waveform will oscillate two cycles as illustrated in Figure 9.15. The mathematical expression and the waveform all show that instantaneous power of a resistive load is always positive, or a resistor always dissipates power, indicating that the resistor is an energy consuming element. ● Instantaneous power for capacitive and inductive components: In a purely inductive load circuit, voltage leads current by 908. In a purely capacitive circuit, voltage lags current by 908. Substituting f ¼ Æ90 into the equation of instantaneous power gives p ¼ VI cosðÆ90 Þð1 À cos 2otÞ þ VI sinðÆ90 Þ sin 2ot ¼ ÆVI sin 2ot ð9:3Þ The instantaneous power for inductive and capacitive loads can be obtained from (9.3) as follows: ● Instantaneous power for an inductive load: p L ¼ VI sin 2ot ● Instantaneous power for a capacitive load: p C ¼ ÀVI sin 2ot The diagrams of instantaneous power for inductive and capacitive loads are illustrated in Figure 9.16. As seen from (9.3) and waveforms in Figure 9.16, both the instantaneous powers of inductive and capacitive loads are sinusoidal quantities with a double frequency 2o. They have an average value of zero over a complete cycle since the positive and negative waveforms will cancel each other out. When instantaneous power is positive, the component stores energy; when instantaneous power is 0 v i t p R VI Figure 9.15 The waveform of instantaneous power for a R load 278 Understandable electric circuits negative, the component releases energy. Therefore, the inductor and capacitor do not absorb power, they convert or transfer energy between the source and elements. This also indicates that the inductor and capacitor are energy storage elements. Instantaneous power for R, L and C components ● p R ¼ VI À VI cos 2ot ● p L ¼ VI sin 2ot ● p C ¼ ÀVI sin 2ot 9.3.2 Active power P (or average power) The active power is also known as average power, which is the product of the RMS voltage and RMS current in an AC circuit. It is actually the average power dis- sipation on the resistive load, i.e. the average power within one period of time (one full cycle) for a sinusoidal power waveform in an AC circuit. The active power is also called true or real power since the power is really dissipated by the load resistor, and it can be converted to useful energy such as heat 0 v t p L i Store / Release / Store / Release (a) 0 v t p C i Release / Store / Release / Store (b) Figure 9.16 The waveforms of instantaneous power for L and C loads Methods of AC circuit analysis 279 or light energy, etc. Electric stoves and lamps are examples of this kind of resistive load. The instantaneous power always varies with time and is difficult to measure, so it is not very practical to use. Since it is the actual power dissipated in the load, average or active power P is used more often in AC sinusoidal circuits. Average power is easy to measure by an AC power metre (an instrument to measure AC power) in an AC circuit. Average power is the average value of instantaneous power in one period of time. It can be obtained from integrating for instantaneous power in one period of time. Note: If you haven’t learned calculus, then just keep in mind that P ¼ VI cos f is the equation for average power, and skip the following mathematical derivation process. P ¼ 1 T Z T 0 pðtÞdt ¼ 1 T Z T 0 ½VI cos fð1 À cos 2otÞ þ VI sin fsin 2otdot P ¼ 1 T VI cos fðotÞ T 0 þ VI sin f 1 T Z T 0 sin 2 ot dot ¼ VI cos f ð9:4Þ where f is a constant, and ot is a variable, so the first part of the integration is a constant VI cos f. The integration of the second part is zero (integrating for sine function), since the average power value for a cosine function in one period of time is zero. Therefore, active or average power P is a constant. It consists of the product of RMS values of voltage and current VI and cos f where cos f is called power factor and it will be discussed at the end of this section. When active power P 4 0, the element absorbs power; when active power P 5 0, the element releases power. ● When f ¼ 0 , the voltage and current are in phase, the circuit is a purely resistive circuit, and P R ¼ VI cos 0 ¼ VI ð,cos 0 ¼ 1Þ Therefore, P R ¼ VI ¼ I 2 R ¼ V 2 R or P R ¼ VI ¼ V m ffiffiffi 2 p I m ffiffiffi 2 p ¼ 1 2 V m I m ● When f ¼ 90 , the voltage leads the current by 908, the circuit is a purely inductive circuit, and P L ¼ VI cos 90 ¼ 0 ð,cos 90 ¼ 0Þ, i.e. P L ¼ 0. 280 Understandable electric circuits ● When f ¼ À90 , the current leads the voltage by 908, the circuit is a purely capacitive circuit, and P C ¼ VI cosðÀ90 Þ ¼ 0 ½,cosðÀ90 Þ ¼ 0, i.e. P C ¼ 0. Active power P (or average power, real power and true power) The active power is the average value of the instantaneous power that is actually dissipated by the load. P ¼ VI cos f When f ¼ 0 P R ¼ VI ¼ 1 2 V m I m ¼ I 2 R ¼ V 2 R When f ¼ 90 P L ¼ 0 When f ¼ À90 P C ¼ 0 9.3.3 Reactive power Q Since the effect of charging/discharging in a capacitor C and storing/releasing energy from an inductor L is that energy is only exchanged or transferred back and forth between the source and the component and will not do any real work for the load. So the average power dissipated on the load is zero. The reactive power Q can describe the maximum velocity of energy transferring between the source and the storage element L or C. The first part in (9.4) is active or average power. The integration of the second part of (9.4) is zero, and that is the reactive power. While energy is converting between the source and energy store elements, the load will do not do any actual work, and average power dissipated on the load will be zero. Also, because the physical meaning of the reactive power is the maximum velocity of energy con- version between the energy storing element and the source, the peak value of the second part is reactive power, denoted as Q. It can be expressed mathematically as Q ¼ VI sin f, and measured in volt-amperes reactive (Var). ● When f ¼ 0 , the circuit is a purely resistive circuit: Q R ¼ VI sin 0 ¼ 0 ð,sin 0 ¼ 0Þ Quantity Quantity symbol Unit Unit symbol Active power P Watt W Methods of AC circuit analysis 281 ● When f ¼ 90 , the circuit is a purely inductive circuit: Q L ¼ VI sin 90 ¼ VI ð,sin 90 ¼ 1Þ Substituting V ¼ IX L or I ¼ V X L into Q L gives Q L ¼ VI ¼ I 2 X L ¼ V 2 X L ● When f ¼ À90 , the circuit is a purely capacitive circuit: Q C ¼ VI sinðÀ90 Þ ¼ ÀVI ½sinðÀ90 Þ ¼ À1 Substituting V ¼ IX C or I ¼ V X C into Q C gives Q C ¼ ÀVI ¼ ÀI 2 X C ¼ À V 2 X C Since Q L is positive (Q L 40) and Q C is negative (Q C 50), the inductor absorbs (consumes) reactive power, and the capacitor produces (releases) reactive power. Reactive power Q Q is the maximum velocity of energy conversion between the source and energy storing element. Q ¼ VI sin f When f ¼ 0 Q R ¼ 0 When f ¼ 90 Q L ¼ VI ¼ I 2 X L ¼ V 2 X L When f ¼ À90 Q C ¼ ÀVI ¼ ÀI 2 X C ¼ À V 2 X C 9.3.4 Apparent power S When the voltage V across a load produces a current I in the circuit of Figure 9.17, the power produced in the load is the product of voltage and current VI. If the load Z includes both the resistor and storage element inductor or capacitor, then VI will be neither a purely active power nor a purely reactive power. Since VI is the expression of the power equation, it is called apparent power. Apparent power is Quantity Quantity symbol Unit Unit symbol Reactive power Q Volt-amperes reactive Var 282 Understandable electric circuits the maximum average power rating that a source can provide to the load or max- imum capacity of an AC source and is denoted as S. The mathematical expression of apparent power is the product of the source current and voltage, i.e. S ¼ IV and is measured in VA (volt-amperes). Substituting I ¼ V/Z or V ¼ IZ into apparent power S equation gives: S ¼ I 2 Z ¼ V 2 Z Usually the power listed on the nameplates of electrical equipment is the apparent power. Apparent power S S is the maximum average power rating that a source can provide to an AC circuit. S ¼ IV ¼ I 2 Z ¼ V 2 =Z where S represents apparent power, measured in VA. Different types of power in AC circuits are summarized in Table 9.2. Z V + - I Figure 9.17 Apparent power Table 9.2 Powers in AC circuits Power General expression R L C Instantaneous power p ¼ VI cos f (1 – cos 2ot) þ VI sin fsin 2ot p R ¼ VI – VIcos 2ot p L ¼ VIsin 2ot p L ¼ –VIsin 2ot Active power P ¼ VI cos f P R ¼ VI ¼ 1/2(V m I m ) ¼ I 2 R ¼ V 2 /R P L ¼ 0 P C ¼ 0 Reactive power Q ¼ VI sin f Q R ¼ 0 Q L ¼ VI ¼ I 2 X L ¼ V 2 /X L Q C ¼ –VI ¼ –I 2 X C ¼ V 2 /X C Apparent power S ¼ VI ¼ I 2 Z ¼ V 2 /Z Methods of AC circuit analysis 283 9.3.5 Power triangle We have discussed three different powers in AC circuits, the active power, reactive power and apparent power. Now the question is what are the relationships between these three powers. These three powers are actually related to one another in a right triangle, is called the power triangle, and can be derived as follows. For a series resistor, inductor and capacitor circuit, if the circuit is more induc- tive ðX ¼ X L À X C > 0; f > 0Þ, then the impedance triangle, voltage triangle and current triangle can be illustrated as shown in Figure 9.18(a–c) (refer to section 9.1). If we multiply all quantities on each side of the voltage triangle by the current I, it will yield VI = S, IV X = Q and V R I = P and this can be illustrated as a power triangle as shown in Figure 9.18(d). If the circuit load is more capacitive (X C > X L , f < 0), the circuit triangles will be opposite to the inductive circuit triangles as shown in Figure 9.19. The impedance triangle indicates that it has an angle f between resistance R and impedance Z of the circuit. It is called the impedance angle; f is also in the power triangle. Later on, we’ll introduce the power factor cos f, and f is also called the power factor angle. f Z R X f • V • I V R • • V X f • I I R • I X • (a) (b) (c) f S Q P (d) Figure 9.18 Circuit triangles for a more inductive circuit f R X Z S Q P f f f • V V R • X V • I R • I X • • I (a) (b) (c) (d) Figure 9.19 Circuit triangles for a more capacitive circuit 284 Understandable electric circuits The relationship between different powers in the power triangle can be obtained from the Pythagoras’ theorem, i.e. S ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi P 2 þ Q 2 p If expressed by complex numbers, it will give: _ S ¼ P þ jQ This is known as the phasor power. The phasor apparent power can also be expressed as _ S ¼ _ V _ I ¼ _ I 2 Z ¼ _ V 2 Z The impedance angle f can be obtained from circuit triangles (either inductive or capacitive circuit) and can be expressed as f ¼ tan À1 Q P ¼ tan À1 X R ¼ tan À1 _ V X _ V R ¼ tan À1 _ I X _ I R Active power P and reactive power Q can be expressed with the impedance angle f and obtained from the power triangle in Figures 9.18 or 9.19 as: P ¼ S cos f and Q ¼ S sin f Power triangle f S P Q P ¼ S cos f; Q ¼ S sin f; S ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi P 2 þ Q 2 p ; Impedance angle: f ¼ tan À1 Q P ¼ tan À1 X R ¼ tan À1 ð _ V X = _ V R Þ ¼ tan À1 ð _ I X = _ I R Þ Phasor power: _ S ¼ _ V _ I ¼ _ I 2 Z ¼ _ V 2 =Z; _ S ¼ P þ jQ 9.3.6 Power factor (PF) The ratio of active power P and apparent power S is called the power factor PF, and represented by cos f. It also can be obtained from the power triangle as Methods of AC circuit analysis 285 PF ¼ P S cos f For a purely resistive circuit ðf ¼ 0 Þ, the reactive power Q is zero, so the apparent power S is equal to the active power P, i.e. S ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi P 2 þ Q 2 p ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi P 2 þ 0 2 p ¼ P and the power factor is 1, i.e. cos f ¼ P=S ¼ P=P ¼ 1 This is the maximum value for the power factor cos f. For a purely reactive load ðf ¼ Æ90 Þ, active power P in the circuit is zero, so the power factor is also zero, i.e. cos f ¼ P=S ¼ 0=S ¼ 0. Therefore, the range of the power factor cos f is between 0 and 1, and the impedance angle f is between 08 and +908. The power factor is an important factor in circuit analysis. The circuit source will produce active power P to the load, and the amount of the active power P can be determined by the power factor cos f. This is indicated in the equation of P ¼ S cos f. If the power factor cos f of the load is the maximum value of 1, the active power produced by the source is the maximum capacity of the source, and all the energy supplied by the source will be consumed by the load (P ¼ S, ,cos f ¼ 1 ). If the power factor cos f decreases, the active power P produced by the source will also decrease accordingly ðP # ¼ S cos f #Þ. So increasing the power factor can increase the real power in a circuit. But how to increase the power factor of a circuit? A method called power-factor correction can be used. This method can increase the power factor and does not affect the load voltage and current. Since most of the loads of the electrical sys- tems are inductive loads (such as the loads that are driven by a motor), an inductive load in parallel with a capacitor (Figure 9.20(b)) can increase the power factor of the load. (c) (b) C R L I S P Qʹ Q C fʹ f Q R (a) L Figure 9.20 Increasing the power factor 286 Understandable electric circuits The power triangle in Figure 9.20(c) indicates that when a capacitor C is in parallel with the inductive load, the reactive power Q in the circuit will be reduced to Q 0 (Q 0 ¼ Q 7 Q C ). Therefore, the impedance angle will reduce from f to f 0 , and the power factor cos f will increase to cos f 0 . Since f#! cos f", for instance cos 30 ¼ 0:866 is > cos 60 ¼ 0:5, the total current I will also decrease, since I #¼ P=ðV cos f "Þ ðP ¼ S cos f ¼ VI cos fÞ. This can reduce the source current and line power loss (I 2 R). This is why increasing the power factor has a significant meaning. Power factor (cos f) ● cos f ¼ P=S ð0 cos f 1; cos f À dimensionlessÞ: ● When cos f ¼ 1: All energy supplied by the source is consumed by the load. ● Power-factor correction: An inductive load in parallel with a capacitor can increase cos f. 9.3.7 Total power When calculating the total power in a complicated series–parallel circuit, determine the active power P and reactive power Q in each branch first, and the sum of all the active powers is the total active power P T . The difference between Q LT and Q CT is the total reactive power Q T . Q LT is the sum of all reactive powers for the inductors and Q CT is the sum of all reactive powers for the capacitors. The total apparent power S can be determined by using Q T and P T using the Pythagoras’ theorem, i.e. . S T ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi P 2 T þ Q 2 T q The total power factor can be determined by using the total active and reactive power, i.e. PF T ¼ cos f T ¼ P T S T Total power ● Total active power: P T ¼ P 1 þ P 2 þ Á Á Á þ P n ● Total reactive power: Q T ¼ Q LT À Q CT ¼ ðQ L 1 þ Q L 2 þ Á Á ÁÞ À ðQ C 1 þ Q C 2 þ Á Á ÁÞ where Q LT is the total reactive power for inductors and Q CT the total reactive power for capacitors. Methods of AC circuit analysis 287 ● Total apparent power: S T ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi P 2 T þ Q 2 T p ● Total power factor: PF T ¼ cos f T ¼ P T =S T Example 9.5: Determine the total power factor cos f in the circuit of Figure 9.21 and plot the power triangle for this circuit. Solution: ● Total power factor PF T ¼ cos f T ¼ P T ? S T ? ; S T ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi P T ? 2 þ Q T ? 2 p (the symbol ‘?’ indicates an unknown). Total active power: P T ¼ P 1 þ P 2 þ P 3 ¼ ð10 þ 30 þ20ÞW ¼ 60 W Total reactive power: Q T ¼ Q LT À Q CT ¼ ð70 À 15ÞVar ¼ 55 Var Total apparent power: S T ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi P 2 T þ Q 2 T p ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 60 2 þ 55 2 p ¼ 81:39 VA Total power factor: PF T ¼ cos f T ¼ P T S T ¼ 60 W 81:39 VA % 0:74 ● Impedance angle: f ¼ cos À1 f T ¼ cos À1 0:74 % 42:3 ● The power triangle is shown in Figure 9.22. Example 9.6: Determine the following values in the circuit shown in Figure 9.23: ● the total power P T , Q T and S T for the circuit ● power factor cos f ● power triangle ● source current I ● the capacitance C needed to increase the power factor cos f to 0.87 ● the source current I 0 after increasing the power factor Z 1 Z 2 Z 3 E = 10 V ∠ 0 ° P 1 = 10 W P 2 = 30 W Q C = 15 Var P 3 = 20 W Q L = 70 Var Figure 9.21 Circuit for Example 9.5 42.3 ° = f P T = 60 W Q T = 55 Var S T = 81.39 VA Figure 9.22 The power triangle for Example 9.5 288 Understandable electric circuits Solution: (a) Lamp: P 1 ¼ 5  100 W ¼ 500 W Heating: P 2 ¼ I 2 R ¼ ð10 AÞ 2 ð5 OÞ ¼ 500 W Q 2 ¼ I 2 X L ¼ ð10 AÞ 2 ð5 OÞ ¼ 500 Var Electric stove: P 3 ¼ 6 kW ¼ 6 000 W; f ¼ cos À1 0:75 % 41:4 Q 3 ¼ P 3 tan f ¼ ð6 000 WÞðtan 41:4 Þ % 5 290 Var; f ¼ tan À1 ðQ=PÞ ð Þ Total power: P T ¼ P 1 þ P 2 þ P 3 ¼ ð500 þ 500 þ 6 000ÞW ¼ 7 000 W Q T ¼ Q 1 þ Q 2 þ Q 3 ¼ 0 þ 500 Var þ 5 290 Var ¼ 5 790 Var S T ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi P T þ Q T p ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 7 000 2 þ 5 790 2 p % 9 084:3 VA (b) Power factor PF T ¼ cos f T ¼ P T S T ¼ 7 000 9 084:3 % 0:77 (c) Power triangle (as shown in Figure 9.24): f ¼ cos À1 0:77 % 39:7 (d) Source current I: ,S T ¼ EI ; I ¼ S T E ¼ 9 084:3 VA 110 V % 82:6 A Therefore, _ I ¼ 82:6ff À 39:7 A (Voltage leads current or current lags voltage in the inductive load, so f ¼ À39:7 .) (e) The capacitance C that needs to increase the power factor to 0.87 can be determined by the following way: C ¼ 1 2pfX C ? ) Q C ¼ À V 2 X C ) X C ¼ À V 2 Q C ? ) X C ¼ 1 2pfC Q T ¼ Q C þ Q 0 T ? ) Q 0 T ¼ P T tan f 0 ? ) f 0 ¼ cos À1 0:87 (as shown in Figure 9.25) 10 A 60 kW 0.75 PF C • • • • • • f = 60 Hz 5 lamps 100 W/each R = 5Ω X L = 5Ω I E =110 V∠0 ° • Lamps Heating Electric stove Figure 9.23 Circuit for Example 9.6 P T =7000 W Q T =5790 Var S T =9084.3 VA f = 39.7 ° Figure 9.24 Power triangle for Example 9.6 Methods of AC circuit analysis 289 Therefore, to increase the power factor to 0.87, the power factor angle should be reduced to f 0 ¼ cos À1 0:87 % 29:5 . The reactive power can be determined from the above expression as Q 0 T ¼ P T tan f 0 ¼ 7 000 tan 29:5 % 3 960 Var The new power factor angle f 0 is shown in Figure 9.25. Q C can be obtained from Figure 9.25: Q C ¼ Q T À Q 0 T ¼ 5 790 À 3 960 ¼ 1 830 Var X C ¼ ÀV 2 Q C ¼ À E 2 Q C ¼ À110 2 V 1 830 Var % 6:61 O (The voltage across X L and R is equal to E.) Therefore C ¼ 1 2pfX C ¼ 1 2pð60 HzÞð6:61 OÞ % 0:0004 F ¼ 400 mF That is the capacitance C needed to increase the power factor to 0.87 should be 400 µF. (f) The source current I 0 after increasing the power factor can be determined by the following expression: P ¼ S cos f ¼ IE cos f So, I 0 ¼ P T E cos f 0 ¼ 7000 W 110 Vcos 29:5 % 73:1 A Comparing with the original source current I = 82.6 A from step (d), after a capacitor is in parallel and the power factor is increased, the source current is I 0 = 73.1 A. So the source current can decrease 9.5 A (I 7I 0 = 82.6 A 7 73.1 A = 9.5 A). This can reduce the line power loss (I 2 R) and utilize the capacity of the source more efficiently. 9.4 Methods of analysing AC circuits All analysis methods that we have learned for analysing DC circuits with one or two more sources can also be used for analysing AC circuits, such as the branch P Q c Q T S T =9084.3 VA Q′ T =3960 Var 29.5 ° = f Figure 9.25 New power factor angle 290 Understandable electric circuits current analysis, mesh analysis, node voltage analysis, superposition theorem, Thevenin’s and Norton’s theorems, etc. But the phasor form will be used to represent the circuit quantities. Since these analysis methods have been discussed in detail for DC circuits (chapters 4 and 5), some examples will be presented to use these methods in AC circuits or networks. Reviewing chapters 4 and 5 before reading the following contents is highly recommended. 9.4.1 Mesh current analysis The procedure for applying the mesh current analysis method in an AC circuit: 1. Identify each mesh and label the reference directions for each mesh current clockwise. 2. Apply KVL around each mesh of the circuit, and the numbers of KVL equa- tions should be equal to the numbers of mesh (windowpanes). Sign each self- impedance voltage as positive and each mutual-impedance voltage as negative in KVL equations. ● Self-impedance: An impedance that only has one mesh current flowing through it. ● Mutual-impedance: An impedance that is located on the boundary of two meshes and has two mesh currents flowing through it. 3. Solve the simultaneous equations resulting from step 2, and determine each mesh current. Note: ● Convert the current source to the voltage source first in the circuit, if there is any. ● If the circuit has a current source, the source current will be the same with the mesh current, so the number of KVL equations can be reduced. The procedure for applying the mesh current analysis method in an AC circuit is demonstrated in the following example. Example 9.7: Use the mesh current analysis method to determine the mesh current I 1 in the circuit of Figure 9.26. Solution: Convert the current source to the voltage source (connect R and X to Z) as shown in Figure 9.26(b). There, _ E 2 ¼ _ IR 3 ¼ ð2:5 Aff0 Þð4 OÞ ¼ 10 Vff 0 : 1. Label all the reference directions for each mesh current _ I 1 and _ I 2 (clockwise), as shown in Figure 9.26(b). 2. Write KVL around each mesh (windowpane), and the number of KVL is equal to the number of meshes (there are two meshes in Figure 9.26(b)). Sign each self-impedance voltage as positive, and each mutual-impedance voltage as negative in KVL ðS _ V ¼ S _ EÞ. Mesh 1: ðZ 1 þ Z 2 Þ _ I 1 À Z 2 _ I 2 ¼ À _ E 1 Mesh 2: ÀZ 2 _ I 1 þ ðZ 2 þ Z 3 Þ _ I 2 ¼ À _ E 2 Methods of AC circuit analysis 291 Substitute the following values of Z 1 , Z 2 and Z 3 into the above equations: Z 1 = (6 7 j6)O, Z 2 = 2 O and Z 3 = (4 þ j8)O so, ð8 À j6Þ _ I 1 À 2 _ I 2 ¼ À10 V À2 _ I 1 þ ð6 þ j8Þ _ I 2 ¼ À10 V 3. Solve the simultaneous equations resulting from step 2 using the determinant method, and determine the mesh current _ I 1 : _ I 1 ¼ À10ff 0 À2 À10ff 0 6 þ j8 8 À j6 À2 À2 6 þ j8 % 1:18ff À 151:9 A 9.4.2 Node voltage analysis The following is the procedure for applying the node analysis method in an AC circuit. 1. Label the circuit: ● Label all the nodes and choose one of them to be the reference node. ● Assign an arbitrary reference direction for each branch current (this step can be skipped if using the inspection method). (a) (b) R 1 = 6 Ω X C = 6 Ω R 2 = 2 Ω R 3 = 4 Ω X L = 8 Ω – + E 1 = 10 V∠ 0 ° • R 1 = 6 Ω X C = 6 Ω R 3 = 4 Ω X L = 8 Ω Z 1 R 2 = 2 Ω Z 3 Z 2 + – – + I 2 • • 1 I E 2 = 10V∠ 0 ° • E 1 = 10 V∠ 0 ° • I = 2.5 A ∠ 0 ° • Figure 9.26 Circuits for Example 9.7 292 Understandable electric circuits 2. Apply KCL to all n71 nodes except for the reference node (n is the number of nodes). ● Method 1: Write KCL equations and apply Ohm’s law to the equations (assign a positive sign (þ) to the self-impedance voltage and entering node current, and negative sign (7) for the mutual-impedance voltage and exiting node current). ● Method 2: Convert voltage sources to current sources and write KCL equations using the inspection method. 3. Solve the simultaneous equations and determine each nodal voltage. The procedure for applying the node voltage analysis method in an AC circuit is demonstrated in the following example. Example 9.8: Write node equations for the circuit in Figure 9.27(a). 1. Label nodes a, b, c and d, and choose ground d to be the reference node as shown in Figure 9.27(a). 2. Convert two voltage sources to current sources from Figure 9.27(a) to Figure 9.27(b), and write KCL equations to n 71 ¼ 4 71 ¼ 3 nodes by inspection (method 2). KCL equations are shown in Table 9.3. 3. Three equations can solve three unknowns that are node voltages. 9.4.3 Superposition theorem The following is the procedure for applying the superposition theorem in an AC circuit: 1. Turn off all power sources except one, i.e. replace the voltage source with the short circuit (placing a jump wire), and replace the current source with an open circuit. Redraw the original circuit with a single source. 2. Analyse and calculate this circuit by using the single source method, and repeat steps 1 and 2 for the other power sources in the circuit. 3. Determine the total contribution by calculating the algebraic sum of all con- tributions due to single sources. a b c d –j1Ω + – a b c d 5j A = = + – –j1 Ω –j1 Ω –j1 Ω 1 Ω 1 Ω 5 Ω 2.5 Ω 5 Ω 2.5 Ω j1 Ω j1 Ω 2 V ∠0 º 5 ∠0 º 1∠–90 º 5 V ∠0 º i 1 –2j A = = 2 ∠0 º 1∠90 º i 2 (a) (b) Figure 9.27 Circuits for Example 9.8 Methods of AC circuit analysis 293 _ V a _ V b _ V c _ I N o d e a 1 5 þ 1 À j 1 þ 1 j 1 _ V a 7 ð 1 = j 1 Þ _ V b 7 ð 1 = À j 1 Þ _ V c ¼ À 2 j N o d e b ð 1 = À j 1 Þ _ V a þ 1 2 : 5 þ 1 j 1 þ 1 À j 1 _ V b 7 ð 1 = À j 1 Þ _ V c ¼ 5 j À ð À 2 j Þ N o d e c ð À 1 = À j 1 Þ _ V a 7 ð 1 = À j 1 Þ _ V b þ 1 1 þ 1 À j 1 þ 1 À j 1 _ V c ¼ À 5 j A f t e r s i m p l i f y i n g 0 : 2 _ V a þ j _ V b 7 j _ V c ¼ À j 2 j _ V a þ 0 : 4 _ V b 7 j _ V c ¼ j 7 À j _ V a 7 j _ V b þ ð 1 þ 2 j Þ _ V c ¼ À j 5 T a b l e 9 . 3 K V L E q u a t i o n s f o r e x a m p l e 9 . 8 (The result should be positive when the reference polarity of the unknown in the single source circuit is the same with the original circuit, otherwise it should be negative.) The procedure for applying the superposition theorem in an AC circuit is demonstrated in the following example. Example 9.9: Determine _ V c in circuit as shown in Figure 9.28(a) by using the superposition theorem. Solution: 1. Choose _ E to apply to the circuit first, and use an open circuit to replace the current source _ I as shown in Figure 9.28(b), and calculate _ V 0 c : _ V 0 C ¼ _ E Z 2 Z 1 þ Z 2 ¼ À10Vff0 7:5 Off À 90 10 O À j7:5 O ¼ À75ff À 90 12:5ff À 36:87 V ¼ À6ff À 53:13 V 2. When the current source _ I is applied to the circuit only and the voltage source _ E is replaced by a jump wire, the circuit is as shown in Figure 9.28(c). Cal- culate _ V 00 c in Figure 9.28(c): _ V 00 C ¼ _ IðZ 1 ==Z 2 Þ Z 1 ==Z 2 ¼ Z 1 Z 2 Z 1 þ Z 2 ¼ 10ðÀj7:5Þ 10 À j7:5 O % 75ff À 90 12:5ff À 36:87 O ¼ 6ff À 53:13 O _ V 00 C ¼ _ IðZ 1 ==Z 2 Þ ¼ ð2ff0 AÞð6ff À 53:13 OÞ ¼ 12ff À 53:13 V R = 10 Ω Z 1 Z 2 X C = 7.5 Ω I = 2A ∠0 ° V C - + - + • • E = 10 V ∠0 ° • + R = 10 Ω Z 1 Z 2 X C = 7.5 Ω = + - - + ' V C • • E = 10 V ∠0 ° R = 10 Ω Z 1 Z 2 X C = 7.5 Ω + - ' ' V C • • I = 2A ∠0 ° (a) (b) (c) Figure 9.28 Circuits for Example 9.9 Methods of AC circuit analysis 295 3. Calculate the sum of voltages _ V 0 C and _ V 00 C : _ V C ¼ _ V 0 C þ _ V 00 C ¼ À6ff À 53:13 V þ 12ff À 53:13 V ¼½À6 cosðÀ53:13 Þ À 6j sinðÀ53:13 Þ þ 12 cosðÀ53:13 Þ þ 12j sinðÀ53:13 ÞV %½À3:6 þ j4:8 þ 7:2 À j9:6V ¼ ð3:6 À j4:8ÞV ¼ 6ff À 53:13 V 9.4.4 Thevenin’s and Norton’s theorems The following is the procedure for applying Thevenin’s and Norton’s theorems in an AC circuit. 1. Open and remove the load branch (or any unknown current or voltage branch) in the network, and mark the letter a and b on the two terminals. 2. Determine the equivalent impedance Z TH or Z N : It should be equal to the equivalent impedance when you look at it from the a and b terminals when all sources are turned off or equal to zero. (A voltage source should be replaced by a short circuit, and a current source should be replaced by an open circuit.) i.e. Z TH ¼ Z N ¼ Z ab 3. ● Determine Thevenin’s equivalent voltage V TH : It equals the open circuit vol- tage from the original linear two-terminal network of a and b, i.e. V TH ¼ V ab . ● Determine Norton’s equivalent current I N : It equals the short circuit current for the original linear two-terminal network of a and b, i.e. I N ¼ I sc . 4. Plot Thevenin’s or Norton’s equivalent circuits, and connect the load (or unknown current or voltage branch) to a and b terminals of the equivalent circuit. Then the load voltage or current can be calculated. The procedure for applying Thevenin’s and Norton’s theorems method in an AC circuit is demonstrated in the following example. Example 9.10: Determine the current _ I L in the load branch of Figure 9.29(a) by using Thevenin’s theorem, and use Norton’s theorem to check the answer. Solution: 1. Open the load branch and remove Z L , and label a and b on the terminals of the load branch as shown in Figure 9.29(b). 2. Determine Thevenin’s equivalent impedance Z TH (the voltage source _ E is replaced by a short circuit) in Figure 9.29(b): Z TH ¼Z ab ¼ Z 3 þ Z 4 þ Z 1 ==Z 2 Z TH ¼ 1 À j1 þ Àj2:5ð2:5 þ j2:5Þ Àj2:5 þ ð2:5 þ j2:5Þ O ¼ 1 À j1 þ 6:25 À j6:25 2:5 O ¼ð1 À j1 þ 2:5 À j2:5ÞO ¼ ð3:5 À j3:5ÞO % 4:95ff À 45 O 296 Understandable electric circuits 3. Determine Thevenin’s equivalent voltage _ V TH by using Figure 9.29(c) to calculate the open circuit voltage across terminals a and b: _ V TH ¼ _ V ab ¼ _ V cd Since _ I ¼ 0 for Z 3 and Z 4 in Figure 9.29(c), voltages across Z 3 and Z 4 are also zero ; _ V TH ¼ _ V cd ¼ _ E Z 2 Z 1 þ Z 2 ¼ 5ff0 V 2:5 þ j2:5 Àj2:5 þ ð2:5 þ j2:5Þ O %5ff0 ð1:414ff45 Þ % 7:07ff45 V 4. Plot Thevenin’s equivalent circuit as shown in Figure 9.29(d). Connect the load Z L to a and b terminals of the equivalent circuit and calculate the load current _ I L . 0 ° ∠ - + 5V = • E • -j2.5 Ω j2.5 Ω 2.5 Ω Z L = (1.5 + j3.5) Ω 1 Ω -j Ω I L (a) a b -j2.5 Ω j2.5 Ω 2.5 Ω R TH = Z ab Z 1 Z 4 Z 2 Z 3 1 Ω -j Ω a b c d • • j2.5 Ω 2.5 Ω -j2.5 Ω Z 1 Z 2 Z 4 Z 3 1 Ω -j Ω E = 5 V∠0 ° V TH (b) (c) Figure 9.29 Circuits for Example 9.10(a)–(c) Methods of AC circuit analysis 297 _ I L ¼ _ V TH Z TH þ Z L ¼ 7:07ff45 V ð3:5 À j3:5ÞO þ ð1:5 þ j3:5ÞO ¼ 7:07ff45 V 5 O % 1:4ff45 A 5. Determine Norton’s equivalent circuit in Figure 9.29(a) as seen by Z L . ● Norton’s equivalent impedance Z N : Z N ¼ Z TH ¼ 3.5 7 j3.5 = 4.95 ff7458 ● Norton’s equivalent current I N : It is equal to the short circuit current for the original two-terminal circuit of a and b (as shown in Figure 9.29(e, I). _ I N ¼ _ I SC ¼ _ I Z 2 Z 2 þ ðZ 3 þ Z 4 Þ (the current-divider rule). There, _ I ¼ _ E Z 1 þ Z 2 ==ðZ 3 þ Z 4 Þ ¼ 5ff0 V Àj2:5 þ ð2:5 þ j2:5Þð1 À j1Þ ð2:5 þ j2:5Þ þ ð1 À j1Þ O ¼ 5ff0 V ðÀ j2:5 þ ðj5Þ=ð3:5 þ j1:5ÞÞO % 1:54ff68:2 A Therefore, _ I N ¼ _ I Z 2 Z 2 þ ðZ 3 þ Z 4 Þ ¼ 1:54ff68:2 A ð2:5 þ j2:5ÞO ½2:5 þ j2:5 þ ð1 À j1ÞO % 1:43ff90 A 6. Use Norton’s theorem to check the load current _ I L : Determine the load current _ I L on the terminals of a and b in Figure 9.29(e, II) by using Norton’s equivalent circuit. _ I L ¼ _ I N Z N Z N þ Z L ¼ 1:43ff90 A 4:95ff À 45 O ½ð3:5 À j3:5Þ þ ð1:5 þ j3:5ÞO ¼ 1:43ff90 A 4:95ff À 45 5 % 1:4ff45 A Therefore, _ I L is the same by Norton’s theorem as the method by using Thevenin’s theorem (checked). - + a b • • Z TH = 4.95 Ω∠-45 ° Z L = (1.5 + j3.5) Ω I L V TH = 7.07 ∠45 ° V Figure 9.29(d) Thevenin’s equivalent circuit for Example 9.10 298 Understandable electric circuits a b Z N = 4.95 ∠-45 0 Ω Z L = (1.5 + j3.5) Ω I N = 1.43 ∠90.1 0 A • I L • a b • • • -j Ω 1 Ω -j 2.5 Ω j 2.5 Ω 2.5 Ω E = 5 ∠0 ° V Z 1 Z 4 I N Z 2 Z 3 I (I) (II) Figure 9.29(e) Norton’s equivalent circuit for Example 9.10 Summary Impedance and admittance X L ¼ oL; X C ¼ 1 oC ; j ¼ 1 Àj Component Impedance Z ¼ _ V= _ I Admittance Y ¼ 1/Z Conductance and susceptance R Z R ¼ R Y R ¼ G Conductance: G ¼ 1/R L Z L ¼ j X L Y L ¼ 7jB L Inductive susceptance: B L ¼ 1/X L C Z C ¼ 7j X C Y C ¼ jB C Capacitive susceptance: B C ¼ 1/X C Z ¼ z fff ¼ R þ jX Y ¼ y ff f y ¼ G þ jB Reactance: X ¼ X L À X C and susceptance: B ¼ B C À B L z ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi R 2 þ X 2 p y ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi G 2 þ B 2 p f ¼ tan À1 X R f y ¼ tan À1 B G Methods of AC circuit analysis 299 ● Impedance, voltage, current and power triangles ● For a more inductive circuit: V • X I • V • V • R I • R I • X I • f f f • f S Q P R X Z ● For a more capacitive circuit: f f f f V • X I • X I • R V • R I • V • X R Z P S Q Impedance angle: f ¼ tan À1 X R ¼ tan À1 _ V X _ VR ¼ tan À1 _ I X _ IR ¼ tan À1 Q P ● Characteristics of impedance and admittance: ● The inductive load: X 40 (X L 4X C ), f 40, B 50 (B L 4B C ), f y 50 ● The capacitive load: X 50 (X C 4X L ), f 50, B 40 (B C 4B L ), f y 40 ● The resistive load: X ¼ 0 (X C ¼ X L ), f ¼ 0, B ¼ 0 (B L ¼ B C ), f y ¼ 0. Impedances in series and parallel ● Impedances in series: Z eq ¼ Z 1 þ Z 2 þ Á Á Á þ Z n ● Impedances in parallel: Z eq ¼ 1 ð1=Z 1 Þþð1=Z 2 ÞþÁÁÁþð1=Z n Þ ¼Z 1 ==Z 2 ==ÁÁÁ==Z n Z eq ¼ 1 Y eq Y eq ¼Y 1 þY 2 þÁÁÁþY n Two impedances in parallel: Z eq ¼ Z 1 Z 2 Z 1 þ Z 2 ¼ Z 1 ==Z 2 ● Voltage-divider rule for impedance: _ V 1 ¼ Z 1 Z 1 þ Z 2 _ E _ V 2 ¼ Z 2 Z 1 þ Z 2 _ E Current-divider rule for impedance: _ I 1 ¼ Z 2 Z 1 þ Z 2 _ I T _ I 2 ¼ Z 1 Z 1 þ Z 2 _ I T ● The phasor forms of KVL and KCL: S _ I ¼ 0 _ I in ¼ _ I out S _ V ¼ 0 _ V 1 þ _ V 2 þ Á Á Á þ _ V n ¼ _ E 300 Understandable electric circuits Power of AC circuits ● Power: P ¼ S cos f, Q ¼ S sin f, S ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi P 2 þ Q 2 p Power General form R L C Instantaneous power p ¼ ui ¼VI cos f (1 – cos 2ot) þ VI sin fsin 2ot p R ¼ VI – VI cos 2ot p L ¼ VI sin 2ot p C ¼ –VIsin 2ot Active power P ¼ VI cos f P R ¼ VI ¼ 1/2V m I m ¼ I 2 R ¼ V 2 /R P L ¼ 0 P C ¼ 0 Reactive power Q ¼ VI sin f Q R ¼ 0 Q L ¼ VI ¼ I 2 X L ¼ V 2 /X L Q C ¼ –VI ¼ –I 2 X C ¼ V 2 /X C Apparent power S ¼ VI ¼ I 2 Z ¼ V 2 /Z Quantity Quantity Symbol Unit Unit symbol Instantaneous Power p Watt W Active Power P Watt W Reactive power Q Volt-amperes reactive Var Apparent power S Volt-Amperes VA f S P Q Phasor power: _ S ¼ R þ jQ ¼ _ V _ I ¼ _ I 2 Z ¼ _ V 2 =Z ● Power factor ● Power factor: PF ¼ cos f ¼ P S ð0 cos f 1Þ ● Power-factor correction: A capacitor in parallel with the inductive load can increase the power factor (the power factor angle f# ! cos f") cos f" ! line current _ I # _ I #¼ P Vcosf " ! line power loss ð _ I 2 RÞ # ! utilize capacity of the source more efficiently. ● Total power ● Total active power: P T ¼ P 1 þ P 2 þ Á Á Á þ P n ● Total reactive power: Q T ¼ Q LT À Q CT ¼ ðQ L1 þ Q L2 þ Á Á Á Þ À ðQ C1 þ Q C2 þ Á Á Á Þ (Q LT is the total reactive power of inductors, and Q CT is the total reactive power of capacitors.) ● Total apparent power: S T ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi P 2 T þ Q 2 T p S T ¼ S 1 þ S 2 þ Á Á Á þ S n ● Total power factor: PF T ¼ cos f T ¼ P T S T Methods of AC circuit analysis 301 ● Analysis methods for AC sinusoidal circuits All analysis methods that are used to analyse DC circuits with one or two more sources can also be used to analyse AC circuits. Experiment 9: Sinusoidal AC circuits Objectives ● To become familiar with the operation of an oscilloscope for measuring the sinusoidal AC voltage. ● To become familiar with the operation of an oscilloscope for measuring the phase difference of two waveforms. ● To verify theoretical calculations of the AC series–parallel circuits through experiment. Background information ● X L ¼ oL ¼ 2pfL; X C ¼ 1 oC ¼ 1 2pfC ● Z R ¼ R, Z L ¼ jX L , Z C ¼ 7jX C ● Use the oscilloscope to measure the phase difference f (with dual-channel CH I and CH II): Example L9.1: There are two sinusoidal waveforms A and B with complete cycles of 2p(3608) as shown in Figure L9.1. If these two waveforms appear on the screen of the oscilloscope and occupy six horizontal grids, determine the phase difference of waveforms A and B in Figure L9.1. Solution: Each grid is 608 ð360 =6 grids ¼ 60 =gridsÞ. Since there is one grid difference between waveform A and B as shown in Figure L9.1, the phase differ- ence of waveforms A and B is 608. (If the distance between A and B is 0.5 grids, the phase difference will be 308, i.e. 0.5 6608 ¼ 308.) Use the oscilloscope to measure the current in the inductive or capacitive branch (indirect measurement): Connect a small resistor, called a sensing resistor, in series with the inductor or the capacitor. Measure the voltage across the sensing resistor, and then calculate the branch current using Ohm’s law. Since the sensing resistance is very small, its impact on the circuit measurement and calculation may be negligible. 0 B A 2π π Figure L9.1 Phase difference 302 Understandable electric circuits Equipment and components ● Multimeter ● Breadboard ● Function generator ● Oscilloscope ● Z meter or LCZ meter ● Switch ● Resistors:15 O (two) and 510 O ● Inductor: 1.1 mH ● Capacitor: 3 600 pF Procedure 1. Connect a circuit as shown in Figure L9.2 on the breadboard. Use the multi- meter (ohmmeter function) and Z meter or LCZ meter to measure the values of the resistor, inductor and capacitor, and record in Table L9.1. 2. Use the measured R, L and C values to calculate X C , X L and Z eq of the circuit shown in Figure L9.2 (at frequency of 80 kHz) and record the results in Table L9.2. E = 3 sin ωt V f = 80 kHz L = 1.1 mH R = 510 Ω C =3 600 pF A B C Figure L9.2 Experiment circuit Table L9.1 R R L R C L C Nominal value 510 O 15 O 15 O 1.1 mH 3 600 pF Measured value Table L9.2 X C X L Z eq * Formula for calculation Calculated value * Refer to chapter 9, section 9.2 Methods of AC circuit analysis 303 3. Adjust the frequency of the function generator to 80 kHz. Connect the oscillo- scope probe CH I to the point A in the circuit of Figure L9.2, connect the probe ground of the oscilloscope to the ground of the function generator, and then measure the output sinusoidal voltage of the function generator. Resistor branch 4. Adjust the output voltage of the function generator to 6 Vpeak–peak value (E p7p ¼ 6 V). Connect the oscilloscope probe CH II to point B in the circuit of Figure L9.2 (choose DUAL channel coupling for the oscilloscope), then measure the voltage across the resistor V R (peak value) and record in Table L9.3. 5. Use the measured V R value to calculate the current I R in the resistive branch (Ohm’s law) and record it in Table L9.3. Then use the oscilloscope to observe and determine the phase difference f R of resistor voltage V R relative to source voltage E, and record in Table L9.3. Inductor branch 6. Connect a 15 O resistor R L (sensing resistor) to the inductive branch as shown in Figure L9.3. 7. Connect the oscilloscope probe CH II to the point D in the circuit of Figure L9.3, and measure the peak voltage on resistor R L and record the result as V RL in Table L9.3. 8. Use the measured V RL to calculate the current I L in the inductor branch (Ohm’s law) and record it in Table L9.3. Then use the oscilloscope to observe and determine the phase difference o L of V RL relative to the source voltage E. Record the result in Table L9.3. Note: The oscilloscope probe CH I is still connected to point A of the circuit in Figure L9.3. Capacitor branch 9. Connect a 15 o resistor R C to the circuit as shown in Figure L9.4. 10. Connect the oscilloscope probe CH II to point E of the circuit in Figure L9.4, measure the peak voltage on resistor R C and record the measurement as V RC in Table L9.3. Table L9.3 Resistive branch Inductive branch Capacitive branch Parameter V R I R f R V RL I L f L V RC I C f C Measured value 304 Understandable electric circuits 11. Use the measured V RC to calculate the capacitor current I C (Ohm’s law) and record it in Table L9.3. Then use the oscilloscope to observe and determine the phase difference c C of capacitor voltage V RC relative to source voltage E and record in Table L9.3. Phasor form 12. Calculate the branch currents _ I R ; _ I L and _ I C in the circuit of Figure L9.2 in phasor form (use peak values) and record in Table L9.4. 13. Convert the measured values I R , I L and I C in Table L9.3 to the phasor form and record in Table L9.4 (as the measured value). Compare the measured values and calculated values. Are there any significant differences? If so, explain the reasons. Conclusion Write your conclusions below: Table L9.4 _ I R _ I L _ I C Formula for calculations Calculated value Measured value Methods of AC circuit analysis 305 Chapter 10 RLC circuits and resonance Objectives After completing this chapter, you will be able to: ● understand concepts and characteristics of series and parallel resonance ● determine the following quantities of series and parallel resonant circuits: resonant frequency, resonant current, resonant voltage, resonant impe- dance, bandwidth and quality factor ● plot the frequency response curves of current, voltage and impedance for series and parallel resonant circuits ● understand characteristics of the selectivity in series and parallel of reso- nant circuits ● understand the actual parallel resonant circuits ● understand the applications of the resonant circuits The resonant phenomena that will be introduced in this chapter have a wide range of applications in electrical and electronic circuits, particularly in communication systems. Resonant circuits are simple combinations of induc- tors, capacitors, resistors and a power source. However, since the capacitor or inductor voltage/current in a resonant circuit could be much higher than the source voltage or current, a small input signal can produce a large output signal when resonance appears in a circuit. This is why the resonant circuit is one of the most important circuits in electronic communication systems. Resonance may also damage the circuit elements if it is not used properly. So it is very important to analyse and study resonant phenomena and to know its pros and cons. 10.1 Series resonance 10.1.1 Introduction Resonance may occur in a series resistor, inductor and capacitor (RLC) circuit, as shown in Figure 10.1, when the capacitor reactance X C is equal to the inductor reactance X L . When the magnitudes of inductive reactance X L and capacitive reactance X C are equal (X L ¼ X C ), or when reactance X is zero (X ¼ X L 7X C ¼ 0), the equivalent or total circuit impedance Z is equal to the resistance R, i.e. _ Z ¼ R þjðX L ÀX C Þ ¼ R Under the above condition, resonance will occur in the RLC series circuit. That is, when resonance occurs in a series RLC circuit, the energy of the reactive components in the circuit will compensate each other (X L ¼ X C ), and the equivalent impedance of the series RLC circuit will be the lowest (Z ¼ R). This is the characteristic of the series resonant circuit. Series resonance X L ¼ X C , X ¼ 0, Z ¼ R 10.1.2 Frequency of series resonance The angular frequency of the series resonant circuit can be obtained from X L ¼ X C or oL ¼ 1 oC Solving for o gives o r ¼ 1= ffiffiffiffiffiffiffi LC p . (The subnotation ‘r’ stands for resonance.) Since o ¼ 2pf , solving for f gives the series resonant frequency as f r ¼ 1 2p ffiffiffiffiffiffiffi LC p This is a very important equation for the series resonance. The resonant fre- quency f r is dependent on the circuit elements inductor (L) and capacitor (C), meaning that it may produce or remove resonance by adjusting the inductance L or capacitance C in the RLC series circuit. Frequency of series resonance Resonant frequency: f r ¼ 1=2p ffiffiffiffiffiffiffi LC p Resonant angular frequency: o r ¼ 1= ffiffiffiffiffiffiffi LC p R C L V S V R V L V C I • • • • • Figure 10.1 An RLC series circuit 308 Understandable electric circuits 10.1.3 Impedance of series resonance As previously mentioned, when series resonance occurs, the circuit’s equivalent impedance is at the minimum (Z ¼ R). This is illustrated in Figure 10.2, which is the response curve of the impedance Z versus frequency f in the series reso- nant circuit. When f ¼ f r , the impedance Z is at the lowest point on the curve. 10.1.4 Current of series resonance When resonance occurs in a series RLC circuit, the impedance of the circuit is equal to the resistance (Z ¼ R), and the resonant current will be _ I ¼ _ V Z ¼ _ V R Therefore, when f ¼ f r , X L ¼ X C , the only opposition to the flow of the current is resistance R, i.e. the impedance is minimum and current is maximum in a series resonant circuit. Figure 10.3 illustrates the response curve of current I versus frequency f in the series resonant circuit, and the current is at the highest point on the curve when f ¼ f r . I and Z of series resonance ● Impedance is minimum at series resonance: Z ¼ R ● Current is maximum at series resonance: _ I ¼ _ V= _ Z ¼ _ V=R Z f f r 0 Figure 10.2 The response curve of Z vs. f for series resonance f f r 0 I • Figure 10.3 The response curve of I vs. f for series resonance RLC circuits and resonance 309 10.1.5 Phasor diagram of series resonance An RLC series resonant circuit is equivalent to a purely resistive circuit since Z ¼ R. The capacitor and inductor voltages in the series resonant circuit are equal in magnitude but are opposite in phase since X L ¼ X C , _ V L ¼ jX L _ I L and _ V C ¼ ÀjX C _ I C . The resistor voltage is equal to the source voltage ( _ V R ¼ _ E) since X ¼ 0 when series resonance occurs. Thus, the current _ I and source voltage _ E are also in phase (since _ V R and _ I in phase), and the phase difference between _ E and _ I is zero (f ¼ 0). A phasor diagram of the series resonant circuit is illustrated in Figure 10.4. Phasor relationship of series resonance ● _ V L and _ V C are equal in magnitude but opposite in phase. ● _ I and _ E are in phase, and f ¼ 0. 10.1.6 Response curves of X L , X C and Z versus f The response curves of the inductive reactance X L , capacitive reactance X C and impedance Z versus frequency f are illustrated in Figure 10.5. + j + 0 V R = E V L V C I • • • • • Figure 10.4 Phasor diagram of the series resonant circuit f X L > X C 0 f r ( Z = R) X L < X C X L X C Z Figure 10.5 Response curves of X L , X C and Z vs. f 310 Understandable electric circuits ● X L and f are directly proportional ðX L ¼ 2pfLÞ, i.e. as frequency increases, and X L increases. ● X C and f are inversely proportional ðX C ¼ 1=2pfCÞ, i.e. as frequency increases X C decreases. ● When frequency f is zero in the circuit, X L ¼0, X C and Z approach infinite, ðZ ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi R 2 þð2pfL Àð1=2pfCÞÞ 2 q ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi R 2 þð0 À1Þ 2 q ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi R þ1 2 p )1Þ. The response curves of X L , X C and Z versus f show that when the circuit frequency is below the resonant frequency f r , the inductive reactance X L is lower than the capacitive reactance X C and the circuit appears capacitive. When the circuit frequency is above the resonant frequency f r , the inductive reactance X L is higher than the capacitive reactance X C , and the circuit appears more inductive. Only when the circuit frequency is equal to the resonant fre- quency f r , the resonance occurs in the circuit. Impedance Z is equal to the circuit resistance R and has a minimum value, and the circuit appears purely resistive. These can be summarized as follows: Characteristics of series resonance ● When f 5 f r , X L 5 X C : the circuit is more capacitive. ● When f 4 f r , X L 4 X C : the circuit is more inductive. ● When f ¼ f r , X L ¼ X C , I ¼ I max , Z ¼ Z min ¼ R: the circuit is purely resistive and resonance occurs. 10.1.7 Phase response of series resonance The phase response of the series resonant circuit can also be obtained from Figure 10.5. ● When the frequency of the circuit is above the resonant frequency f r , the circuit is more inductive X L 4 X C , voltage leads current, and the phase difference is between zero and positive 908 (0 f 908). ● When the frequency of the circuit is below the resonant frequency f r , the circuit is more capacitive X L 5 X C , the voltage lags current, and the phase difference is between zero and negative 908 (7908 f 0). ● When the frequency of the circuit is equal to the resonant frequency f r , X L ¼ X C , Z ¼R, voltage and current are in phase, and the phase difference is zero (f ¼ 0). The phase response of the series resonant circuit can be illustrated in Figure 10.6. The following characteristics of the series resonant circuit can also be obtained from Figure 10.6. ● When the frequency increases from the resonant frequency f r to infinite, the phase angle approaches positive 908. RLC circuits and resonance 311 ● When the frequency decreases from the resonant frequency f r to zero, the phase angle approaches negative 908. The expression of the phase angle is f ¼ tan À1 X R ¼ tan À1 X L ÀX C R ¼ tan À1 2pfL Àð1=2pfCÞ R Phase response of series resonance ● When f !1; f !þ90 ● When f !0; f !À90 10.1.8 Quality factor There is an important parameter known as quality factor in the resonant cir- cuit, which is denoted as Q. The quality factor is defined as the ratio of stored energy and consumed energy in physics and engineering, so it is the ratio of the reactive power stored by an inductor or a capacitor and average power con- sumed by a resistor in a resonant circuit, i.e. Quality factor Q ¼ Reactive power=average power ð10:1Þ The quality factor can be used to measure the energy that a circuit stores and consumes. The lower the energy consumption of a resistor (power loss) in a circuit, the higher the quality factor, and the better the quality of the resonant circuit. If substituting the equations of the reactive power and average power into the quality factor equation (10.1), the quality factor of the series resonance will be obtained as follows: Q ¼ I 2 X L I 2 R ¼ X L R ¼ oL R where R is the total or equivalent resistance in the series circuit. f 0 f r +90° –90° f Figure 10.6 Phase response of the series resonant circuit 312 Understandable electric circuits Similarly, the quality factor Q can also be expressed by the capacitive reactance and the resistance as Q ¼ X C R ¼ 1 oCR The quality factor can be used to judge the quality of an inductor (or coil). A coil always contains a certain amount of winding resistance R w , which is the resistance of the wire in the winding. The quality factor Q for a coil is defined as the ratio of the inductive reactance and the winding resistance, i.e. Q ¼ X L R w The lower the winding resistance R w of a coil, the higher the quality of the coil. Note: Both the quality factor and reactive power are denoted by the letter Q, so be careful not to confuse them. The quality factor is a dimensionless para- meter, and the unit of reactive power is Var, which can be used to distinguish between these two quantities. Quality factor Q ● Quality factor: is the ratio of the reactive power and average power. ● Quality factor of the series resonance: Q ¼ X L =R ¼ X C =R. ● Quality factor of the coil: Q ¼ X L =R w . (The lower the R w , the higher the quality of the coil.) 10.1.9 Voltage of series resonant Multiplying current _ I for both the denominator and nominator of the quality factor equation Q ¼ X L =R, gives Q ¼ X L R ¼ _ IX L _ IR ¼ _ V L _ E Similarly, for Q ¼ X C =R Q ¼ _ IX C _ IR ¼ _ V C _ E Therefore, when the resonance occurs in an RLC series circuit: _ V L ¼ _ V C ¼ _ EQ ð10:2Þ The quality factor Q is always greater than 1, so the inductor or capacitor voltage may greatly exceed the source voltage in a series resonant circuit, as can RLC circuits and resonance 313 be seen from the equation (10.2). This means that a lower input voltage may produce a higher output voltage; therefore, the series resonance is also known as the voltage resonance. That is one of the reasons that series resonant circuits have a wide range of applications. When choosing the storage elements L and C for a series resonant circuit, the affordability of their maximum voltage should be taken into account, or else the high resonant voltage may damage circuit components. The concept of circuit resonance is similar to resonance in physics, which is defined as a system oscillating at maximum amplitude at resonant frequency, so a small input force can produce a large output vibration. There are many examples of resonance in daily life, such as pushing a child in a playground swing to the resonant frequency, which makes the swing go higher and higher to the maximum amplitude with very little effort. Another example is bouncing a basketball. Once the ball is bounced to the resonant frequency, it will yield a smooth response, and the ball will reach maximum height since a small force produces a large vibration. Resonance may also cause damage. For example, a legend says that when a team of soldiers walking a uniform pace passed through a bridge, the bridge collapsed since the uniform pace reached resonant frequency that resulted in a small force producing a large vibration. Voltage of series resonance ● A lower input voltage may produce a higher output voltage. ● Inductor or capacitor voltage may greatly exceed the supply voltage _ V L ¼ _ V C ¼ _ EQ ðQ > 1Þ Example 10.1: A series resonant circuit is shown in Figure 10.7. Determine the total equivalent impedance, quality factor, and inductor voltage of this circuit. Solution: Z ¼ R T ¼ R þR w ¼ ð2 þ0:5ÞO ¼ 2:5ff0 O f r ¼ 1 2p ffiffiffiffiffiffiffi LC p ¼ 1 2p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð2:5 mHÞð0:25 mFÞ p % 6 366 Hz X L ¼ 2pfL ¼ 2pð6 366Þð2:5 mHÞ % 100 O Q ¼ X L R T ¼ 100 O 2:5 O ¼ 40 _ V L ¼ jX L _ I ¼ _ E Z jX L ¼ 2:5ff0 V 2:5ff0 O Â100ff90 O ¼ 100ff90 V 314 Understandable electric circuits This example shows that the inductor voltage of the series resonant circuit is indeed greater than the supply voltage. ð _ V L ¼ 100ff90 VÞ > ð _ E ¼ 2:5ff0 VÞ 10.2 Bandwidth and selectivity 10.2.1 The bandwidth of series resonance When an RLC series circuit is in resonance, its impedance will reach the minimum value and the current will reach the maximum value. The curve of the current versus frequency of the series resonant circuit is illustrated in Figure 10.8. As displayed in the diagram, the current reaches the maximum value I max as the frequency closes in on the resonant frequency f r , which is located at the centrer of the curve. The characteristic of the resonant circuit can be expressed in terms of its bandwidth (BW) or pass-band. The bandwidth of the resonant circuit is the difference between two frequency points f 2 and f 1 . BW¼ f 2 Àf 1 where f 2 and f 1 are called critical, cutoff or half-power frequencies. R = 2 Ω L = 2.5 mH R w = 0.5 Ω C = 0.25 μF + – E = 2.5 V ∠0° • Figure 10.7 Circuit for Example 10.1 I f f r 0 f 1 f 2 I max I max 0.707 P max 2 1 BW Figure 10.8 Bandwidth of a series resonant circuit RLC circuits and resonance 315 As shown in Figure 10.8, the bandwidth of the resonant circuit is a fre- quency range between f 2 and f 1 when current I is equivalent to 0.707 of its maximum value I max , or 70.7 per cent of the maximum value of the curve. The reason to define the term ‘half-power’ frequency can be derived from the following mathematical process. The power delivered by the source at the points f 1 and f 2 can be determined from the power formula P ¼ I 2 R P f 1 ¼ I 2 f 1 R ¼ ð0:707I max Þ 2 R % 0:5I 2 max R ¼ 0:5P max and P f 2 ¼ I 2 f 2 R ¼ ð0:707I max Þ 2 R % 0:5I 2 max R ¼ 0:5P max Therefore, at both points f 2 and f 1 , the circuit power is only one-half of the maximum power that it is produced by the source at resonance frequency f r , where f 2 is the upper critical frequency, and f 1 is the lower critical frequency. Bandwidth (pass-band) ● Bandwidth (BW ¼ f 2 7 f 1 ) is the range of frequencies at I ¼ 0.707I max . ● f 2 and f 1 are critical, or cutoff or half-power frequencies: P f 1;2 ¼ 0:5P max . 10.2.2 The selectivity of series resonance Figure 10.8 shows the frequency range between f 2 and f 1 at which the current is near its maximum value, and the series resonant circuits can select frequencies in this range. The curve in the Figure 10.8 is called the selectivity curve of the series resonant circuit. The selectivity is the capability of a series resonant cir- cuit to choose the maximum current that is closer to the resonant frequency f r . The steeper the selectivity curve, the faster the signal attenuation (reducing), the higher the maximum current value, and the better the circuit selectivity. For example, in Figure 10.9, the selectivity curve 1 has a bandwidth of BW 1 and a maximum current I 1max , which has a better current selectivity than selectivity curve 2 or 3. This means that the series resonant circuit of curve 1 has a higher quality and can be expressed as Q ¼ f r =BW, where Q is the quality factor of the series resonant circuit. The bandwidth BW is an important characteristic for the resonant circuit. A series resonant circuit with a narrower bandwidth has a better current selectivity. A series resonant circuit with a wider bandwidth is good for passing the signals. Sometimes in order to take into account both aspects, the selec- tivity curve between narrow and wide curves may be chosen, such as the selectivity curve 2 (BW 2 ) in Figure 10.9. Therefore, the concepts of bandwidth and selectivity may apply to different circuits with different design choices. 316 Understandable electric circuits Selectivity of the series resonance The capability of the circuit to choose the maximum current I max closer to the resonant frequency f r . 10.2.3 The quality factor and selectivity The quality factor Q in the resonant circuit is a measure of the quality and selectivity of a resonant circuit. The higher the Q value, the narrower the bandwidth ðBW#¼ f r =Q "Þ, the higher the maximum current, and the better the current selectivity, which is desirable in many applications. As mentioned earlier, the disadvantage of the narrower BW or higher Q is that the ability for passing signals in the circuit will be reduced. The lower the Q value, the wider the bandwidth ðBW"¼ f r =Q #Þ, and the better the ability to pass signals; however, it will have a poor current selectivity. This is the reason that Q is denoted as the ‘quality’ factor since it represents the quality of a resonant circuit. Example 10.2: Given a series resonant circuit shown in Figure 10.10(a), deter- mine the bandwidth BW and current _ I (phasor-domain) of this circuit with three resistors that are 50, 100, and 200 O, and plot their selectivity curves. Solution: When R ¼ 50 O: Q ¼ X L R ¼ 2 kO 50 O ¼ 40; BW 1 ¼ f r Q ¼ 50 Hz 40 ¼ 1:25 Hz _ I ¼ _ E Z ¼ _ E R ¼ 10ff0 V 50 O ¼ 0:2ff0 A I f 0 BW 1 BW 3 BW 2 I 1max I 3max I 2max 3 2 1 Figure 10.9 Selectivity of a series resonant circuit RLC circuits and resonance 317 When R ¼ 100 O: Q ¼ X L R ¼ 2 kO 100 O ¼ 20; BW 2 ¼ f r Q ¼ 50 Hz 20 ¼ 2:5 Hz _ I ¼ _ E Z ¼ _ E R ¼ 10ff0 V 100 O ¼ 0:1ff0 A When R ¼ 200 O: Q ¼ X L R ¼ 2 kO 200 O ¼ 10; BW 3 ¼ f r Q ¼ 50 Hz 10 ¼ 5 Hz _ I ¼ _ E Z ¼ _ E R ¼ 10ff0 V 200 O ¼ 0:05ff0 A Example 10.2 shows that the selectivity curve of a resonant circuit depends greatly upon the amount of resistance in the circuit. When resistance R in a series resonant circuit has a smaller value, the selectivity curve of the circuit is X L = 2 kΩ R = 50 Ω + E = 10 V∠0° X C = 2 kΩ f r = 50 Hz 0 BW 1 = 1.25 Hz BW 3 = 5 Hz BW 2 = 2.5 Hz f r = 50 Hz 0.2 A 0.1 A 0.05 A I f (a) (b) • – • Figure 10.10 (a) Circuit for Example 10.2; (b) Selectivity curve for Example 10.2 318 Understandable electric circuits steeper, the quality factor Q has a higher value, the current at the resonant frequency f r has a higher value, and the selectivity is better. However, the pass- band (BW) of the circuit with a smaller R value is narrower, and the ability to pass signal will be poor. Quality factor and selectivity ● Quality factor: a measure of the quality and selectivity of a resonant circuit Q ¼ f r =BW: ● Q "¼ X L =R#)BW#¼ f r =Q": the steeper the selectivity curve, the better the current selectivity, but the worse the ability to pass signals. ● Q #¼ X L =R")BW"¼ f r =Q#: the flatter the selectivity curve, the worse the current selectivity, but the better the ability to pass signals. The analysis method of the series resonant circuit can also be applied to the parallel resonant circuits. 10.2.3.1 Series resonance summary 10.3 Parallel resonance 10.3.1 Introduction Resonance may occur in a parallel resistor, inductor and capacitor (RLC) circuit, as shown in Figure 10.11, when the circuit inductive susceptance B L is equal to the capacitive susceptance B C . jwL jwC 1 I T I R I L I C E R • • • • • Figure 10.11 A parallel RLC circuit Characteristics Series resonance Condition of resonance X L ¼ X C , X ¼ 0, Z ¼ R Resonant frequency f r ¼ 1= 2p ffiffiffiffiffiffiffi LC p À Á Impedance Z ¼ R minimum (admittance Y maximum) Current _ I T ¼ _ V=R (maximum) Bandwidth BW ¼ f 2 7 f 1 ¼ f r /Q Quality factor Q ¼ X L =R ¼ X C =R Relationship of voltage and quality factor _ V L ¼ _ V C ¼ _ EQ RLC circuits and resonance 319 The analysis method of the parallel resonance is similar to series resonance. When the magnitudes of the capacitive susceptance B C and the inductive sus- ceptance B L are equal (B C ¼ B L ), or when the susceptance B is zero (B ¼ B C 7 B L ¼ 0), the circuit input equivalent (total) admittance Y is equal to the circuit conductance G, i.e. Y ¼ G þj B ¼ G Under the above condition, resonance will occur in the RLC parallel circuit. That is, when the resonance occurs in an RLC parallel circuit, the energy of the reactive components in the circuit will compensate each other (B C ¼ B L ), and the equivalent admittance of the parallel RLC circuit is at the lowest (Y ¼ G). This is the characteristic of the parallel resonant circuit. Parallel resonance B C ¼ B L , B ¼ 0, Y¼ G 10.3.2 Frequency of parallel resonance The angular frequency of the parallel resonant circuit can be obtained from Y ¼ G þjðB C ÀB L Þ ¼ 1 R þj oC À 1 oL From B C ¼ B L or oC ¼ 1=oL, solving for o gives o r ¼ 1= ffiffiffiffiffiffiffi LC p . Since o ¼ 2pf , the parallel resonant frequency is f r ¼ 1=ð2p ffiffiffiffiffiffiffi LC p Þ. You may have noticed that the parallel resonant angular frequency o r and resonant frequency f r are the same with those in the series resonant circuit. The resonant frequency f r is dependent on the circuit elements L and C, meaning that if adjusting the inductance L or capacitance C in the RLC parallel circuit, resonance may be produced or removed. Frequency of parallel resonance ● Resonant frequency: f r ¼ 1= 2p ffiffiffiffiffiffiffi LC p À Á ● Resonant angular frequency: o r ¼ 1= ffiffiffiffiffiffiffi LC p 10.3.3 Admittance of parallel resonance As previously mentioned, when parallel resonance occurs, the equivalent admittance Y of the circuit is at the minimum (Y ¼ G), B ¼ B C ÀB L ¼ 0 so the circuit equivalent impedance Z is at a maximumðZ "¼ 1=Y #Þ. This is shown in Figure 10.12, which is the response curve of the impedance Z versus the fre- quency f in the parallel resonant circuit. When f ¼ f r , the impedance Z is at the highest point on the curve and this is opposite to the series resonance. 320 Understandable electric circuits 10.3.4 Current of parallel resonance When resonance appears in a parallel RLC circuit, the impedance of the circuit is equal to the resistance (Z ¼ R), and the total current in the circuit will be _ I T ¼ _ V Z ¼ _ V R Therefore, when f ¼ f r , B C ¼ B L , Y ¼ G, the admittance Y is at the minimum , Y ¼ G þ j (B C 7 B L ), the impedance Z is at the maximum, and the current is at the minimum in the parallel resonant circuit, _ I T #¼ _ V=Z"¼ _ V=R. Figure 10.13 illustrates the response curve of current I versus frequency f in the parallel resonant circuit and current is at the lowest point on the curve when f ¼ f r . This is also opposite of series resonance. I and Z of parallel resonance ● Impedance is maximum at parallel resonance: Z ¼ R ðB ¼ 0; Y ¼ GÞ; ,Y ¼ G þjðB C ÀB L Þ ● Current is minimum at parallel resonance: _ I ¼ _ V= _ Z ¼ _ V=R Z f f r 0 Figure 10.12 The response curve of Z vs. f for parallel resonance f f r 0 I • Figure 10.13 The response curve of I vs. f for parallel resonance RLC circuits and resonance 321 10.3.5 Phasor diagram of parallel resonance An RLC parallel resonant circuit is equivalent to a purely resistive circuit since Y ¼ G and Z ¼ R. The capacitor and inductor branch currents in the parallel resonant circuit are equal in magnitude but opposite in phase, since B L ¼ B C (B L ¼ 1/X L ; B C ¼ 1/X C ) and _ I L ¼ V L jX L ¼ Àj V L X L ; _ I C ¼ V C ÀjX C ¼ j V C X C þj ¼ À1 j i.e. _ I L ¼ À _ I C . The resistor voltage is equal to the source voltage ð _ V R ¼ _ EÞ in the parallel resonant circuit of Figure 10.13. The total current ð _ I T Þ and the source voltage _ E are in phase (since _ V R and _ E are in phase) and the phase difference between _ E and _ I T is zero, i.e. the admittance angle f y ¼ 0. A phasor diagram of the parallel resonant circuit is illustrated in Figure 10.14. Phasor relationship of parallel resonance ● _ I L and _ I C are equal in magnitude but opposite in phase, _ I L ¼ À _ I C . ● _ I T and _ E are in phase, and f y ¼ 0. 0 I T I C I L E = V R • • • • • Figure 10.14 Phasor diagram of the parallel resonant circuit 10.3.6 Quality factor From the previous description, we know that the quality factor is the ratio of the reactive power stored by an inductor or a capacitor and the average power dissipated by a resistor in a circuit, i.e. quality factor Q ¼ reactive power/ average power. If we substitute the expressions of the reactive power and average power in Figure 10.11 into the quality factor equation, the quality factor of a parallel resonance will be obtained as follows: Q ¼ _ E 2 =X L _ E 2 =R ¼ R X L ð, _ V L ¼ _ V R ¼ _ EÞ 322 Understandable electric circuits Similarly, the quality factor Q can be expressed by the capacitive reactance and the resistance as Q ¼ R X C The quality factor of a parallel resonant circuit is inverted with the series resonant circuit. Recall the quality factor of a series resonant circuit: Q ¼ X L R ¼ X C R Quality factor Q Quality factor of the parallel resonance: Q ¼ R=X L ¼ R=X C 10.3.7 Current of parallel resonance Dividing the voltage _ E for both the denominator and the numerator of the quality factor equation Q ¼ _ E 2 =X L _ E 2 =R gives Q ¼ _ E=X L _ E=R ¼ _ I L _ I T Similarly For Q ¼ _ E 2 =X C _ E 2 =R ; Q ¼ _ E=X C _ E=R ¼ _ I C _ I T Therefore, when resonance occurs in an RLC parallel circuit _ I L ¼ _ I C ¼ _ I T Q ð10:3Þ Usually the quality factor Q is always greater than 1, the inductor or capacitor branch current may greatly exceed the total supply current in a parallel reso- nant circuit, and this can be seen from the equation (10.3). This means that a lower input current may produce a higher output current, and therefore the parallel resonance is also known as current resonance. It is similar to series resonance, and there are benefits and disadvantages to using parallel reso- nance. When choosing the storage elements L and C for a parallel resonant RLC circuits and resonance 323 circuit, the affordability of their maximum current should be taken into account, or else the higher resonant current may damage circuit components. Current of parallel resonance ● A lower input current may produce a higher output current. ● The inductor or capacitor current may greatly exceed the supply current _ I L ¼ _ I C ¼ _ I T Q ðQ > 1Þ 10.3.8 Bandwidth of parallel resonance The characteristic of the parallel resonant circuit can be expressed in terms of its bandwidth (BW) or pass-band. Recall that BW¼ f 2 Àf 1 or BW¼ f r =Q. The bandwidth of the parallel resonant circuit is illustrated in Figure 10.15. When the RLC parallel circuit is in resonance, its current reaches the minimum value. The BW of the parallel resonant circuit is a frequency range between the critical or cutoff frequencies f 2 and f 1 , when the current is equivalent to 0.707 of its maximum value I max , or 70.7 per cent of the maximum value of the curve. 10.3.8.1 Parallel resonance summary I f f r 0 f 1 f 2 BW I max 0.707I max Figure 10.15 The bandwidth of the parallel resonance Characteristics Parallel resonance Conditions of resonance B L ¼ B C , B ¼ 0, Y ¼ G Resonant frequency f r ¼ 1= 2p ffiffiffiffiffiffiffi LC p À Á Impedance Z ¼ R maximum (admittance Y minimum) Current _ I T ¼ _ V=R (minimum) Bandwidth BW¼ f 2 Àf 1 ¼ f r =Q Quality factor Q ¼ R=X L ¼ R=X C Relationship of current and quality factor _ I L ¼ _ I C ¼ _ I T Q 324 Understandable electric circuits 10.4 The practical parallel resonant circuit In practical electrical or electronic system applications, the parallel resonant circuit is usually is formed by an inductor (coil) in parallel with a capacitor. Since a practical coil always has internal resistance (winding resistance), an actual parallel resonant circuit will look like the one illustrated in Figure 10.16. C L R I L I C V S I • • • • Figure 10.16 A practical parallel circuit 10.4.1 Resonant admittance The input equivalent admittance of the practical parallel circuit shown in Figure 10.16 is Y ¼ 1 R þjX L þj 1 X C Multiplying (R 7 jX L ) to the numerator and denominator of the first term in the above expression gives Y ¼ R R 2 þX 2 L Àj X L R 2 þX 2 L þj 1 X C or Y ¼ R R 2 þX 2 L þj 1 X C À X L R 2 þX 2 L ð10:4Þ The parallel resonance occurs when the circuit admittance Y is equal to the circuit conductance G (Y ¼ G), so when the resonance occurs for the practical parallel circuit in Figure 10.16, the resonant admittance should be Y ¼ G ¼ R R 2 þX 2 L ð,Y ¼ G þjBÞ RLC circuits and resonance 325 10.4.2 Resonant frequency According to the parallel resonant conditions, resonance occurs when the capacitive susceptance B C is equal to the inductive susceptance B L , i.e. B C ¼ B L Thus, (10.4) gives X L R 2 þX 2 L ¼ 1 X C or oL R 2 þðoLÞ 2 ¼ oC ð10:5Þ The resonance frequency and angular frequency for the circuit in Figure 10.16 can be obtained from (10.5) as follows: Resonance angular frequency: o r ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi L ÀCR 2 L 2 C r ¼ 1 ffiffiffiffiffiffiffi LC p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 À CR 2 L r Resonance frequency: f r ¼ 1 2p ffiffiffiffiffiffiffi LC p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 À CR 2 L r ðo ¼ 2pf Þ This indicates that resonance will occur in the circuit of Figure 10.16 only when 1 À CR 2 L > 0; 1 > CR 2 L ; R 2 < L C ; or R < ffiffiffiffi L C r If 1 ÀðCR 2 =LÞ < 0; resonance will not occur. Practical parallel resonance ● Resonant admittance: Y ¼ R R 2 þX 2 L ● Resonant angular frequency: o r ¼ 1 ffiffiffiffiffiffiffi LC p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 À CR 2 L r ● Resonant frequency: f r ¼ 1 2p ffiffiffiffiffiffiffi LC p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 À CR 2 L r ● When 1 ÀðCR 2 =LÞ > 0, or R < ffiffiffiffiffiffiffiffiffi L=C p resonance occurs. 326 Understandable electric circuits 10.4.3 Applications of the resonance As previously mentioned, resonant circuits are used in a wide range of appli- cations in communication systems, such as filters, tuners, etc. The purpose of resonant circuits are the same 7 to select a specific frequency (resonant fre- quency f r ) and reject all others, or select signals over a specific frequency range that is between the cutoff frequencies f 1 and f 2 . The key circuit of a communication system is a tuned amplifier (tuning circuit). Figure 10.17 is a simplified radio tuning circuit for a radio circuit. The combination of a practical parallel resonant circuit and an amplifier can select the appropriate signal to be amplified. The input signals in the radio tuner circuit have a wide frequency range, because there are many different radio signals from different radio stations. When adjusting the capacitance of the variable capacitor in the practical par- allel resonant circuit (i.e. adjusting the switch of the radio channel), the circuit resonant frequency f r will consequently change. Once f r matches the desired input signal frequency with the highest input impedance, the desired input signal will be passed, and this is the only signal that will be amplified. After it is amplified by the amplifier in the circuit, this signal of the corresponding station can be clearly heard. Figure 10.18 is a simplified series radio tuning circuit. It is similar to the parallel tuning circuit. When adjusting the capacitance of the variable capa- citor in the series resonant circuit, the circuit resonant frequency f r will change. Once f r matches the desired input signal frequency with the highest current, the desired input signal will be passed and amplified. C L R W f f r 0 Z • Figure 10.17 A simplified parallel radio tuner f f r 0 L R C I V S • • Figure 10.18 A simplified series radio tuner RLC circuits and resonance 327 Summary Series/parallel resonance Characteristics Series resonance Parallel resonance Conditions of resonance X L ¼ X C , X ¼ 0, Z ¼ R B L ¼ B C , B ¼ 0, Y ¼ G Phasor relationship ● _ V L and _ V C are equal in magnitude but opposite in phase. ● _ I and _ E in phase f ¼ 0. ● _ I L and _ I C are equal in magnitude but opposite in phase. ● _ I T and _ E in phase j y ¼ 0. Phasor diagram +j + 0 • V R = E • V L • V C • • I 0 • I T • I C • I L E = V R • • Resonant frequency f r ¼ 1=ð2p ffiffiffiffiffiffiffi LC p Þ f r ¼ 1=2p ffiffiffiffiffiffiffi LC p Impedance Z ¼ R minimum (admittance Y maximum) Z f f r 0 Z ¼ R maximum (admittance Y minimum) Z f f r 0 Current _ I ¼ _ V=R (maximum) f f r 0 • I _ I T ¼ _ V=R (minimum) f f r 0 • I Bandwidth BW¼ f 2 Àf 1 ¼ f r =Q BW¼ f 2 Àf 1 ¼ f r =Q Quality factor Q ¼ X L =R ¼ X C =R or Q ¼ f r =BW Q ¼ R=X L ¼ R=X C or Q ¼ f r =BW Relationship of voltage/current and Q _ V L ¼ _ V C ¼ _ EQ _ I L ¼ _ I C ¼ _ I T Q 328 Understandable electric circuits ● Bandwidth (pass-band): the frequency range corresponding to _ I ¼0:707 _ I max , and BW¼ f 2 7f 1 . ● f 2 and f 1 : critical frequencies or cutoff frequencies or half-power point frequencies. P f 1;2 ¼ 0:5P max ● Quality factor: a measure of the quality and selectivity of a resonant circuit. ● Selectivity: the capability of the circuit to choose the maximum current closer to the resonant frequency f r . ● In a series resonant circuit, V L or V C may greatly exceed the supply voltage E, i.e. a lower input voltage may produce a higher output voltage. ● In a parallel resonant circuit, I L or I C may greatly exceed the total current I T , i.e. a lower input current may produce a higher output current. Experiment 10: Series resonant circuit Objectives ● Observe and analyse the characteristics of a resonant circuit by using the oscilloscope and function generator. ● Determine the resonant frequency f r of the series resonant circuit by experiment. ● Observe and analyse the frequency versus current curve and verify that the circuit current reaches the maximum in a RLC series resonant circuit by experiment. Background information ● Impedance of RLC series circuit: Z ¼ R þjðX L ÀX C Þ ● Resonance: X L ¼ X C ; V L ¼ V C ; _ V R ¼ _ E ● Current reaches maximum at the series resonance: _ I ¼ _ V R =R f f r 0 • I ● Resonant frequency: f r ¼ 1=2p ffiffiffiffiffiffiffi LC p ● Quality factor: Q ¼ X L =R ¼ X C =R Equipment and components ● Multimeter ● Breadboard RLC circuits and resonance 329 ● Function generator ● Oscilloscope ● LCZ meter or Z meter ● Resistor: 300 O ● Inductor: 12 mH ● Capacitor: 0.0013 µF Procedure 1. Use multimeter (ohmmeter function) to measure the winding resistance R w of the 12-mH inductor. Record the value in Table L10.1. 2. Calculate the quality factor Q of the series resonant circuit in Figure L10.1 using Q ¼ X L =ðR þR w Þ (inductor has a winding resistance R W ). Record the value in Table L10.1. 3. Compute the resonant frequency f r and record the value in Table L10.1. 4. Set the sinusoidal output voltage of the function generator to 4 V (peak value), and then adjust the frequency of the function generator to the cir- cuit resonant frequency f r . 5. Construct a circuit as shown in Figure L10.1 on the breadboard. 6. Use the oscilloscope CH I to measure the peak source voltage E from the function generator in Figure L10.1. Note that the ground of the oscillo- scope, function generator and circuit should be connected together. 7. Use the oscilloscope CH II to measure the voltage across resistor V R (peak value). Record the value in Table L10.1. Is the value of V R and supply voltage E approximately equal? Table L10.1 R w Q f r V R V C V L Value R = 300 Ω C = 0.0013 μF L = 12 mH V e = 4 sin ωt Figure L10.1 Series RLC resonant circuit 330 Understandable electric circuits 8. Exchange the position of the capacitor and resistor in the circuit as shown in Figure L10.2, and so that ground of the oscilloscope, function generator and circuit (capacitor) are connected together. Then measure the voltage across the capacitor (V C ), and record the value in Table L10.1. 9. Exchange the position of the capacitor and inductor in the circuit as shown in Figure L10.3, so that the ground of the oscilloscope, function generator and circuit (inductor) are connected together. Then measure the voltage across the inductor (V L ), and record the value in Table L10.1. 10. Calculate the currents and voltages of the RLC series resonant circuit at different frequencies given in Table L10.2. Record the values in Table L10.2. R = 300 Ω C = 0.0013 μF L = 12 mH V e = 4 sin ωt Figure L10.2 The circuit to measure V C R = 300 Ω C = 0.0013 μF L = 12 mH V e = 4 sin ωt Figure L10.3 The circuit to measure V L Table L10.2 F I V R V C V L 0.3 f r ¼ 0.6 f r ¼ f r ¼ 1.4 f r ¼ 1.7 f r ¼ RLC circuits and resonance 331 11. Adjust the frequency of function generator to 0.3 f r , 0.6 f r , f r , 1.4 f r and 1.7 f r , respectively. Repeat steps 7–9 and record the results in Table L10.3. 12. Does the circuit current reach the maximum at resonant frequency f r ? If not, explain the reason. 13. Based on the data in Table L10.2, plot the current versus frequency curve (with current I in the vertical axis and frequency f in the horizontal axis). Conclusion Write your conclusions below: Table L10.3 f V R V C V L 0.3 f r ¼ 0.6 f r ¼ f r ¼ 1.4 f r ¼ 1.7 f r ¼ 332 Understandable electric circuits Chapter 11 Mutual inductance and transformers Objectives After completing this chapter, you should be able to: ● understand the concept of mutual inductance ● understand the dot convention concept ● know the basic construction of a transformer ● know different types of transformers ● understand the characteristics of transformers ● determine the turns ratio of an ideal transformer ● calculate the current, voltage, impedance and power of the primary and secondary of a transformer ● understand the concept of impedance matching of a transformer ● know applications of transformers The concept of self-inductance has been introduced in chapter 6. This chapter will introduce the mutual inductance and transformer. A transformer is a device that is built based on the principle of mutual inductance and can be used to increase or decrease the voltage or current, and transfer electric energy from one circuit to another. It also can be used for impedance matching. Transformers have a very wide range of applications in power systems, tele- communications, radio, instrumentation and many other electrical and elec- tronics fields. 11.1 Mutual inductance 11.1.1 Mutual inductance and coefficient of coupling As discussed in chapter 6 (section 6.3), when a changing current flows through a coil (inductor), it will produce an electromagnetic field around the coil, and as a result an induced voltage v L will flow across it. The changing current in a coil that produces the ability to generate an induced voltage is called self-inductance. Mutual inductance is the ability of a coil to produce an induced voltage due to the changing of the current in another coil nearby. In Figure 11.1, a coil L 1 is placed close to another coil L 2 . When AC cur- rent i 1 flows through the first coil L 1 , the changing of alternating current will produce a changing electromagnetic field and flux f 1 , resulting in a self- induced voltage v 1 across the first coil L 1 . Since the two coils are very close, there is also a portion of magnetic flux, f 1–2 , that is produced by changing the electromagnetic field linked to the coil L 2 , and consequently produces the induced voltage v 2 across the second coil L 2 . The phenomenon of a portion of the flux of a coil linking to another coil is called inductive coupling, and this is the principle of mutual inductance. Mutual inductance is denoted by L M and can be expressed mathematically using the following formula: L M ¼ k ffiffiffiffiffiffiffiffiffiffi L 1 L 2 p : Mutual inductance An inducted voltage in one coil due to a current change in a nearby coil. L M ¼ k ffiffiffiffiffiffiffiffiffiffi L 1 L 2 p There are three factors that affect mutual inductance: inductances of the two coils L 1 , L 2 and the coupling coefficient k. The coefficient of coupling k determines the degree of the coupling between the two coils, and it is the ratio of f 1–2 and f 1 : k ¼ f 1À2 f 1 f 1 is the magnetic flux generated by the current i 1 in the first coil L 1 , and f 1–2 is the portion of the magnetic flux that is generated by the current i 1 in the first coil L 1 and linked to the second coil as shown in Figure 11.2(a). f 1–2 is called the crossing link flux. e i 1 ϩ v 1 ϩ Ϫ Ϫ v 2 L 1 L 2 Figure 11.1 Magnetic coupling 334 Understandable electric circuits The induced voltage generated by a changing current (AC) that flows through the self-inductance coil L 1 is given by v 1 ¼ L(di 1 )/dt (chapter 6). So when the AC current i 1 flows through the second coil L 2 , the induced voltage in the coil L 2 is given by v 2 ¼ L M (di 1 )/dt, or _ V 2 ¼ jL M _ I 1 in the phasor form. In practice, not all of the magnetic flux generated by current i 1 will pass through L 1 and L 2 , and the portion of the magnetic flux that does not link with L 1 and L 2 is known as a leakage flux. The closer the two coils are placed (or if the two coils have a common core as shown in Figure 11.3(b)), the higher the cross-linking flux f 1–2 and the lower the leakage flux. The full-coupling occurs when k ¼ (f 1–2 )/(f 1 ) ¼ 1, i.e. f 1–2 ¼ f 1 , when all of the flux link coils 1 and 2, and there will be no leakage flux. If the gap between the two coils is large, it will cause the cross-linking flux to decrease, the leakage flux to increase, and the coupling coefficient k to decrease. k is in the range between 0 and 1 (0 k 1). Coefficient of coupling ● The coefficient of the coupling: k ¼ ðf 1À2 Þ=f 1 (0 k 1). ● f 1 : The flux generated by the current i 1 in the first coil L 1 . ● f 1–2 : The flux generated by the current i 1 in the coil L 1 cross-linking to coil L 2 . 11.1.2 Dot convention The polarity of the induced voltage across the mutually coupled coils can be determined by the dot convention method. This method can be used to indicate whether the induced voltage in the second coil is in phase or out of phase with the voltage in the first coil. The dot convention method places two small phase dots (.) or asterisks (*), one on the coil L 1 and the other on the coil L 2 , to indicate that polarities of the induced voltage v 1 in the coil L 1 and v 2 in the adjacent coil L 2 are same at these points, as shown in Figure 11.3. This means that the dotted terminals of coils (a) (b) L 1 L 2 L M k e L 1 L 2 k L M i1 f 1 f 12 Figure 11.2 Mutual inductance Mutual inductance and transformers 335 should have the same voltage polarity at all time, and dotted terminals are known as corresponding terminals. Dot convention Dotted terminals of coils have the same voltage polarity. 11.2 Basic transformer 11.2.1 Transformer A transformer is an electrical device formed by two coils that are wound on a common core. You may have seen transformers on top of utility poles. A trans- former uses the principle of mutual inductance to convert AC electrical energy from input to output. Recall that mutual inductance is the ability of a coil to produce induced voltage due to the changing of current in another coil nearby. Figure 11.4 shows two simplified transformer circuits. A changing current from the AC voltage source in the first coil produces a changing magnetic field, inducing a voltage in the second coil. The first coil is called primary winding, and the second coil connected to the load Z L is called secondary winding. Structurally, the transformers are categorized as two main types: the air- core and iron-core transformers. The symbols for themare shown in Figure 11.4 (a and b), respectively (inside the dashed lines). Ϫ Ϫ Ϫ ϩ ϩ ϩ * * * * v 1 v 2 v 1 ϩ Ϫ Ϫ Ϫ Ϫ ϩ ϩ ϩ . . . v 1 v 2 v 1 v 2 . (a) (b) Figure 11.3 Dot convention e Z L L 1 L 2 e Z L (a) (b) Figure 11.4 Implified transformer circuits (a) air-core; (b) iron-core 336 Understandable electric circuits Transformer A transformer uses the principle of mutual inductance to convert AC electrical energy from input to output. 11.2.2 Air-core transformer The air-core transformers are usually used in high-frequency circuits, such as in instrumentation, radio and TV circuits. An air-core transformer does not have a physical core, so it can be obtained by placing the two coils L 1 and L 2 close to each other, or by winding both the coils L 1 and L 2 on a hollow cylindrical-shaped core with isolating material as illustrated in Figure 11.5(b). The circuit of an air-core transformer is shown in Figure 11.5(a), there R 1 and R 2 represent the primary and secondary winding resistors of the transformer. The air-core transformer is also known as a linear transformer. When the core of the transformer is made by the insulating material with constant per- meability, such as air, plastic, wood, etc., it is a liner transformer. Air-core transformer A transformer uses the principle of mutual inductance to convert AC electrical energy from input to output. 11.2.3 Iron-core transformer Iron-core transformers are usually used in audio circuits and power systems. The coils of the iron-core transformer are wound on the ferromagnetic material that are laminated sheets insulated to each other, as illustrated in Figure 11.6. (a) (b) L 1 L 2 v s Z L L 1 L 2 R 1 R 2 Figure 11.5 Air-core transformer Mutual inductance and transformers 337 When two coils are wound on a common core, it will have higher cross- linking flux and lower leakage flux. The ferromagnetic materials can provide an easy path for the magnetic flux. Furthermore, if two coils are wound on a common core, the flux generated in the coil L 1 will almost all link with the coil L 2 . This means that the coupling coefficient k is close to 1, and this is the reason that iron-core transformer is usually considered as the ideal transformer (k ¼ 1). 11.2.4 Ideal transformer The coupling coefficient k of an ideal transformer is 1, i.e. ideal full-coupling, neglecting winding resistance and magnetic losses in the coils of the transfor- mer. Figure 11.6(a) is a circuit of an ideal transformer with the voltage source, and the load, and the portion within the dashed line is the symbol of the ideal transformer. An iron-core transformer is considered the ideal transformer because it uses ferromagnetic materials with high permeability as its core. Also the pri- mary and secondary windings are wound on a common core, which have near zero leakage flux and can achieve a full-coupling (k ¼ 1). ● Transformer parameters: The parameters of an ideal transformer in Figure 11.6(a) are listed in Table 11.1. f e Z S ϭ Z L ϩ Ϫ Ϫ v S v P ϩ i P i S Z N P N S Z P (a) (b) Figure 11.6 Iron-core transformer Table 11.1 Parameters Name v P Primary voltage v S Secondary voltage N p Number of turns on the primary coil N S Number of turns on the secondary coil i p Primary current i S Secondary current Z P Primary impedance Z S ¼ Z L Secondary or load impedance 338 Understandable electric circuits ● Turns ratio n: The turns ratio of a transformer is the ratio of the number of turns, i.e. the number of turns on the secondary coil N S to the number of turns on the primary coil N P , which can be derived from the voltage ratio of the secondary and primary voltages. From Faraday’s law described in chapter 6, v L ¼ N df dt we can get: ● the primary voltage v P ¼ N P df dt ● the secondary voltage v S ¼ N S df dt Dividing v S by v P gives the transformer’s turns ratio n: v S v P ¼ N S N P ¼ n If the transformer is an ideal transformer, i.e. the transformer has no power loss itself, the input power is equal to the output power, i.e. p P ¼ p S or v P i P ¼ v S i S , so i P =i S ¼ v S =v P ¼ n ð11:1Þ v P ¼ v S /n can be obtained from (11.1), and also i P ¼ n i S . The primary impe- dance can be obtained by substituting v P and i P into Z P as follows: Z P ¼ v P i P ¼ v S =n n i S ¼ 1 n 2 Z L or n 2 ¼ Z L =Z P ; n ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffi Z L =Z P p where the secondary impedance is the load Z L Z L ¼ v S i S Turns ratio ● Instantaneous form: n ¼ N S =N P ¼ v S =v P ¼ i P =i S ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffi Z L =Z P p , or Z L ¼ n 2 Z P ● Phasor form: n ¼ N S =N P ¼ _ V S = _ V P ¼ _ I P = _ I S ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffi Z L =Z P p Power ● Instantaneous form: p S ¼ i S v S ; p P ¼ i P v P ● Phasor form: _ P S ¼ _ I S _ V S ; _ P P ¼ _ I P _ V P ● Conversion of the voltage, current and impedance: The expressions of the transformer’s turns ratio indicate that a transformer can be used to convert voltage, current and impedance. Mutual inductance and transformers 339 ● Voltage conversion: ● From the primary to the secondary, multiplying by n: v S = nv P ● Fromthe secondary to the primary, multiplying by 1/n: v P =(1/n)v S ● Current conversion: ● Fromthe primary to the secondary, multiplying by 1/n: i S ¼ (1/n)(i P ) ● From the secondary to the primary, multiplying by n: i P ¼ n i S ● Impedance conversion: ● From the primary to the secondary, multiplying by 1/n 2 : Z P ¼ (1/n 2 )(Z L ) ● From the secondary to the primary, multiplying by n 2 : Z L ¼ n 2 Z P (The converted impedance is also called the reflected impedance, meaning the reflection of the primary impedance results in the secondary impedance.) Transformer parameters conversion ● Voltage conversion: v S ¼ nv P ; v P ¼ ð1=nÞðv S Þ ● Current conversion: i S ¼ ð1=nÞði P Þ; i P ¼ n i S ● Impedance conversion: Z L ¼ n 2 Z P ; Z P ¼ ð1=n 2 ÞðZ L Þ Example 11.1: The number of turns on the primary is 40 for an ideal trans- former, and the number of turns on the secondary is 100. _ V P ¼ 50 V, _ I P ¼ 5 A and Z L ¼ 2 O. Determine the transformer’s turns ratio, secondary voltage, secondary current, primary impedance (reflected from the secondary) and the primary power (the amplitude only). Solution: ● n ¼ N S N P ¼ 100 40 ¼ 2:5 ● _ V S ¼ n _ V P ¼ ð2:5Þð50 VÞ ¼ 125 V ● _ I S ¼ _ I P n ¼ 5 A 2:5 ¼ 2 A ● Z P ¼ Z L n 2 ¼ 2 O 2:5 2 ¼ 0:32 O ● _ P S ¼ _ I S _ V S ¼ð2 AÞð125 VÞ ¼ 250 W ¼ 0:25 kW 11.3 Step-up and step-down transformers 11.3.1 Step-up transformer A step-up transformer is a transformer that can increase its secondary voltage. Since a step-up transformer always has more secondary winding turns than the primary, the secondary voltage of a step-up transformer (v S ) is always higher than the primary voltage (v P ), i.e. v S 4v P . The value of the secondary voltage 340 Understandable electric circuits depends on the turns ratio (n). The equation of n ¼ N S /N P ¼ v S /v P indicates that to have a higher secondary voltage, the number of turns on the secondary winding must be greater than that of the primary, i.e. N S 4N P as illustrated in Figure 11.7(a), meaning the turns ratio n ¼ (N S /N P ) 41. This is an important characteristic of a step-up transformer. Step-up transformer ● v S 4 v P ● N S 4 N P ● n 4 1 11.3.2 Step-down transformer A step-down transformer is a transformer that can decrease its secondary voltage. Since a step-downtransformer always has less turns onthe secondary winding than the primary, the secondary voltage of a step-down transformer (v S ) is always lower than the primary voltage (v P ), i.e. v S 5 v P . The value of the secondary voltage depends on the turns ratio (n). The equation n ¼ N S /N P ¼ v S /v P indicates that to have a voltage that is lower in secondary than primary, the number of turns on the secondary coil must be lesser than primary, i.e. N S 5N P as illustrated in Figure 11.7(b), meaning the turns ratio n ¼ (N S /N P ) 5 1. This is an important char- acteristic of a step-down transformer, which is opposite of a step-up transformer. Step-down transformer ● v S 5 v P ● N S 5 N P ● n 5 1 Example 11.2: If a transformer has 125 turns of secondary windings and 250 turns of primary windings, calculate its turns ratio and determine if it is a step- up or a step-down transformer. (a) (b) v P N P v S N S v P v S N P N S Figure 11.7 (a) Step-up and (b) step-down transformers Mutual inductance and transformers 341 Solution: Since N S ¼ 125, N P ¼ 250 and N S 5N P or n ¼ N S =N P ¼ 125=250 ¼ 1=2 ¼ 0:551, i.e. n 51 it is a step-down transformer. 11.3.3 Applications of step-up and step-down transformers As mentioned in the previous section, transformers can be used to convert voltage, current and impedance. In the power system, the basic usage of transformers is stepping up or stepping down the voltage or current, which will require converting voltage or current from primary to secondary winding. The functions of step-up and step-down transformers are to increase or decrease the voltage of their secondary windings, and have important appli- cations in the power transmission system. A simplified power transmission system is illustrated in Figure 11.8. The voltage generated from the generator of a power plant needs to rise to a very high value through the step-up transformer so that it can be delivered through long-distance transmission lines. This can reduce the loss of energy or power created due to the winding resistance in the line (I 2 R w ¼ P Loss ) for a long-distance-line transmission, and improve the efficiency of the electricity transmission. Decreasing the current to reduce the power loss on the transmission line may reduce the output power (P ¼ IV) of the transmission system. If the vol- tage is increased through the step-up transformer before the transmission, it can maintain the same output power, but reduce the power loss on the line, i.e.: ~ P ¼ ðI # ÞðV " Þ ) ðI 2 # ÞðRÞ ¼ P Loss # If a step-up transformer is used to increase the voltage by 100 V, then the current will reduce by 100 A ½v S "¼ ðn " Þðv P Þ; i S #¼ ð1=n "Þði P Þ, and the loss of the power due to the winding resistance in the line will reduce to 10 000 W, since I 2 R w ¼ P Loss and I 2 # R w ) P Loss #. The local distribution stations require step-down transformers to reduce the very high voltage by the long distance transmission and can send it to commercial or residential areas. Power plant 2 2 k V 500 kV 6 6 k V 4800 kV 220 kV /110 kV Industries Step-down transformer Step-down transformer Step-down transformer Step-up transformer Commercial areas Residential areas Figure 11.8 Power transmission system 342 Understandable electric circuits 11.3.4 Other types of transformers There are other types of commonly used transformers listed as follows: ● Center-tapped transformer: It has a tap (connecting point) in the middle of the secondary winding, and it can provide two balanced output voltages with the same value, as shown in Figure 11.9(a). ● Multiple-tapped transformer: It has multiple taps in the secondary wind- ing, and it can provide several output voltages with different values, as shown in Figure 11.9(b). ● Adjustable (or variable) transformer: The output voltage of adjustable (or variable) transformer across the secondary winding is adjustable. The secondary winding of the adjustable transformer can provide an output voltage that may be variable in a range of zero to the maximum values. An adjustable transformer is shown in Figure 11.9(c). ● Auto-transformer: It is a transformer with only a single winding, which is a commoncoil for boththe primary andthe secondary coils, anda portionof the common coil acts as part of both the primary and secondary coils, as shown in Figure 11.9(d). An auto-transformer can be made smaller and lighter. Figure 11.9 (b) Multiple-tapped transformer Figure 11.9 (c) Adjustable transformer Figure 11.9 (a) Center-tapped transformer Mutual inductance and transformers 343 11.4 Impedance matching In addition to stepping-up and stepping-down voltages, a transformer has another important application, matching the load and source impedance in a circuit to achieve the maximum power transfer from the source to the load. It is known as impedance matching. 11.4.1 Maximum power transfer The theory of maximum power transfer in the DC circuits was introduced in chapter 5, i.e. the maximum power delivered from a source to a load in a circuit can be achieved when the load resistance is equal to the internal resistance of the source (R L ¼ R S ), or when the load resistance is equal to the Thevenin/ Norton equivalent resistance of the network (R L ¼ R TH ¼ R N ). This theory can also be applied to an AC circuit by replacing the resistance with impedance. When the load impedance Z L is equal to the source internal impedance Z S , the power received by the load from the source reaches the maximum, this is shown in Figure 11.10. Maximum power transfer When Z L = Z S , the power delivered from the source to the load reaches the maximum. Figure 11.9 (d) Auto-transformer e Z S Z L Figure 11.10 Impedance matching 344 Understandable electric circuits 11.4.2 Impedance matching In the practical circuits (or Thevenin’s equivalent circuits), the internal resis- tance of the source is fixed, usually is not matching with the load impedance, and also not adjustable. In this case, a transformer with an appropriate turns ratio n can be placed between the load and source to make the load impedance and the source internal resistance equal, and to achieve the maximum power transfer, i.e. n ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffi Z L =Z P p . Impedance matching Place a transformer with n ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffi Z L =Z P p between the source and the load to achieve maximum power transfer. Example 11.3: A simplified amplifier circuit is illustrated in Figure 11.11(a). The circuit within dashed lines is Thevenin’s equivalent circuit for the amplifier circuit, and its internal resistance is 100 O. How do we deliver the maximum power to the speaker if the resistance of the speaker is 4 O (so the speaker can have the maximum volume)? Solution: ● Since the load impedance (Z L ¼ R L ¼ 4 O) does not match with the source internal impedance (Z S ¼ R S ¼ 100 O) currently, the maximum power cannot be delivered to the speaker if the source and load are connected directly. ● Choose an audio transformer with the appropriate turns ratio n, i.e. n ¼ ffiffiffiffiffi Z L Z P r ¼ ffiffiffiffiffiffiffiffi 4 100 r ¼ 0:2 ¼ 1 5 (a) (b) 100 Ω e 4 Ω e 100 Ω n = 1/5 4 Ω Figure 11.11 Circuits for Example 11.3 Mutual inductance and transformers 345 ● Therefore, if placing an impedance matching transformer with the turns ratioof 1/5 betweenthe amplifier andspeaker as illustratedinFigure 11.11(b), the speaker will have the maximum volume. Summary Mutual inductance ● Mutual inductance: An inducted voltage in one coil due to a change cur- rent in the other nearby coil. ● L M ¼ k ffiffiffiffiffiffiffiffiffiffi L 1 L 2 p . ● Coefficient of coupling: k ¼ f 1À2 f 1 ð0 k 1Þ. ● Dot conversion: Dotted terminals of coils have the same voltage polarity. Basic transformer ● Transformer: It uses the principle of mutual inductance to convert AC electrical energy from input to output. ● The parameters of an ideal transformer (k ¼ 1): ● Turns ratio: n ¼ N S =N P ¼ v S =v P ¼ i P =i S ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffi Z L =Z P p or Z L ¼ n 2 Z P In phasor form: n ¼ N S =N P ¼ _ V S = _ V P ¼ _ I P = _ I S ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffi Z L =Z P p ● Power: p S ¼ i S v S ; p P ¼ i P v P _ P S ¼ _ I S _ V S _ P P ¼ _ I P _ V P ● Transformer conversion (V, I and Z): ● Voltage conversion: v S ¼ n v P ; v P ¼ ð1=nÞðv S Þ ● Current conversion: i S ¼ ð1=nÞði P Þ; i P ¼ n i S ● Impedance conversion: Z L ¼ n 2 Z P ; Z P ¼ ð1=n 2 ÞðZ L Þ ● Step-up transformer: ● v S 4 v P ● N S 4 N P ● n 4 1 ● Step-down transformer: ● v S 5 v P Parameters Name v P Primary voltage v S Secondary voltage N P Number of turns on the primary coil N S Number of turns on the secondary coil i P Primary current i S Secondary current Z P Primary impedance Z S ¼ Z L Secondary or load impedance 346 Understandable electric circuits ● N S 5N P ● n 5 1 ● Impedance matching: Place a transformer with the turns ratio n ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffi Z L =Z P p between the source and the load to achieve maximum power transfer from the source to the load. Experiment 11: Transformer Objectives ● Experimentally verify the equation for calculating the turns ratio of the transformer. ● Experimentally verify the theory of impedance matching transformer. ● Analyse experimental data, circuit behaviour and performance, and com- pare them to the theoretical equivalents. Background information ● The turns ratio of the transformer: n ¼ N S =N P ¼ v S =v P ¼ i P =i S ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffi Z L =Z P p ● Impedance matching: Place a transformer with the turns ratio n ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffi Z L =Z P p between the source and the load (if Z in 6¼ Z L ) to achieve maximum power transfer from the source to the load. ● Formula for the transformer’s impedance matching: Z L ¼ n 2 Z P Equipment and components ● Multimeter ● Breadboard ● Function generator ● A small transformer with lower secondary voltage (12–25 V) ● A small audio transformer ● Continuously adjustable auto-transformer ● Several resistors ● A small audio speaker Procedure Part I: Turns ratio of a transformer 1. Choose a small transformer with lower secondary voltage (12–25 V) in the laboratory, and measure the primary (R P ) and secondary (R S ) resistances of the transformer using a multimeter (ohmmeter function). Record the values in Table L11.1. Table L11.1 Primary resistance R P Secondary resistance R S Calculated turns ratio n ¼ V S /V P Measured turns ratio n Mutual inductance and transformers 347 2. Calculate the turns ratio n of the transformer using the primary and sec- ondary voltages of the transformer (from the nameplate). Record the values in Table L11.1. 3. Plug the primary of the adjustable auto-transformer into the AC power outlet, and connect the secondary adjustable auto-transformer to the pri- mary of the small transformer, as shown in Figure L11.1(a). Note: The continuously adjustable auto-transformer is used as a variable AC voltage source. If the primary voltage of the small transformer is not too high, a function generator can be used to replace the adjustable auto-transformer. 4. Adjust the output voltage of the auto-transformer until it is slightly lower than the rated primary voltage of the small transformer, then measure the primary and secondary voltages (RMS values) of the small transformer using a multimeter (voltmeter function). Calculate the turns ratio n of the small transformer using measured primary and secondary voltages, and record the value in the column ‘Measured turns ratio’ in Table L11.1. 5. Connect a suitable load resistor R L to the secondary of the small trans- former as shown in Figure L11.1(b). Determine the value of resistor R L by calculating the secondary current I S and power P S , and make sure that I S and P S will not exceed the rated current and power of the small transfor- mer after connecting R L to the secondary. Calculate the primary current I P (I P = P P /V P ) and record the value in Table L11.2. v P v S Figure L11.1 (a) Auto-transformer and small transformer V P V S R L I S I P Figure L11.1 (b) The circuit for step 5 Table L11.2 Primary current I P Secondary current I S Turns ratio: n ¼ I P /I S 348 Understandable electric circuits 6. Measure the secondary current I S using either the direct or indirect method. Record the value in Table L11.2. 7. Calculate turns ratio n of the transformer using primary and secondary currents in Table L11.2, and record the value. Compare the turns ratio n of the transformer in Tables L11.1 and L11.2. Are there any significant dif- ferences? If so, explain the reasons. Part II: Impedance matching 1. Set the frequency of the function generator to 2 kHz. Then connect a small audio speaker to the two terminals of the function generator as shown in Figure L11.2(a) 2. Adjust the output voltage of the function generator to a suitable value so that the speaker reaches a comfortable listening volume. 3. Measure the voltage across the two terminals of the speaker in Figure L11.2(a). Record the value in Table L11.3 in ‘Without transformer’ row. 4. Calculate the turns ratio ðn ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffi Z L =Z P p Þ of a transformer that can be used as the impedance matching. Record the value in Table L11.3. ● The output impedance of the function generator Z P usually is about 600 O. ● The impedance of the small speaker Z L usually is about 8 O. 5. Find a small audio transformer that has a turns ratio n closer to the cal- culated n in step 4, and connect it between the function generator and speaker as shown in Figure L11.2(b). Table L11.3 Voltage across the speaker Without transformer: With transformer: Transformer turns ratio: n ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffi Z L =Z P p (a) (b) e e Figure L11.2 Circuit for Part II Mutual inductance and transformers 349 6. Measure the voltage across the speaker in Figure. L11.2(b). Record the value in Table L11.3 in the row ‘With transformer’. 7. Compare the volume of the speaker with and without the transformer, and explain the reason in the conclusion. Conclusion Write your conclusions below: 350 Understandable electric circuits Chapter 12 Circuits with dependent sources Objectives After completing this chapter, you should be able to: ● understand the concept of the dependent source ● define four types of dependent sources ● know how to convert dependent sources equivalently ● understand the methods for analysing circuits with dependent sources The DC and AC circuits we have discussed in the previous chapters have independent voltage/current sources (Figure 12.1) and will not be affected by other voltages and currents in the circuit. The circuits we will analyze in this chapter have dependent (or controlled) sources, in which the source voltage or current is a function of other voltage or current in the circuit. Dependent sources are a useful concept in modelling and analysing electronic components, such as transistors, amplifiers, filters, etc. V I i v Figure 12.1 Independent sources 12.1 Dependent sources 12.1.1 Dependent (or controlled) sources As the name suggests, when a source voltage or current is controlled by or dependent on other voltage or current in the circuit, it is called a dependent (or controlled) source. The dependent sources can be categorized into the following four types according to whether it is controlled by a circuit voltage or current, as well as whether the dependent source itself is a voltage source or current source: ● voltage-controlled voltage source (VCVS) ● voltage-controlled current source (VCCS) ● current-controlled voltage source (CCVS) ● current-controlled current source (CCCS) The above dependent sources can be represented by the symbols in Figure 12.2. In this figure, k 1 , k 2 , k 3 and k 4 are called control coefficients or gain parameters. A voltage-controlled source has a voltage across its two terminals that equals to a control coefficient k multiplied by a controlling voltage or current elsewhere in the same circuit. A current-controlled source has a current in its branch that equals to a control coefficient k multiplied by a voltage or current elsewhere in the same circuit. 5 _ I in the circuit of Figure 12.3(a) represents a CCCS. Its control coefficient k is 5, and the current _ I is a controlling current through the 5 Oresistor branch in the same circuit. 8 _ V in the circuit of Figure 12.3(b) is a VCVS. Its control coefficient is 8, and the voltage _ V is a controlling voltage across resistor R 2 in the same circuit. + – + – k 1 V k 2 I k 3 V k 4 I (VCVS) (VCCS) (CCVS) (CCCS) + – – V I (a) (b) Figure 12.2 Dependent sources: (a) controlling sources and (b) controlled sources 352 Understandable electric circuits After you take an analogue electronics course, you will understand that a good example for modelling a CCCS is a transistor circuit. Based on the property of a bipolar transistor, a current amplifier, its large collect current i c is proportional to the small base current i b according to the relationship i c ¼ bi b . In this equation, the current gain b is the same as the control coefficient k in the dependent source. Dependent (controlled) sources The source voltage or current is a function of the other voltage or current in the same circuit. 12.1.2 Equivalent conversion of dependent sources Equivalent conversion of dependent sources is the same as the equivalent conversion of independent sources. For instance, the voltage-controlled source in Figure 12.4(a) can be converted equivalently to a current-controlled source as shown in Figure 12.4(b), and vice versa. Internal resistance R S of the source does not change before and after the conversion, just apply Ohm’s law to convert the source. 5 V 2 Ω 3 Ω 5 Ω 5 R 1 + - + - • V 8V • • I • I • E 1 E 2 R 2 • Figure 12.3 Circuits with dependent sources (a) (b) ⇔ R s b + - a V S • R S a b I S • Figure 12.4 Equivalent conversion: (a) R S ¼ R S ; _ V S ¼ _ I S R S and (b) R S ¼ R S ; _ I S ¼ _ V S =R S Circuits with dependent sources 353 Example 12.1: The VCVS in Figure 12.5(a) can be converted equivalently to a VCCS as shown in Figure 12.5(b). Example 12.2: The CCCS in Figure 12.6(a) can be converted equivalently to a CCVS as shown in Figure 12.6(b). Equivalent conversion of dependent sources The same as the equivalent conversion of independent sources: ● controlled current source!controlled voltage source: R S ¼ R S ; V S ¼ I S R S ● controlled voltage source!controlled current source: R S ¼ R S ; I S ¼ V S R S (a) (b) R = 2 kΩ b + - a + - 10∠0° V • V = V∠0° R = 2 kΩ a b - + I = 5∠0° mA • • V V∠0° Figure 12.5 Circuit for Example 12.1. (a) Voltage-controlled voltage source (VCVS) and (b) voltage-controlled current source (VCCS), _ I ¼ _ V=R ¼ 10 Vff0 =2 kO ¼ 5ff0 mA ⇒ b a 5 kΩ 10∠0° mA • I = I∠0° A + - a b 5 kΩ I∠0° A V = 50∠0° V • • I (a) (b) Figure 12.6 Circuit for Example 12.2. (a) Current-controlled current source (CCCS) and (b) current-controlled voltage source (CCVS), _ V ¼ _ IR S ¼ ð10ff0 mAÞð5 kOÞ ¼ 50ff0 V 354 Understandable electric circuits 12.2 Analysing circuits with dependent sources The analysing methods for circuits with dependent sources are similar to that for circuits with independent sources. The following examples will describe these methods. Example 12.3: Determine the current I in the circuit of Figure 12.7(a). (a) 2 Ω 4 Ω 4 Ω 2 Ω 2 A 2I I 0.5I 3 A (b) 4 Ω 2 Ω a 1 A 0.5I I 2I Figure 12.7 Circuits for Example 12.3 Solution: Simplify and convert the circuit of Figure 12.7(a) to that in Figure 12.7(b). There, 3 A À 2 A ¼ 1 A 2 þ 4 ==4 ¼ 4 Note: This circuit has a CCCS, simplify the circuit without changing the CCCS (both controlling branches and controlled source). Write the KCL equation for the node a in Figure 12.7(b): ÆI ¼ 0 : 1 A À 2I ÀI þ 0:5I ¼ 0 Circuits with dependent sources 355 Current I can be solved from the above equation: À2:5I ¼ À1 A ; I ¼ 0:4 A Example 12.4: Determine the voltage V in the circuit of Figure 12.8. Solution: Applying KVL, P V = 0: À6 þ 2I þ 4V þ 8 þ 3I ¼ 0 That is 2 þ 5I þ 4V ¼ 0 Substituting V¼ 73I into the above equation gives: 2 þ 5I þ 4ðÀ3IÞ ¼ 0 2 þ 5I À 12I ¼ 0 Solving for I: 2 À 7I ¼ 0; I % 0:29 A Solving for V: V ¼ À3I ¼ ðÀ3 Þð0:29 AÞ ¼ À0:87 V Example 12.5: Write node voltage equations for the circuit in Figure 12.9 using the node voltage analysis method. (Write KVL by treating the dependent source as an independent source first, and then represent the control quantity as node voltages.) 6 V 2 Ω 3 Ω + - 4 V + - V I 8 V Figure 12.8 Circuit for Example 12.4 356 Understandable electric circuits Solution: The procedure for applying the node voltage analysis method (chapter 4, section 4) to the above circuit is as follows: 1. Label nodes a, b and c, and choose ground c as the reference node as shown in Figure 12.9. 2. Write KCL equations to n71 = 371 = 2 nodes (nodes a and b) by inspection. Node a: 1 R 1 þ 1 R 2 þ 1 R 3 V a À 1 R 3 V b ¼ 1 R 1 E 1 ð12:1Þ Node b: À 1 R 3 V a þ 1 R 3 þ 1 R 4 V b ¼ 3V ð12:2Þ Substituting the control voltage V ¼ V a ÀV b to (12.2) gives: À 1 R 3 V a þ 1 R 3 þ 1 R 4 V b ¼ 3ðV a ÀV b Þ ð12:3Þ 3. Solving (12.1) and (12.3) can determine the node voltage V a and V b (if R 1 , R 2 , R 3 and E 1 are given). Example 12.6: Use the mesh current analysis method to write mesh equations for the circuit in Figure 12.10. (Write KVL by treating the dependent source as an independent source first, and then represent the controlling quantity as mesh current.) Solution: The procedure for applying the mesh current analysis method (chapter 4, section 3) to the above circuit is as follows: 1. Label all the reference directions for each mesh current I 1 , I 2 and I 2 (clockwise) as shown in Figure 12.10. R 2 R 4 3V (VCCS) E 1 R 1 R 3 a c V - + b Figure 12.9 Circuit for Example 12.5 Circuits with dependent sources 357 2. Apply KVL around each mesh (windowpane), and ensure the number of KVL equations is equal to the number of meshes (there are three meshes in Figure 12.10). Mesh 1: ðR 1 þR 2 ÞI 1 ÀR 1 I 2 ÀR 2 I 3 ¼ E ð12:4Þ Mesh 2: ÀR 1 I 1 þ ðR 1 þR 3 ÞI 2 ÀR 3 I 3 ¼ À7I 0 ð12:5Þ Mesh 3: ÀR 2 I 1 ÀR 3 I 2 þ ðR 2 þR 3 þR 4 ÞI 3 ¼ 0 Substituting the controlling current I 0 = I 1 7I 2 to (12.5) yields ÀR 2 I 1 þ ðR 1 þR 3 ÞI 2 ÀR 3 I 3 ¼ À7ðI 1 ÀI 2 Þ ð12:6Þ 3. Solving all the three simultaneous equations, (12.4), (12.5) and (12.6), resulting from step 2 can determine the three mesh currents I 1 , I 2 and I 3 . Example 12.7: Determine the branch current I for the circuit in Figure 12.11 by using the superposition theorem. (The dependent source will not act separately E + - + I - 7I 0 I 1 R 3 R 4 R 2 R 1 I 0 I 2 I 3 (CCVS) Figure 12.10 Circuit for Example 12.6 (a) (b) (c) + = 20 V I = ? 6 A 4I (CCVS) 4 Ω − + 2 Ω − + 4 Ω 2 Ω 6 A 4I I" I' − + 4 Ω 2 Ω 4I 20 V Figure 12.11 Circuits for Example 12.7 358 Understandable electric circuits in the superposition theorem. Do not change the dependent source in the cir- cuit when another independent source is acting in the circuit.) Solution: The procedure for using the superposition theorem (chapter 5, section 5.1) to the above circuit is as follows: 1. Choose 20 V voltage source applied to the circuit first, replace the 6 A current source with an open circuit as shown in Figure 12.11(b), and calculate I 0 : I 0 ¼ 20 V 4 O þ 2 O % 3:33 A 2. When a 6 A current source is applied to the circuit, replace the 20 V voltage source with a short circuit as shown in Figurec 12.11(c), and calculate I 00 : I 00 ¼ À 6 A 2 O 2 O þ 4 O ¼ À2 A ðthe current À divider ruleÞ (The 6 A current is negative due to its assumed direction to be opposite to I 00 .) 3. Calculate the sum of currents I 0 and I 00 : I ¼ I 0 þI 00 ¼ 3:33 A þ ðÀ2 AÞ ¼ 1:33 A Example 12.8: Determine the voltage across the two terminals a and b in Figure 12.12(a) by using Thevenin’s theorem. 5 V 2 Ω 3 Ω R L I b a 2I (CCCS) (a) 2 Ω + – 3 Ω I = 0 a b 2I = 0 5 V CCCS (b) 2 Ω 3 Ω + – CCVS a b I (2I × 3) = 6I (c) Figure 12.12 (a–c) Circuits for Example 12.8 Circuits with dependent sources 359 Solution: The procedure for using Thevenin’s theorem (chapter 5, section 5.2) to the above circuit is as follows: 1. Open and remove the load branch resistor R L , and mark a and b on the terminals of the load branch as shown in Figure 12.12(b). 2. Determine Thevenin’s equivalent voltage V TH : Since the branches a and b are open, I = 0, and the CCCS is also 0 (2I = 0) in the circuit of Figure 12.12(b), so: V TH ¼ V ab ¼ 5 V 3. Determine Thevenin’s equivalent resistance R TH : Replace the 5 V voltage source with a short circuit and convert CCCS to CCVS as shown in Figure 12.12(c). R TH ¼ R ab ¼ V ab I ¼ 6I þ ð2 þ 3 ÞI I ¼ 11 4. Plot Thevenin’s equivalent circuit as shown in Figure 12.12(d). Analysing circuits with dependent sources Analysing circuits with dependent sources is similar to the methods of analysis for circuits with independent sources. Summary ● Dependent (controlled) sources: The source voltage or current is a function of other voltage or current in the circuit. ● VCVS ● VCCS ● CCVS ● CCCS (d) R L a b V TH = 5 V R TH = 11 Ω Figure 12.12 (d) Thevenin’s equivalent circuit 360 Understandable electric circuits ● Equivalent conversion of dependent sources is the same as the equivalent conversion of independent sources: ● controlled current source!controlled voltage source: R S ¼ R S ; _ V S ¼ _ I S R S ● controlled voltage source!controlled current source: R S ¼ R S ; _ I S ¼ _ V R S ● The method of analysis for circuits with dependent sources is similar to the methods of analysis for circuits with independent sources (ideal sources). Circuits with dependent sources 361 Appendix A Greek alphabet Uppercase/lowercase Letter Uppercase/lowercase Letter A a Alpha N n Nu B b Beta Ä x Xi À g Gamma O o Omicron D d Delta Å p Pi E " Epsilon P r Rho Z z Zeta S s or B Sigma H Z Eta T t Tau Y y Theta Y u Upsilon I i Iota È f Phi K k Kappa X w Chi à l Lambda É c Psi M m Mu o Omega Appendix B Differentiation of the phasor For a sinusoidal function f ðtÞ ¼ F m sinðot þcÞ, taking the derivative of f(t) with respect to t gives df ðtÞ dt ¼ F m o cosðot þcÞ and df ðtÞ dt ¼ F m o sinðot þc þ 90 Þ ¼ J m ½oF m e jðotþcþ90 Þ ¼ J m ðoF m e jot e jc e j90 Þ ¼ J m ðjoFe jot Þ There F¼F m e jc and e j90 ¼ j (From Euler’s formula, e j90 ¼ cos 90 þ jsin 90 ¼ j) Therefore, the phasor of df ðtÞ=dt is joF (there e jot is the rotating factor), i.e. df ðtÞ dt , joF Therefore, the derivative of the sinusoidal function with respect to time can be obtained by its phasor F multiplying with jo; this is equivalent to a phasor that rotates counterclockwise by 908on the complex plane (since þj ¼ þ90 ). Bibliography Robert L. Boylestad. Introductory Circuit Analysis. 7th edition. New York, NY: Merrill (an imprint of Macmillan Publishing Company), 1994. Allan H. Robbins and Wilhelm C. Miller. Circuit Analysis: Theory and Prac- tice. Albany, NY: Delmar Publishers, 1995. Thomas L. Floyd. Principles of Electric Circuits. 7th edition. Upper Saddle River, NJ: Prentice Hall, 2003. Charles K. Alexander and Mathew N.O. Sadiku. Fundamentals of Electric Circuits. Boston, MA and London: McGraw Hill, 2000. David A. Bell. Electric Circuits: Principles, Applications, and Computer Ana- lysis. Englewood Cliffs, NJ: Prentice Hall, 1995. Nigel P. Cook. Introductory DC/AC Circuits. 4th edition. Upper Saddle River, NJ: Prentice Hall, 1999. S.A. Boctor. Electric Circuit Analysis. Englewood Cliffs, NJ: Prentice Hall, 1987. Robert T. Paynter. Introductory Electric Circuits. Upper Saddle River, NJ: Prentice Hall, 1999. Richard J. Fowler. Electricity: Principle and Application. 6th edition. New York, NY: McGraw-Hill, 2003. Electronics Technician Common Core Learning Guider. Burnaby, BC: Ministry of Education, Skills and Training and the Centre for Curriculum, Transfer and Technology, 1997. David Buchla. Experiments in Electronics Fundamentals and Electric Circuits Fundamentals. 6th edition. Upper Saddle River, NJ: Prentice Hall, 2004. Allan H. Robbins and Wilhelm C. Miller. Circuit Analysis: Theory and Practice (Laboratory Manual). Albany, NY: Delmar Publishers, 1995. David A. Bell. Laboratory Manual for Electric Circuits. Englewood Cliffs, NJ: Prentice Hall, 1995. Mark Edward Hazen. Laboratory Manual for Fundamentals of DC and AC Circuits. Orlando, FL: Saunders College Publishing, 1990. Lorne MacDonald. Basic Circuit Analysis for Electronics: Through Experi- mentation. 3rd edition. Chico, CA: Technical Education Press, 1998. Qiu, Guangyuan. Electric Circuits (Revised edition). Beijing: People’s Educa- tion Press. 1983. Qian, Jianping. Electric Circuits. 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Web resources http://www.job-search-engine.com/keyword/electric-electronic-engineering? page=11 http://www.ieeeusa.org/careers/yourcareer.html http://en.wikipedia.org/wiki http://www.glenbrook.k12.il.us/gbssci/phys/Class/circuits/u9l3b.html http://en.wikipedia.org/wiki/Electrical_resistivity#Table_of_resistivities http://en.wikipedia.org/wiki/Siemens_(unit) http://www.kpsec.freeuk.com/breadb.htm http://www.hiviz.com/kits/instructions/breadboard.htm http://ourworld.compuserve.com/homepages/Bill_Bowden/resistor.htm http://en.wikipedia.org/wiki/Electronic_color_code http://www.electronics-radio.com/articles/test-methods/meters/multimeter- resistance-measurement.php http://www.ee.nmt.edu/*rhb/ee101/labs/2lab/2lab.html http://www.nanotech-now.com/metric-prefix-table.htm http://www.wisc-online.com/objects/index_tj.asp?objID=DCE1802 http://www-ferp.ucsd.edu/najmabadi/CLASS/MAE140/NOTES/analysis-2.pdf http://www.tpub.com/neets/book2/3b.htm http://www.tpub.com/content/neets/14193/css/14193_138.htm http://www.allaboutcircuits.com/vol_1/chpt_13/4.html http://oscilloscope-tutorials.com/Oscilloscope/DisplayControls.asp http://books.google.ca/books?id=ZzHudUMb7WAC&pg=PA26&lpg=PA26 &dq=Dependent+voltage+Source&source=web&ots=78V2Df3xV5&sig= ZKXLpVNMsqZyxiRlYdI858fd84s&hl=en&sa=X&oi=book_result&resnum =2&ct=result#PPA26,M1 http://books.google.ca/books?id=avEjv8zAhQkC&pg=PT615&lpg=PT615 &dq=control+coefficients+voltage+controlled+sources&source=web& ots=QjyOKEW-Dv&sig=H_gLC4hWUTo192z9n9B5djI24w8&hl=en&sa =X&oi=book_result&resnum=9&ct=result#PPT615,M1 The publisher has no responsibility for the persistence or accuracy of URLs for external or third-party Internet websites referred to in this book, and does not guarantee that any content on such websites is, or will remain, accurate or appropriate. 366 Understandable electric circuits Index Active power 279–81, 285–6 Adjustable transformer 343 Admittance 266, 269–72, 299–300 Air-core transformer 337 Alternating current (ac) 227, 229, 260, 263 Ampere 3, 8 Amplitude 230, 232 Ammeter 9, 17, 24, 91 Angular frequency 231–2, 257 Angular velocity 231–2, 257 Apparent power 282–3, 301 Applied voltage 13, 24 Autotransformer 343 Average power 279–81 Average value 236–7, 258 Balanced bridge 90–1, 94 Bandwidth 315–16, 329 Blocking ac 256 Branch 41, 47, 56 Branch current analysis 108–9, 122 Breadboard 26–7 Breakdown voltage 170–1, 190 Capacitance 167–8, 190 Capacitive reactance 254–5, 267 Capacitive susceptance 255, 270, 299 Capacitors 164–7, 190 ac response 254–7 charging 165–6, 190, 209–10 discharging 166–7, 190, 209–10 in parallel 176–7 in series 174–5 Capacitors in series–parallel 178 Center-tapped transformer 343 Charging equations 209 Charging process of an RC circuit 199–201 Characteristics of a capacitor 191, 256, 259 admittance 269–71, 300 impedance 271, 300 an inductor 191, 252, 259 a resistor 191 Chassis ground 70–1, 92 Circuit ground 70 Circuits quantities and their SI units 54 Circuit symbol 5–7 Closed-loop circuit 36–7 Coefficient of the coupling 335, 346 Complex number 240–2, 258–9 Common ground 70–1, 92 Conductance 17, 25, 259, 299 Conductance form of Ohm’s law 20, 25 Controlled source 352–3, 360 Conversion between rectangular and polar forms 241–2, 258 Conversion of dependent sources 353–4 Conventional current flow version 10, 24 Critical frequencies 315–16 Current 8 direction 9–10, 24 source 50–1 sources in series 107–8, 122 triangle 267, 284, 300 Current-controlled current source (CCCS) 352, 360 Current-controlled voltage source (CCVS) 352, 360 Current divider rule (CDR) 76–7, 93–4, 300 Current source !Voltage source 102, 122, 361 Circuit symbol 6, 24 Cutoff frequency 315–16 DC Blocking 172, 252 D and p configuration 83–4, 94 Delta to wye conversion (D !Y) 84–6, 94 Dependent sources 352–4, 360 Dielectric constant 169–70 Differentiation of the sine function in phasor notation 246, 259 Direct current (dc) 228, 257 Discharging equations 207, 222 Discharging process of the RC circuit 204–5 Dot convention 335–6, 346 Double-subscript notation 70–1, 93 Earth ground 70–1, 92 Effective value 237, 239, 258 Electric circuit 4, 24 Electric current 8, 24 Electric power 33 Electrolytic capacitor 192 Electron flow version 10, 24 Electromagnetic field 179–80, 190 Electromotive force (EMF) 11, 13, 24 Energy 32, 56 Energy storage element 166 Energy releasing equations for RL circuit 218 Energy storing equations for RL circuit 214 Energy stored by a capacitor 173 Energy stored by an inductor 185–6 Equivalent parallel capacitance 177 Equivalent parallel inductance 189 Equivalent resistance 65 Equivalent series capacitance 175 Equivalent (total) series resistance 65, 92 Equivalent series inductance 188 Equivalent parallel resistance 74–5, 93 Euler’s formula 241–2 Excitation 197 Factors affecting capacitance 169, 190 Factors affecting inductance 184–5, 191 Factors affecting resistance 15–16, 24 Faraday’s law 180, 190 First-order circuit 195–6, 220, 222 Frequency 229–30, 257 Frequency of series resonance 308 Function generator 260 Half-power frequency 315–16 I–V characteristic 19–20 Ideal current source 50–1, 56 Ideal transformer 338–9, 346 Ideal voltage source 48, 56 Impedance 265–6, 271, 300 angle 284–5, 300 matching 344–5, 347 in series 272–3, 300 in parallel 272–3, 300 triangle 284, 300 Inductance 182–3, 190 Inductive reactance 251–2 Inductive susceptance 251–2, 270, 299 Inductors 182, 190 ac response 250–1, 253 in series 188 in series–parallel 189 in parallel 188–9 Initial conditions 198–9, 221 Initial state 197 Input 197 Iron-core transformer 337–8 Instantaneous power 276–9, 301 Instantaneous value 236–7, 244, 258 International system of units (SI) 53–4 368 Understandable electric circuits Integration of the sine function, in phasor notation 246, 259 Kirchhoff’s current law (KCL) 41–3, 56, 274, 300 Kirchhoff’s voltage law (KVL) 36–8, 56, 274, 300 KVL extension 40 LCZ meter 192 Leakage current 170, 190 Lenz’s law 181–2, 190 Linear circuits 128, 155 Linear network 134 Linear two-terminal network with the sources 135, 155 Linearity property 128 Load 5, 24 Load voltage 13, 24 Loop 47, 56 Maximum power 148–9 Maximum power transfer 147–8, 156, 344 Mesh 47, 56 Mesh current analysis 113, 122–3, 291 Metric prefix 54–5, 57 Milestones of the electric circuits 3–4, 23 Multimeter 28–9, 58–60 Multiple-tapped transformer 343 Mutual inductance 333–4, 346 Millman’s theorem 151–2, 156 Mutually related ref. polarity of V and I 22–3, 25 Natural response 197–8 Network 134 Network with the power supplies 134 Node 47, 56 Node voltage analysis 116–17, 123, 292–3 Norton’s theorem 135–6, 155–6, 296 Ohmmeter 16–17, 24 Ohm’s law 19–20, 25, 250, 252 for a capacitor 172 for an inductor 183 Operations on complex numbers 241 Oscilloscope 260–4 Output 197 Parallel circuit 71–3, 93 Parallel current 73–4, 93 Parallel power 75, 93 Parallel resonance 319, 322, 326, 328 Parallel voltage 73, 93 Pass-band 316 Passing dc 184 Peak value 235–7, 257–8 Peak-peak value 235–7, 258 Period 229–30, 257 Phasor 239–40, 242–4, 259 domain 244, 259 diagram 243, 322 notation 239–40, 259 power 285, 301 Phase difference 232–4, 258 Phase shift 230–2, 257 Polar form 240–2, 258 Potential difference 11–2, 24 Power 32–3, 56–7 of ac circuits 276–7, 301 source 5, 24 triangle 284–5, 300 Power factor 285–7, 301 Power-factor correction 286–7, 301 Practical parallel circuit 325, 327 Quality factor 312–13, 317, 322–3, 329 Resonant circuit 307 RC circuit 199 RC time constant 208–9 Reactance 259, 267, 270, 299 Reactive power 281–2, 301 Real current source 52–3, 56 Real power 279–81 Real voltage source 48–9, 56 Index 369 Rectangular form 240–2, 258 Reference direction of current 20–1, 25 Reference polarity of voltage 21–2, 25 Reference direction of power 34–5, 56 Resistance 14, 24 Resistivity 15–16, 24 Resistors 14 ac response 248–50 colour code 26–7 Requirements of a basic circuit 5, 24 Response 196 Right-hand spiral rule 179 RL circuit 211–2, 216 RL time constant 218–19 RMS value 237–9, 258 Rotating factor 244–5, 259 Schematic 5–6, 24 Selectivity 316–17, 329 Self-inductance (inductance) 182–3, 190 Series circuit 63–4, 92–3 Series current 66, 92 Series–parallel circuit 79, 94 Series power 66 Series resonance 307–8, 317, 328 Short circuit 50 SI units 53–4, 57 SI prefixes 54–5, 57 Single-subscript notation 70–1, 93 Source equivalent conversion 102, 122 Source-free response 197, 220 Source voltage 13, 24 Steady-state 196, 220 Step-down transformer 341–2, 346–7 Step response 196–7, 220 Step-up transformer 340–1, 346 Substitution theorem 152–3, 156 Supernode 46, 56 Superposition theorem 128–9, 155, 293–5 Susceptance 259, 269–70, 299 Switching circuit 198 Symbols and units of electrical quantities 25 t = 0– 198, 221 t = 0+ 198, 221 Thevenin’s theorem 135–6, 155–6, 296 Time constant 208–9, 218, 221–2 Time domain 244, 259 Total power 287–8, 301 Total series resistance 65, 92 Total series voltage 65, 92 Transformer 336–7 Transformer parameters conversion 340, 346 Transient state 196, 220 True power 279, 281 Turns ratio 339, 346 Two-terminal network 134–5 Viewpoints 139 Voltage 11–13, 24 divider 273 drop 13, 24 rise 13, 24 source 48–50 sources in parallel 105–6, 122 sources in series 104–5, 122 triangle 284, 300 Voltage-controlled current source (VCCS) 352, 360 Voltage-controlled voltage source (VCVS) 352, 360 Voltage divider rule (VDR) 67–9, 92, 300 Voltage source !Current source 102, 122 Voltmeter 13, 17, 24, 58 Wheatstone bridge 89 Winding resistance 186–7, 191 Wires 5, 24 Work 31, 56 Wye and delta configurations 83–4 Wye to delta conversion (Y!D) 86–7 Y !D 86–7, 94 Y or T configuration 83–4, 94 Z meter 192 370 Understandable electric circuits IET CIRCUITS, DEVICES AND SYSTEMS SERIES 23 Understandable Electric Circuits Other volumes in this series: Volume 2 Volume 3 Volume 4 Volume 5 Volume 6 Volume 8 Volume 9 Volume 10 Volume 11 Volume 12 Volume 13 Volume 14 Volume 15 Volume 16 Volume 17 Volume 18 Volume 19 Volume 20 Volume 21 Volume 22 Analogue IC design: the current-mode approach C. Toumazou, F.J. Lidgey and D.G. Haigh (Editors) Analogue-digital ASICs: circuit techniques, design tools and applications R.S. Soin, F. Maloberti and J. France (Editors) Algorithmic and knowledge-based CAD for VLSI G.E. Taylor and G. Russell (Editors) Switched currents: an analogue technique for digital technology C. Toumazou, J.B.C. Hughes and N.C. Battersby (Editors) High-frequency circuit engineering F. Nibler et al. Low-power high-frequency microelectronics: a unified approach G. Machado (Editor) VLSI testing: digital and mixed analogue/digital techniques S.L. Hurst Distributed feedback semiconductor lasers J.E. Carroll, J.E.A. Whiteaway and R.G.S. Plumb Selected topics in advanced solid state and fibre optic sensors S.M. Vaezi-Nejad (Editor) Strained silicon heterostructures: materials and devices C.K. Maiti, N.B. Chakrabarti and S.K. Ray RFIC and MMIC design and technology I.D. Robertson and S. Lucyzyn (Editors) Design of high frequency integrated analogue filters Y. Sun (Editor) Foundations of digital signal processing: theory, algorithms and hardware design P. Gaydecki Wireless communications circuits and systems Y. Sun (Editor) The switching function: analysis of power electronic circuits C. Marouchos System on chip: next generation electronics B. Al-Hashimi (Editor) Test and diagnosis of analogue, mixed-signal and RF integrated circuits: the system on chip approach Y. Sun (Editor) Low power and low voltage circuit design with the FGMOS transistor E. Rodriguez-Villegas Technology computer aided design for Si, SiGe and GaAs integrated circuits C.K. Maiti and G.A. Armstrong Nanotechnologies M. Wautelet et al. Understandable Electric Circuits Meizhong Wang The Institution of Engineering and Technology Published by The Institution of Engineering and Technology, London, United Kingdom First edition † 2005 Higher Education Press, China English translation † 2010 The Institution of Engineering and Technology First published 2005 Reprinted 2009 English translation 2010 This publication is copyright under the Berne Convention and the Universal Copyright Convention. All rights reserved. Apart from any fair dealing for the purposes of research or private study, or criticism or review, as permitted under the Copyright, Designs and Patents Act 1988, this publication may be reproduced, stored or transmitted, in any form or by any means, only with the prior permission in writing of the publishers, or in the case of reprographic reproduction in accordance with the terms of licences issued by the Copyright Licensing Agency. Enquiries concerning reproduction outside those terms should be sent to the publisher at the undermentioned address: The Institution of Engineering and Technology Michael Faraday House Six Hills Way, Stevenage Herts, SG1 2AY, United Kingdom www.theiet.org While the author and publisher believe that the information and guidance given in this work are correct, all parties must rely upon their own skill and judgement when making use of them. Neither the author nor publisher assumes any liability to anyone for any loss or damage caused by any error or omission in the work, whether such an error or omission is the result of negligence or any other cause. Any and all such liability is disclaimed. The moral rights of the author to be identified as author of this work have been asserted by him in accordance with the Copyright, Designs and Patents Act 1988. British Library Cataloguing in Publication Data A catalogue record for this product is available from the British Library ISBN 978-0-86341-952-2 (paperback) ISBN 978-1-84919-114-2 (PDF) Typeset in India by MPS Ltd, A Macmillan Company Printed in the UK by CPI Antony Rowe, Chippenham Contents Preface 1 Basic concepts of electric circuits Objectives 1.1 Introduction 1.1.1 Why study electric circuits? 1.1.2 Careers in electrical, electronic and computer engineering 1.1.3 Milestones of electric circuit theory 1.2 Electric circuits and schematic diagrams 1.2.1 Basic electric circuits 1.2.2 Circuit schematics (diagrams) and symbols 1.3 Electric current 1.3.1 Current 1.3.2 Ammeter 1.3.3 The direction of electric current 1.4 Electric voltage 1.4.1 Voltage/electromotive force 1.4.2 Potential difference/voltage 1.4.3 Voltmeter 1.5 Resistance and Ohm’s law 1.5.1 Resistor 1.5.2 Factors affecting resistance 1.5.3 Ohmmeter 1.5.4 Conductance 1.5.5 Ohm’s law 1.5.6 Memory aid for Ohm’s law 1.5.7 The experimental circuit of Ohm’s law 1.5.8 I–V characteristic of Ohm’s law 1.5.9 Conductance form of Ohm’s law 1.6 Reference direction of voltage and current 1.6.1 Reference direction of current 1.6.2 Reference polarity of voltage 1.6.3 Mutually related reference polarity of current/voltage Summary Experiment 1: Resistor colour code xiii 1 1 1 1 2 3 4 4 5 8 8 9 9 10 10 11 13 14 14 15 16 17 17 18 19 19 20 20 20 21 22 23 25 vi 2 Understandable electric circuits Basic laws of electric circuits Objectives 2.1 Power and Energy 2.1.1 Work 2.1.2 Energy 2.1.3 Power 2.1.4 The reference direction of power 2.2 Kirchhoff’s voltage law (KVL) 2.2.1 Closed-loop circuit 2.2.2 Kirchhoff’s voltage law (KVL) 2.2.3 KVL #2 2.2.4 Experimental circuit of KVL 2.2.5 KVL extension 2.2.6 The physical property of KVL 2.3 Kirchhoff’s current law (KCL) 2.3.1 KCL #1 2.3.2 KCL #2 2.3.3 Physical property of KCL 2.3.4 Procedure to solve a complicated problem 2.3.5 Supernode 2.3.6 Several important circuit terminologies 2.4 Voltage source and current source 2.4.1 Voltage source 2.4.1.1 Ideal voltage source 2.4.1.2 Real voltage source 2.4.2 Current source 2.4.2.1 Ideal current source 2.4.2.2 Real current source 2.5 International units for circuit quantities 2.5.1 International system of units (SI) 2.5.2 Metric prefixes (SI prefixes) Summary Experiment 2: KVL and KCL Series–parallel resistive circuits Objectives 3.1 Series resistive circuits and voltage-divider rule 3.1.1 Series resistive circuits 3.1.1.1 Total series voltage 3.1.1.2 Total series resistance (or equivalent resistance) 3.1.1.3 Series current 3.1.1.4 Series power 3.1.2 Voltage-divider rule (VDR) 3.1.3 Circuit ground 31 31 31 31 32 32 34 36 36 36 38 38 40 41 41 41 41 44 44 46 47 47 48 48 48 50 50 52 53 53 54 55 57 63 63 63 63 65 65 66 66 67 70 3 4.2.4.2.2.1 RY and RD 3.1.2. mesh current analysis Summary Experiment 4: Mesh current analysis and nodal voltage analysis The network theorems Objectives 5.Contents Parallel resistive circuits and the current-divider rule 3.1.1 Procedure for applying mesh current analysis 4.4 Current sources in series 4.2 Method for analysing series–parallel circuits 3.2 vii 71 71 73 73 74 75 76 79 80 81 83 83 84 86 87 89 90 91 92 95 101 101 101 101 104 104 105 106 107 108 109 113 114 116 117 121 122 123 127 127 128 128 128 5 .2.2.1 Voltage source.2.3 Current sources in parallel 4.1.4.2.1.2 Sources in series and parallel 4.1.5 Balanced bridge 3.1 Equivalent resistance 3.3.1.2 Voltage sources in parallel 4.3.1 Procedure for applying the node voltage analysis 4.2 Current-divider rule (CDR) 3.1 Voltage sources in series 4.4.1 Procedure for applying the branch circuit analysis 4.1.4 Wye (Y) and delta (D) configurations and their equivalent conversions 3.4.1 Source equivalent conversion 4.3.2.2.3 Wye to delta conversion (Y!D) 3.1.2 Parallel current 3.1.1.4 Nodal voltage analysis 4.3 Equivalent parallel resistance 3.1 Parallel voltage 3.1 Parallel resistive circuits 3.2 Delta to wye conversion (D!Y) 3.3 Mesh current analysis 4.4. current source and their equivalent conversions 4.1 Superposition theorem 5.3.1 Introduction 5.2 Steps to apply the superposition theorem 3.2.1 Wye and delta configurations 3.1.4 Total parallel power 3.4 Using D!Y conversion to simplify bridge circuits 3.5 Node voltage analysis vs.1.3 Series–parallel resistive circuits 3.6 Measure unknown resistors using the balanced bridge Summary Experiment 3: Series–parallel resistive circuits 4 Methods of DC circuit analysis Objectives 4.2 Branch current analysis 4.4.4. 3.4 Millman’s and substitution theorems 5.1.2 Steps to apply Thevenin’s and Norton’s theorems 5.3.3 Maximum power transfer 5.1 Electromagnetic field 6.3 Viewpoints of the theorems 5.1 The construction of a capacitor 6.7 Leakage current 6.1 Capacitors in series 6.2.3 Self-inductance 6.3 Capacitors in series–parallel 6.3.3.1.3.4.3.2.1.4.4.2.3.1.2.2 Thevenin’s and Norton’s theorems 5.1.3.1 Inductors in series 6.6 Factors affecting capacitance 6.3 Inductor 6.1 Millman’s theorem 5.1.3 Energy storage element 6.2 Capacitors in parallel 6.10 Energy stored by a capacitor 6.6 The energy stored by an inductor 6.4.8 Breakdown voltage 6.1.3 Lenz’s law 6.3.2 Inductors in parallel 6.2 Inductor 6.1.1.2 Substitution theorem Summary Experiment 5A: Superposition theorem Experiment 5B: Thevenin’s and Norton’s theorems 133 133 135 139 147 151 151 152 155 156 158 163 163 164 164 165 166 166 167 169 170 170 171 173 174 174 176 178 179 179 179 180 181 182 182 183 184 185 186 188 188 188 189 6 Capacitors and inductors Objectives 6.1 Capacitor 6.1 Electromagnetism induction 6.1.2 Faraday’s law 6.9 Relationship between the current and voltage of a capacitor 6.2.1.7 Winding resistor of an inductor 6.1.3 Inductors in series–parallel .5 Capacitance 6.4 Relationship between inductor voltage and current 6.5 Factors affecting inductance 6.viii Understandable electric circuits 5.4.4 Inductors in series and parallel 6.2 Charging a capacitor 6.2.2 Capacitors in series and parallel 6.1.3.1 Introduction 5.4 Discharging a capacitor 6. 5.1 The discharging process of the RC circuit 7.2 The step response of an RC circuit 7.3 RL time constant t 7.2.1 The transient response 7.1.1.3.4 Three important components of a sine function 8.2 Quantity analysis for the charging process of the RC circuit 7.3 Average value 8.1.3.5 Phase difference of the sine function 8.4.1.2 Quantitative analysis of the energy storing process in an RL circuit 7.3 The source-free response of the RC circuit 7.2 Quantity analysis of the RC discharging process 7.1.4.2 Instantaneous value 8.5.5.1 Introduction to alternating current (AC) 8.1 Energy storing process of the RL circuit 7.2.2 DC and AC waveforms 8.1 The charging process of an RC circuit 7.1 Energy releasing process of an RL circuit 7.4 The step response of an RL circuit 7.2.5 Source-free response of an RL circuit 7.1.2.1 The first-order circuit and its transient response 7.2.2 Quantity analysis of the energy release process of an RL circuit 7.1 Peak and peak–peak value 8.1 The difference between DC and AC 8.1.3 Period and frequency 8.1.4 The RC time constant and charging/discharging 7.Contents Summary Experiment 6: Capacitors 7 Transient analysis of circuits Objectives 7.4 The RL time constant and the energy storing and releasing Summary Experiment 7: The first-order circuit (RC circuit) Fundamentals of AC circuits Objectives 8.3.5.3.3 RC time constant t 7.2.2 Circuit responses 7.3 The initial condition of the dynamic circuit 7.4 Root mean square (RMS) value ix 190 191 195 195 195 195 196 198 199 199 201 204 204 205 208 209 211 212 213 215 215 216 218 219 220 221 227 227 227 227 228 229 230 232 235 235 236 236 237 8 .2 Sinusoidal AC quantity 8. 2.4.3 Reactive power Q 9.3 Characteristics of the impedance 9.3.1 Introduction to phasor notation 8.1.2 Impedance in series and parallel 9.6 Power factor (PF) 9.3 Power in AC circuits 9.4 Thevenin’s and Norton’s theorems Summary Experiment 9: Sinusoidal AC circuits .2 Node voltage analysis 9.1 Impedance of series and parallel circuits 9.1.2.7 Total power 9.3 239 239 240 242 243 244 246 248 248 250 254 257 260 265 265 265 265 266 267 269 272 272 273 274 276 276 279 281 282 284 285 287 290 291 292 293 296 299 302 9 Methods of AC circuit analysis Objectives 9.4.2 Inductor’s AC response 8.1 Instantaneous power p 9.3.5 Power triangle 9.2 Voltage divider and current divider rules 9.4 Methods of analysing AC circuits 9.1 Mesh current analysis 9.1 Resistor’s AC response 8.2.4.2 Active power P (or average power) 9.3.4 Phasor diagram 8.4 Apparent power S 9.1. inductors and capacitors in sinusoidal AC circuits 8.4.3.4.3.1 Impedance 9.3.1.5 Rotating factor 8.3 Capacitor’s AC response Summary Experiment 8: Measuring DC and AC voltages using the oscilloscope 8.3 Superposition theorem 9.4 Resistors.3 Phasor 8.3.x Understandable electric circuits Phasors 8.4.3.2 Complex numbers review 8.2 Admittance 9.4.3.3.3.3.1 Impedance and admittance 9.3 The phasor forms of KVL and KCL 9.4 Characteristics of the admittance 9.3.6 Differentiation and integration of the phasor 8. 1 Series resonance 10.1.1 The bandwidth of series resonance 10.2 Resonant frequency 10.6 Quality factor 10.2 The selectivity of series resonance 10.1.2 Air-core transformer 11.1. XC and Z versus f 10.3.1.4 The practical parallel resonant circuit 10.3 Admittance of parallel resonance 10.5 Phasor diagram of series resonance 10.2 Bandwidth and selectivity 10.4.4 Current of series resonance 10.7 Phase response of series resonance 10.5 Phasor diagram of parallel resonance 10.3 Impedance of series resonance 10.Contents 10 RLC circuits and resonance Objectives 10.1 Mutual inductance 11.2.3.3.1 Introduction 10.3.1 Transformer 11.3.3.1 Parallel resonance summary 10.4.2.2.3 Iron-core transformer 11.3.1.3.1.1.2 Basic transformer 11.1.1 Introduction 10.3 Parallel resonance 10.4 Current of parallel resonance 10.4.3 The quality factor and selectivity 10.3.7 Current of parallel resonance 10.6 Response curves of XL.9 Voltage of series resonant 10.2.2 Dot convention 11.2 Frequency of parallel resonance 10.2.1.3 Applications of the resonance Summary Experiment 10: Series resonant circuit Mutual inductance and transformers Objectives 11.1.8 Quality factor 10.1 Mutual inductance and coefficient of coupling 11.2.8 Bandwidth of parallel resonance 10.3.8.1 Resonant admittance 10.4 Ideal transformer xi 307 307 307 307 308 309 309 310 310 311 312 313 315 315 316 317 319 319 319 320 320 321 322 322 323 324 324 325 325 326 327 328 329 333 333 333 333 335 336 336 337 337 338 11 .2.1 Series resonance summary 10.1.2.2 Frequency of series resonance 10. 3 Step-up and step-down transformers 11.2 Equivalent conversion of dependent sources 12.4.4 Other types of transformers 11.3.3.3.1 Maximum power transfer 11.1.2 Step-down transformer 11.4.4 Impedance matching 11.3 Applications of step-up and step-down transformers 11.2 Analysing circuits with dependent sources Summary Appendix A: Greek alphabet Appendix B: Differentiation of the phasor Bibliography Index .1 Dependent (or controlled) sources 12.1 Step-up transformer 11.xii Understandable electric circuits 11.3.1 Dependent sources 12.2 Impedance matching Summary Experiment 11: Transformer 340 340 341 342 343 344 344 345 346 347 351 351 352 352 353 355 360 363 364 365 367 12 Circuits with dependent sources Objectives 12.1. Tables organizing and summarizing variables. including current. concepts. resistor. in 2005 and reprinted in 2009. inductor. which clearly present the important information. Objectives at the beginning of each chapter to highlight to readers the knowledge that is expected to be obtained in the chapter.Preface The book Understandable Electric Circuits is based on my teaching notes for the circuit analysis course that I have taught for many years at Canadian and Chinese institutions. the network theorems (superposition. Summary at the end of each chapter to emphasize the key points and formulas in the chapter. basic circuit laws/rules (Ohm’s law. voltage/current divider rules). This unique and well-structured book provides understandable and effective introduction to the fundamentals of DC/AC circuits. laws/rules and formulas to emphasize and locate important facts and points. Outlining (boxing) of all important principles. technicians. Suitable readers This book is intended for college and university students. engineers or any other professionals who require a solid foundation in the basics of electric circuits. Thevenin’s/Norton’s theorems. which was published by the ‘Higher Education Press’. dependent/independent sources. values and formulas. Millman’s and substitution theorems). admittance. series/ parallel and wye/delta circuits. mutual inductance and transformers and more. methods of DC/AC analysis (branch current and mesh/node analysis). impedance. capacitor. Laboratory experiments at the end of each chapter are convenient for hands-on practice. power. maximum power transfer. the largest and the most prominent publisher of educational books in China. They also include how to use basic electrical instruments such as the multimeter and oscilloscope. which is convenient for students reviewing before exams. . technologists. transient analysis. KVL/KCL. procedures and examples. voltage. The English version of this book continues in the spirit of its successful Chinese version. RLC circuits and resonance. Key features As an aid to readers. the book provides some noteworthy features: ● ● ● ● ● ● Clear and easy-to-understand written style. They deserve sincere acknowledgement for their time and energy. My daughter Alice Wang (a busy PhD student). who deserves a special acknowledgement for her patience and dedication to editing some chapters and all the experiments in the book. the concepts. Felicity Hull. communications. electronics. for translating several chapters of this book from Chinese to English even though he is very weak after having several operations. an electrical and computer engineer. for taking the time to edit some chapters of this book. Acknowledgements Special thanks to Lisa Reading. and Jo Hughes. Li Wang (an electronics and physics instructor). marketing executive. control and automation. books and journals sales manager. biomedical imaging and more. industrial instrumentation. I would also like to express my gratitude to Ying Nan. Suzanne Bishop. the commissioning editor for books at the Institution of Engineering and Technology. it is my good fortune to have help and support from my family members: My husband. I also appreciate the support from Bianca Campbell. . embedded systems. Clear step-by-step procedures for applying methods of DC/AC analysis and network theorems make this book easy for readers to learn electric circuits themselves. I really appreciate her belief in my ability to write this book. signal processing. power systems (including renewable energy). electronic and computer engineering such as electrical. My special thanks to all of them. laws/rules and theorems are explained in an easy-to-understand style. production controller. In addition. It provides readers with the necessary foundation for DC/AC circuits in related fields. electrical apparatus and machines. power electronics. And finally. marketing manager.xiv Understandable electric circuits It targets an audience from all sectors in the fields of electrical. and also proofreading the entire book. new terms. In addition. I would like to express my sincere gratitude to Ramya Srinivasan (project manager of my production process from MPS Ltd) for her highly efficient work and good guidance/suggestions that have helped to refine the writing of this book. my son Evan Wang has given a hand in editing several chapters. It is also suitable for non-electrical or electronics readers. To make this book more reader friendly. nanotechnology. computers. and her help and support in publishing it. you will be able to: ● ● ● ● ● ● ● ● ● ● understand the purpose of studying electric circuits know the requirements of a basic electric circuit become familiar with circuit symbols become familiar with the schematics of electric circuits understand the concepts of current and voltage understand resistance and its characteristics become familiar with the ammeter. electronic and computer engineering and other related areas. When you start reading this book.1. Our everyday life would be unthinkable without electricity or the use of electronic products. Moreover. electronic and computer engineering has made and continues to make incredible contributions to most aspects of human society – a truth that cannot be neglected. it may have a bigger impact on human .1 Introduction 1.Chapter 1 Basic concepts of electric circuits Objectives After completing this chapter. perhaps you have already chosen the electrical or the electronic fields as your professional goal – a wise choice! Electrical.1 Why study electric circuits? Electrical energy is the great driving force and the supporting pillar for modern industry and civilization. Any complex electrical and electronic device or control system is founded from the basic theory of electric circuits. Only when you have grasped and understood the basic concepts and principles of electric circuits can you further study electrical. voltmeter and ohmmeter know the difference between the electron flow and the conventional current flow know the concept of reference directions of voltage and current know how to apply Ohm’s law 1. 2 Understandable electric circuits civilization in the future. experts forecast that demand for professionals in this field will grow continuously. electronic and computer technology is developing so rapidly that many career options exist for those who have chosen this field. As long as you have gained a solid foundation in electric circuits and electronics. electronic and computer engineering Nowadays. 1. the training that most employers provide in their branches will lead you into a brand new professional career very quickly.1. There are many types of jobs for electrical and electronic engineering technology.2 Careers in electrical. Only a partial list is as follows: ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● Electrical engineer Electronics engineer Electrical design engineer Control and automation engineer Process and system engineer Instrument engineer Robotics engineer Product engineer Field engineer Reliability engineer Integrated circuits (IC) design engineer Computer engineer Power electronics engineer Electrical and electronics engineering professor/lecturer Designer and technologist Biomedical engineering technologist Electrical and electronics technician Hydro technician Electrician Equipment maintenance technician Electronic test technician Calibration/lab technician Technical writer for electronic products Electronic repair Electrical and electronic technicians. Reading this book or other electric circuit book is a first step into the electrical. so you definitely do not want to miss this good opportunity. engineers and experts will be in demand in the future. technologists. Therefore. electrical. This is good news for people who have chosen these areas of study. electronic and computer world that will introduce you to the foundation of the professions in these areas. . this is known as Lenz’s law. He discovered that relative motion of the magnetic field and conductor can produce electric current. and the principle of charge interactions (attraction or repulsion of positive and negative electric charges). Watt is the unit of power. a Baltic German physicist. Coulomb developed Coulomb’s law. ● ● ● ● ● ● ● ● ● Coulomb is the unit of electric charge. He constructed the first electric battery that could produce a reliable. an Italian physicist. Ohm is the unit of resistance. Faraday also discovered that the electric current originates from the chemical reaction that occurs between two metallic conductors. and formulated the most famous electric circuit law – Ohm’s law. and all are important milestones in the field of electric engineering. it was named in the honour of Heinrich Rudolf Hertz (1857–1894). Volt is the unit of voltage.Basic concepts of electric circuits 3 1. a French physicist. The laws and physical quantities that they discovered are named after them. Lenz’s law was named in the honour of Heinrich Friedrich Emil Lenz (1804– 1865). He discovered that the polarity of the induced current that is produced in the conductor of the magnetic field always resists the change of its induced voltage. Hertz is the unit of frequency. a Scottish physicist and mathematician. a Scottish engineer and inventor. which is the definition of the electrostatic force of attraction and repulsion. ´ Ampere is the unit of electric current. He established the relationship between voltage. it was named in the honour of Charles Augustin de Coulomb (1736–1806). it was named in the honour of Georg Simon Ohm (1789–1854).3 Milestones of electric circuit theory Many early scientists have made great contributions in developing the theorems of electrical circuits. it was named in the honour of Andre` Marie Ampere (1775–1836). He . which we know today as the Faraday’s law of electromagnetic induction. We list here only the ones that are described in this book. it was named in the honour of Alessandro Volta (1745–1827). He made great improvements in the steam engine and made important contributions in the area of magnetic fields. a German physicist. it was named in the honour of Michael Faraday (1791–1867). a German physicist and mathematician. Maxwell is the unit of magnetic flux. it was named in the honour of James Watt (1736–1819). an English physicist and chemist.1. it was named in the honour of James Clerk Maxwell (1831–1879). He was one of the main discoverers of electromagnetism and is best known for defining a method to measure the flow of current. a French physicist. The German physicist Wilhelm Eduard Weber (1804–1891) shares the honour with Maxwell (1 Wb ¼ 108 Mx). current and resistance. steady current. Maxwell had established the Maxwell’s equations that represent perfect ways to state the fundamentals of electricity and magnetism. Faraday is the unit of capacitance. Being familiar with them will be beneficial for further study of electric circuits. 1. wires. an electric circuit can be defined as a sum of all electric components in the closed loop of pathway with flowing electric charges. ● ● The majority of the laws and units of measurement stated above will be used in the later chapters of this book. but is never lost. More specifically. Electric circuit A closed loop of pathway with electric charges or current flowing through it. an English physicist. Hertz confirmed Maxwell’s electromagnetic theory.4 Understandable electric circuits was the first person to broadcast and receive radio waves. capacitors. Wires connect the power .1).2. Henry is the unit of inductance. power sources. such as an electric circuit that includes resistors. a Scottish-American scientist. etc.2 Electric circuits and schematic diagrams 1.1 Requirements of a basic circuit A basic electric circuit contains three components: the power supply. inductors. Joule’s law was named after him and states that heat will be produced in an electrical conductor. the load and the wires (conductors) (Figure 1. He discovered selfinduction and mutual inductance. Through the low-frequency microwave experiment. it was named in the honour of Joseph Henry (1797–1878). switches. He made great contributions in discovering the law of the conservation of energy.1 Basic electric circuits An electric circuit is a closed loop of pathway with electric charges flowing through it. Wire Power Supply Wire Load Figure 1. (these electric components will be explained later). it was named in the honour of James Prescott Joule (1818–1889). This law states that energy may transform from one form into another. Joule is the unit of energy. The electric battery and generator are examples of power supply. electric stove. it can convert other forms of energy to electrical energy. A power supply is a device that supplies electrical energy to the load of the circuit. electric fan and speaker are all electric loads. the solar generator converts solar energy into electrical energy.Basic concepts of electric circuits 5 supply and the load. . and consumes electrical energy. Circuit diagrams can make electric circuits easier to understand.2. the wind generator converts wind energy into electrical energy. and carry electric charges through the circuit. The load may be any device that can receive electrical energy and convert it into other forms of energy. and carry electric charges through the circuit.2 is an example of a simple electric circuit – a flashlight (or electric torch) circuit. Load is a device that is connected to the output terminal of an electric circuit. 1. In this circuit the battery is the power supply and the small light bulb is the load. the thermo generator converts heat energy into electrical energy. Wires connect the components in a circuit together. the nuclear power generator converts nuclear energy into electrical energy. It consumes or absorbs electrical energy from the source. the hydroelectric generator converts hydroenergy (the energy of moving water) into electrical energy. Load is a device that is usually connected to the output terminal of an electric circuit. Requirements of a basic circuit ● ● ● Power supply (power source) is a device that supplies electrical energy to a load. Figure 1. and they are connected together by wires. For example: ● ● ● ● ● ● the battery converts chemical energy into electrical energy. it can convert the other energy forms into electrical energy.2 Circuit schematics (diagrams) and symbols Studying electric circuits usually requires drawing or recognizing circuit diagrams. electric motor. light bulb. For example: ● ● ● ● ● electric lamp converts electrical energy into light energy electric stove converts electrical energy into heat energy electric motor converts electrical energy into mechanical energy electric fan converts electrical energy into wind energy speaker converts electrical energy into sound energy Therefore. A schematic is a simplified circuit diagram that shows the interconnection of circuit components. .1. Further study of this book will help you understand all the circuit elements in Table 1. It uses standard graphic circuit symbols according to the layout of the actual circuit connection.1 lists some commonly used electric circuit symbols in this book. Schematics are represented by circuit symbols according to the layout of the actual circuit connection. The different circuit components are represented by different circuit symbols. but when studying more theories of electric circuits.2. power supply. It is not very difficult to draw a realistic pictorial representation of the flashlight circuit as shown in Figure 1. electric motor and other loads can be represented by a circuit symbol – the resistor R.2) is shown in Figure 1. The more common electric circuits are usually represented by schematics. since all of them have the same characteristic of converting electrical energy into other forms of energies and consuming electrical energy. they are represented by the same circuit symbol – the DC power supply E. Table 1. The circuit symbols are the idealization and approximation of the actual circuit components. etc.2 The flashlight circuit analyse and calculate. For example. The schematic of the flashlight circuit (Figure 1.6 Understandable electric circuits (Switch) Figure 1. This is a way to draw circuit diagrams far more quickly and easily. both the battery and the direct current (DC) generator can convert other energy forms into electrical energy and produce DC voltage. inductor. switch. E E The electric lamp. Therefore.3. The most commonly used circuit symbols are the resistor. capacitor. ground. circuits can be more and more complex and drawing the pictorial representation of the circuits will not be very realistic. electric stove. 1 The commonly used circuit schematic symbols Component DC power supply AC power supply Current source Lamp Connected wires Unconnected wires Fixed resistor Variable resistor Capacitor Inductor Switch Speaker Ground Fuse Ohmmeter Ammeter Voltmeter Transformer Ω A V 7 Circuit symbol or Figure 1.Basic concepts of electric circuits Table 1.3 Schematic of the flashlight circuit . power supply and load). If Q represents the amount of charges that is moving past a point at time t.1 Current Although we cannot see electric charges or electric current in the electric circuits. These circuit quantities are very important to study in electric circuits. 1. power. voltage. they are analogous to the flow of water in a water hose or pipe. Electric current is measured by the amount of electric charges that flows past a given point at a certain time interval in an electric circuit.). electric current is a flow of electric charges through an electric circuit (wires. then the current I is: Charge Time Q t Current ¼ or I¼ If you have learned calculus. etc. litres or gallons per minute or hour. . Water current is a flow of water through a water circuit (faucet. pipe or hose. Water is measured in litres or gallons. so you can measure the amount of water that flows out of the tap at certain time intervals. Electric current I ● ● Current is the flow of electric charges through an electric circuit. This section will discuss one of them – the electric current. and they will be used throughout this book. current also can be expressed by the derivative: i ¼ dq=dt.3 Electric current There are several key circuit quantities in electric circuit theory: electric current. i. Current I is measured by the amount of charges Q that flows past a given point at a certain time t: I ¼ Q/t Quantity Charge Time Current Quantity symbol Q t I Unit Coulombs Seconds Amperes Unit symbol C s A Note: Italic letters have been used to represent the quantity symbols and non-italic letters to represent unit symbols.8 Understandable electric circuits 1. etc.e.3. . . and they assumed at that time the current was a flow of .25 1018 . as shown in Figure 1. .25 6 1018 charges (1 C % 6.. A Figure 1..3. . since 1 C is approximately equal to 6.5..4. . . . . ..3 The direction of electric current When early scientists started to work with electricity. .. . . . . .. ..Basic concepts of electric circuits 9 A current of 1 A means that there is 1 C of electric charge passing through a given cross-sectional area of wire in 1 s: 1A ¼ 1C 1s More precisely. .. . 1 A of current actually means there are about 6. There are 6.3. and its symbol is A .2 Ammeter Ammeter is an instrument that can be used to measure current..4 1 A of current Example 1. .25 1018 charges passing through this given cross-sectional area in 1 s Figure 1. .25 6 1018 charges passing through a given cross-sectional area of wire in 1 s. . .. as shown in Figure 1. . It must be connected in series with the circuit to measure current. . . . . . .1: If a charge of 100 C passes through a given cross-sectional area of wire in 50 s. the structure of atoms was not very clear. . .25 6 1018 charges). 6...5 Measuring current with an ammeter 1. what is the current? Solution: Since Q ¼ 100 C and t ¼ 50 s I¼ Q 100 C ¼ ¼ 2A t 50 s 1. . .. . . . 4 Electric voltage 1.4. there are two methods to express the direction of electric current. I I (a) Conventional current flow (b) Electron flow Figure 1. in which the current is defined as a flow of negative charges (electrons) from the negative terminal of a power supply unit to its positive terminal. In this book.6.1 Voltage/electromotive force We have analysed the flow of water in the water circuit to the flow of electric current in the electric circuit.10 Understandable electric circuits positive charges (protons) from the positive terminal of a power supply (such as a battery) to its negative terminal. in which the current is defined as a flow of positive charges (protons) from the positive terminal of a power supply to its negative terminal. The concept of a water circuit can help develop an understanding of another important circuit quantity – voltage. measurement and applications of the electric circuits as long as one method is used consistently. But by the time the real direction of current flow was discovered. One is known as the conventional current flow version. and it will not affect the analysis. scientists discovered that electric current is in fact a flow of negative charges (electrons) from the negative terminal of a power supply to its positive terminal. the conventional current flow version is used. a flow of positive charges (protons) from the positive terminal of a power supply to its negative terminal had already been well established and used commonly in electrical circuitry. it will make no difference as to which method is used. The other is called electrons flow version. Conventional and electron current flow version ● ● Conventional current flow is defined as a flow of positive charges (protons) from the positive terminal of a power supply to its negative terminal. calculation. Electron flow version is defined as a flow of negative charges (electrons) from the negative terminal of a power supply to its positive terminal. Which way does electric charge really flow? Later on. . design. 1.6 The direction of electric current Because the charge or current cannot be seen in electric circuits. These two methods are shown in Figure 1. Currently. as shown in Figure 1. A B Figure 1. Voltage is responsible for the pushing and pulling of electrons or current through an electric circuit. and its unit is volts (non-italic letter V). EMF produced by a voltage source is analogous to water pressure produced by a pump in a water circuit. EMF is symbolized by E. and the negative electrode of the battery repels the electrons. the positive electrode of the battery attracts the negative charges (electrons).4. The higher the voltage. 1. the flashlight will not work. electric pressure or voltage is required for an electric circuit. Just as water pressure is required for a water gun or water circuit. The trigger of a water gun is attached to a pump that squirts water out of a tiny hole at the muzzle.2 Potential difference/voltage Assuming there are two water tanks A and B. When EMF is exerted on a circuit. Once the battery is connected to the load (lamp) by wires.7. If there is no pressure from the gun (the trigger is not pressed). It is the electric pressure or force that is supplied by a voltage source. The battery is one example of a voltage source that produces electromotive force (EMF) between its two terminals. water will flow from tank A to B only when tank A has a higher water level than tank B.Basic concepts of electric circuits 11 The concept of voltage works on the principle of a water gun.2. a current cannot form in a specific direction.7 Water-level difference . Let us further analyse the voltage by using the previous flashlight or torch circuit in Figure 1. Low-pressure squirting produces thin streams of water over a short distance. it moves electrons around the circuit or causes current to flow through the circuit since EMF is actually ‘the electron-moving force’. Electromotive force (EMF) EMF is an electric pressure or force that is supplied by a voltage source. Since electric charges in the wire (conductor) randomly drift in different directions. while high pressure produces a very powerful stream over a longer distance. the greater the current will be. which causes electric current to flow in a circuit. there will be no water out of the muzzle. This causes the electrons to flow in one direction and produce electric current in the circuit. which causes current to flow in a circuit. Voltage is symbolized by V (italic letter). If only a small lamp is connected with wires without a battery in this circuit. and its unit is also volts (V). W = 1J Q = 1C a V = 1V b Figure 1.12 Understandable electric circuits Common sense tells us that ‘water flows to the lower end’. The formula may be expressed as: Voltage ¼ Work Charge or V ¼ W Q If you have learned calculus. This potential difference or voltage is produced by the EMF of the voltage source. so water will only flow when there is a water-level difference. As water will flow between two places in a water circuit only when there is a water-level difference. if a light bulb is continuously kept on. Quantity Voltage Work (energy) Charge Quantity symbol V W Q Unit Volt Joule Coulomb Unit symbol V J C For example. Voltage V (or potential difference) V is the amount of energy or work required to move electrons between two points: V ¼ W/Q. It is the water-level difference that produces the potential energy for tank A. if 1 J of energy is used to move a 1 C charge from point a to b. i.8 Potential difference or voltage . current will flow between two points in an electric circuit only when there is an electrical potential difference.8. to maintain continuous movement of electrons in the circuit.e. voltage can also be expressed by the derivative v ¼ dw/dq. For instance. the two terminals of the lamp need to have an electrical potential difference. and it is the amount of energy or work that would be required to move electrons between two points. This concept can also be used in the electric circuit. it will have a 1 V potential difference or voltage across two points. and work is done when water flows from tank A to B. Work is represented by W and measured in joules (J). as shown in Figure 1. as shown in Figure 1. applied voltage. V Figure 1. The voltage or the potential difference always exists between two points.9. The voltmeter should be connected in parallel with the circuit component to measure voltage. the two are used interchangeably. such as the source voltage. Voltage across the two terminals of the load is called the load voltage. Voltage drop: Voltage across a component when current flows from a higher potential point to a lower potential point. Current will flow between two points in a circuit only when there is a potential difference. Voltage rise: Voltage across a component when current flows from a lower potential point to a higher potential point. Voltage across a component in a circuit is sometimes called voltage drop when current flows from a higher potential point to a lower potential point in the circuit.3 Voltmeter Voltmeter is an instrument that can be used to measure voltage. There are different names representing voltage or potential difference in electric circuits.4. 1. or voltage rise when current flows from a lower potential point to a higher potential point in the circuit. voltage rise. Its symbol is V . What are the differences between them? The EMF can be called source voltage or applied voltage since it is supplied by a voltage source and applied to the load in a circuit. Source voltage or applied voltage (E or VS): EMF can be called source voltage or applied voltage (the EMF is supplied by a voltage source and applied to the load in a circuit).Basic concepts of electric circuits 13 Although voltage and potential difference are not exactly same. load voltage. voltage drop.9 Measuring voltage with a voltmeter . Load voltage (V): Voltage across the two terminals of the load. etc. The resistor and resistance of a circuit have different meanings. The resistance is a measure of a material’s opposition to the flow of current. A variable resistor has a resistance value that can be easily changed or adjusted manually or automatically. This is because the rocks (water resistance) ‘resist’ the flow of water. but all of them belong to one of the two categories. A resistor is a component of a circuit. As mentioned in section 1. For example. Resistance (R): The measure of a material’s opposition to the flow of current.5 Resistance and Ohm’s law 1. shapes and sizes.2. and its unit is ohms (O). materials. motor and other such loads may be represented by resistor R because once this kind of load is connected to an electric circuit. Sometimes resistor R will need to be adjusted to a different level for different applications. A resistor can also be used to maintain a safe current level in a circuit. The resistor (current resistance) ‘resists’ the flow of electrical current. Resistor (R): A two-terminal component of a circuit that limits the flow of current. either fixed or variable. A resistor is a two-terminal component of a circuit that is designed to resist or limit the flow of current. What will happen when we throw some rocks into a small creek? The speed of the water current will slow down in the creek. A fixed resistor has a ‘fixed’ resistance value and cannot be changed. A similar concept may also be used in an electric circuit.5. electric stove. the intensity of light of an adjustable lamp can be adjusted by using resistors. cause resistance and reduce current in the circuit.14 Understandable electric circuits 1. Resistors are of many different types. The higher the value of resistance. The resistance of a conductor is a measure of how difficult it is to resist the current flow. the smaller the current will be. the lamp.1 Resistor Let us use the water current as an example again to explain the resistor. There are a variety of resistors with different resistance values for different applications. Symbols of the resistor ● ● Fixed resistor Variable resistor or . it will consume electrical energy. 68 6 1078 2.10 Factors affecting resistance ● ● ● ● Cross-sectional area of the wire A: More water will flow through a wider pipe than that through a narrow pipe. every conductor that makes up the wires has some level of resistance no matter what kind of material it is made from. i.2 Factors affecting resistance There is no ‘perfect’ electrical conductor. Similarly. Table 1.2 6 1077 .2 Table of resistivities (r) Material Copper Gold Aluminium Silver Iron Brass Nichrome Tin Lead Resistivity r (O Á m) 1. Table 1.5. the less the resistance in the wire and the more the flow of current. the larger the diameter of the wire. The different materials have different resistivity. Temperature T: Resistivity of a material is dependent upon the temperature surrounding the material. Resistivity increases with an increase in temperature for most materials.09 6 1077 2. Resistivity r: It is a measure for the opposition to flowing current through a material of wire.e. or how difficult it is for current to flow through a material.1 6 1076 1.82 6 1078 1.8 6 1077 1.10).0 6 1077 0. length of the conductor (‘). There are four main factors affecting the resistance in a conductor: the cross-sectional area of the wire (A). temperature (T) and resistivity of the material (r) (Figure 1. the more the resistance and the more the time taken for the current to flow.2 lists resistivity of some materials at 20 8C.44 6 1078 2.Basic concepts of electric circuits 15 1. Length ‘: The longer the wire. more or less resistance in the materials. r A Figure 1.59 6 1078 1. the greater the cross-sectional area. 3: There is a copper wire 50 m in length with a cross-sectional area of 0.11 Measuring resistance with an ohmmeter .68 6 1078 O Á m ¼ 1. E R Ω Figure 1. 1.68 6 1076 O Á cm (copper) R¼r ‘ ð1:68  10À6 O Á cmÞð5 000 cmÞ ¼ % 0:0646 O A 0:13 cm2 The resistance of this copper wire is 0.5. The resistor must be removed from the circuit to measure resistance as shown in Figure 1.3 Ohmmeter Ohmmeter is an instrument that can be used to measure resistance.11.0646 O. Example 1.0646 O resistance. T the temperature and r the resistivity (conducting ability of a material for a wire). Its symbol is Ω . A 50-m-long wire only has 0.16 Understandable electric circuits Factors affecting resistance can be mathematically expressed with the following formula: R¼r ‘ A ‘ R ¼ r A where A is the cross-sectional area. Although there is resistance in the copper wire. A ¼ 0. it is very small. What is the resistance of the wire? Solution: ‘ ¼ 50 m ¼ 5 000 cm. r ¼ 1. thus we can say that copper is a good conducting material. Copper and aluminium are commonly used conducting materials with reasonable price and better conductivity.13 cm2. ‘ the length. FactorsÁaffecting resistance À Note: r is a Greek letter pronounced ‘rho’ (see Appendix A for a list of Greek letters).13 cm2. 1 A ‘ G¼ or G ¼ . the greater the conductance G. voltage and resistance . The factors that affect resistance are the same for conductance. It is the ability of a material to pass current rather than resist it.5 Ohm’s law Ohm’s law is a very important and useful equation in electric circuit theory. just as conductance G is the reciprocal of resistance R. the better the conductivity of the material. Ammeter A : An instrument that is used to measure current.4: What is the conductance if the resistance R is 22 O? Solution: G ¼ 1/R ¼ 1/22 O % 0.e. which was derived from spelling ohm backwards and with an upside-down Greek letter omega . Mho actually is the reciprocal of ohm.R ¼ r R r‘ A Increasing the cross-sectional area (A) of the wire or reducing the wire length (‘) can get better conductivity. or how easy rather than how difficult it is for current to flow through a circuit. Some books use a unit mho ( ) for conductance.5. and the resistor must be removed from the circuit to measure the resistance. 1. Conductance is the conductivity of the material. and vice versa.5. it should be connected in series in the circuit. conductance is the reciprocal of resistance. Conductance G G is the reciprocal of resistance: G ¼ 1/R O The SI unit of conductance is the siemens (S).4 Conductance Conductance (G) is a term that is opposite of the term resistance.0455 S or 0. Mathematically. This can be seen from the equation of conductance. but in the opposite way.Basic concepts of electric circuits 17 Voltmeter V : An instrument that is used to measure voltage. Ohmmeter Ω : An instrument that is used to measure resistance.0455 O O 1. It is often preferable and more convenient to use conductance in parallel circuits. it should be connected in parallel with the component. This will be discussed in later chapters. Example 1. i. It precisely expresses the relationship between current. the less the resistance R of the material. (a) V I R (b) I V R (c) I V R Figure 1. (b) I ¼ V/R. Ohm’s law states that current through a conductor in a circuit is directly proportional to the voltage across it and inversely proportional to the resistance in it. R ¼ V =I These three equations can be illustrated in Figure 1. and solving for V and R respectively.e. it is the voltage that causes current to flow. 1. and inversely proportional to R: I ¼ V/R or I ¼ E/R. I ¼ V =R. Replacing voltage. I through a conductor is directly proportional to V.5. current and resistance into the above expression will obtain Ohm’s law: Current ¼ Voltage Resistance Ohm’s law ● ● Ohm’s law expresses the relationship between I. By covering one of the three variables from Ohm’s law in the diagram. i. and resistance is the opposition to the current flow.12 as a memory aid for Ohm’s law. (a) V ¼ IR.6 Memory aid for Ohm’s law Using mathematics to manipulate Ohm’s law. we can write Ohm’s law in several different forms: V ¼ IR. so current flow is the result or effect of voltage.12 Memory aid for Ohm’s law. I¼ V R or I¼ E R Any form of energy conversion from one type to another can be expressed as the following equation: Effect ¼ Cause Opposition In an electric circuit. V and R. we can get the right form of Ohm’s law to calculate the unknown. (c) R ¼ V/I .18 Understandable electric circuits with a simple mathematical equation. this graph of current versus voltage will be a straight line. With Ohm’s law we can confirm that current in the circuit is indeed 0.5 V. Since I–V characteristic shows the relationship between current I and voltage V for a resistor.02 A: I ¼ E=R ¼ 2:5 V =125 O ¼ 0:02 A 0. as shown in Figure 1. The different lines with different slopes on the I–V characteristic can represent the different values of resistors.15. R ¼ V/I ¼ 5 V/0.02 A.Basic concepts of electric circuits 19 1. When voltage V is 5 V and current is 0.14 I–V characteristics (R ¼ 10 O) When voltage V is 10 V and current is 1 A.8 I–V characteristic of Ohm’s law Using a Cartesian coordinate system. a 20-O resistor can be illustrated as in Figure 1.5.13 may prove Ohm’s law.14. If a voltmeter is connected in the circuit and the source voltage is measured.5 V V E R = 125 Ω Figure 1.02 A A 2. For example. R ¼ V/I ¼ 10 V/1 A ¼ 10 O.7 The experimental circuit of Ohm’s law The experimental circuit with a resistor of 125 O in Figure 1. voltage V (x-axis) is plotted against current I (y-axis). So the straight line in Figure 1.1 A ¼ 10 O. Also.1 A 0 V 5V 10 V Figure 1. it is called the I–V characteristic of Ohm’s law. connecting an ammeter and measuring the current in the circuit will result in I ¼ 0. I R 1A 0. . E ¼ 2.5 A.5.14 describes the current–voltage relationship of a 10-O resistor.13 The experimental circuit of Ohm’s law 1. If the resultant mathematical calculation for current through that component or branch is positive (I 4 0). the resistance does not change with the voltage or current. As shown in Figure 1.16 Reference direction of current . If the resultant mathematical calculation for the current of that component is negative (I 5 0). When the relationship of voltage and current is not a straight line.15 I–V characteristics (R ¼ 20 O) The I–V characteristic of the straight line illustrates the behaviour of a linear resistor. the actual current direction is opposite to the assumed or reference direction. the resultant resistor will be a non-linear resistor.5 A 0. the actual current direction is consistent with the assumed or reference direction.6.16. Therefore. the solid line arrows indicate the reference I=2A I = –2 A 10 V 5Ω 10 V 5Ω I>0 I<0 Figure 1. and I ¼ V =RÞ 1. which is the concept of reference direction of current. it is convenient to assume an arbitrarily chosen current direction (with an arrow). the actual current direction through a specific component or branch may change sometimes.1 Reference direction of current When performing circuit analysis and calculations in many situations. 1. i.e. and it may be difficult to determine the actual current direction for a component or branch.25 A 0 V R 5V 10 V Figure 1.15. If the voltage decreases from 10 to 5 V.9 Conductance form of Ohm’s law Ohm’s law can be written in terms of conductance as follows: I ¼ GV ðsince G ¼ 1=R.20 Understandable electric circuits I 0. the resistance still equals 20 O as shown in Figure 1.5.6 Reference direction of voltage and current 1. 17 shows two methods to represent the reference direction of current: ● ● Expressed with an arrow. Figure 1.6. If the resultant calculation is negative (V 5 0). (b) Double subscription indicates the reference I direction 1.17 Reference direction of current I.Basic concepts of electric circuits 21 current directions and the dashed line arrows indicate the actual current directions. the direction of the arrow indicates the reference direction of current. for instance Iab. the actual voltage polarity is consistent with the assumed reference polarity. and arrows represent the actual voltage polarities. Reference direction of current Assuming an arbitrarily chosen direction as the reference direction of current I: ● ● If I 4 0 the actual current direction is consistent with the reference current direction. . (a) Arrow indicates the reference I direction. If the resultant calculation for voltage across a component is positive (V 4 0).18. the positive (þ) and negative (7) polarities represent the reference voltage polarities. indicates the reference direction of current is from point a to b.2 Reference polarity of voltage Similar to the current reference direction. the actual voltage polarity is opposite to the assumed reference polarity. If I 5 0 the actual current direction is opposite to the reference current direction. As shown in Figure 1. the voltage reference polarity is also an assumption of arbitrarily chosen polarity. a I Iab R R b (a) (b) Figure 1. Expressed with a double subscription. 6. positive sign (þ) indicates a higher potential position. along with the current reference direction . indicates that the potential position a is higher than the potential position b. the direction of the arrow points from positive to negative.18 Reference polarity of voltage Reference polarity of voltage Assuming an arbitrarily chosen voltage polarity as the reference polarity of voltage: ● ● If V 4 0 the actual voltage polarity is consistent with the reference voltage polarity. + a R V R V R Vab – b Figure 1. Expressed with a double subscription. If V 5 0 the actual voltage polarity is opposite to the reference voltage polarity. Figure 1.19 shows three methods to indicate the reference polarity of voltage: ● ● ● Expressed with an arrow. then the reference current direction and reference voltage polarity is consistent.3 Mutually related reference polarity of current/voltage If the reference direction of current is assigned by flow from the positive side to the negative side of voltage across a component (the reference arrow pointing from þ to 7). In other words. Expressed with polarities.19 Methods indicating the reference polarity of voltage 1.22 Understandable electric circuits + + +V –V – V>0 V<0 – Figure 1. and negative sign (7) indicates a lower potential position. for instance Vab. and this is shown in Figure 1.20 (a) Mutually related reference polarity of I and V. then the reference I direction and reference V polarity is consistent.Basic concepts of electric circuits 23 is the voltage from positive to negative polarity.20. In this case. Summary Milestones of the electric circuits Name of scientist Charles Augustin de Coulomb Alessandro Volta ´ ` Andre-Marie Ampere Georg Simon Ohm James Watt Friedrich Emil Lenz James Clerk Maxwell Wilhelm Eduard Weber Heinrich Rudolf Hertz Kirchhoff Joseph Henry James Prescott Joule Michael Faraday Nationality French Italian French German Scottish German Scottish German German German Scottish-American British British Name of unit/law Coulomb Volt Ampere Ohm Watt Lenz Maxwell Webber Hertz Kirchhoff Henry Joule Faraday Named for Unit of charge (C) Unit of voltage (V) Unit of current (A) Unit of resistance (O) Unit of power (W) Lenz’s law Unit of flux (maxwell). This is called the mutually related reference direction or polarity of current/voltage. (b) Non-mutually related reference polarity of I and V Mutually related reference polarity of V and I If the reference I direction is assigned by an arrow pointing from þ to 7 of voltage across a component. Maxwell’s magnetic field equation Unit of flux (weber) 1 Wb ¼ 108 Mx Unit of frequency (Hz) Kirchhoff’s current and voltage laws Unit of inductance (H) Unit of energy (J) Unit of capacitance (F) . I R + V – I + R V – (b) (a) Figure 1. if we only know one reference direction or polarity. it is also possible to determine the other. temperature (T) and resistivity (r). represented by the symbol A . Ohmmeter: An instrument used for measuring resistance. Requirements of a basic circuit: ● Power supply (power source): A device that supplies electrical energy to a load. ● Wires: Wires connect the power supply unit and load together. ● Load: A device that is connected to the output terminal of a circuit. Resistance (R): Measure of a material’s opposition to the flow of current. Its symbol is Ω and the resistor must be removed from the circuit to measure the resistance. It should be connected in series in the circuit. length (‘). Current direction: ● Conventional current flow version: A flow of positive charge (proton) from the positive terminal of a power supply to its negative terminal. Schematic: A simplified circuit diagram that shows the interconnection of circuit components. Resistor (R): A two-terminal component of a circuit that limits the flow of current. and is represented by circuit symbols. Its symbol is V and it should be connected in parallel with the component. . Load voltage (V): Voltage across two terminals of the load. Voltage drop: Voltage across a component when current flows from a higher potential point to a lower potential point in a circuit. Voltage (V) or potential difference: The amount of energy or work that would be required to move electrons between two points: V ¼ W/Q (or v ¼ dw/dt). and consumes electrical energy. Factors affecting resistance: R ¼ rð‘=AÞ. The EMF is supplied by a voltage source and applied to the load in a circuit.24 Understandable electric circuits Basic concepts ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● Electric circuit: A closed loop of pathway with electric current flowing through it. Circuit symbols: The idealization and approximation of the actual circuit components. and carry current flowing through the circuit. Electric current (I): A flow of electric charges through an electric circuit: I ¼ Q/t (or I ¼ dq/dt). where cross-sectional area (A). Voltmeter: An instrument used for measuring voltage. ● Electron flow version: A flow of negative charge (electron) from the negative terminal of a power supply unit to its positive terminal. Ammeter: An instrument used for measuring current. Voltage rise: Voltage across a component when current flows from a lower point to a higher point in a circuit. Electromotive force (EMF): An electric pressure or force supplied by a voltage source causing current to flow in a circuit. Source voltage or applied voltage (E or VS): EMF can be called source voltage or applied voltage. Mutually related polarity of voltage and current: If the reference current direction is assigned by an arrow pointing from þ to 7 voltage of the component. voltage V and resistance R. Ohm’s law: It expresses the relationship between current I. Symbols and units of electrical quantities: Quantity Charge EMF Work (energy) Resistance Resistivity Conductance Current Voltage Quantity symbol Q E W R r G I V or E Unit Coulomb Volt Joule Ohm Ohm Á metres Siemens or mho Ampere Volt Unit symbol C V J O OÁm S or A V Experiment 1: Resistor colour code Objectives ● ● ● Become familiar with the breadboard Interpret the colour code for resistors Measure resistors with a multimeter (ohmmeter function) O . ● If V 5 0 actual voltage polarity is opposite to the reference voltage polarity. I¼ V R or I¼ E R ● ● ● ● ● Conductance form of Ohm’s law: I ¼ GV. then the reference current direction and reference voltage polarity is consistent.Basic concepts of electric circuits ● ● 25 Conductance (G): It is the reciprocal of resistance: G ¼ 1/R. ● If I 5 0 actual current direction is opposite to the reference current direction. Reference polarity of voltage: Assuming an arbitrarily chosen voltage polarity as the reference polarity of voltage: ● If V 4 0 actual voltage polarity is consistent with the reference voltage polarity. Reference direction of current: Assuming an arbitrarily chosen current direction as the reference direction of current: ● If I 4 0 actual current direction is consistent with the reference current direction. 100 O (2).26 Understandable electric circuits Equipment and components ● ● ● Breadboard Resistors: 12 O (2). 1. and Figure L1. or for professionals to build temporary electrical or electronic circuits to try out ideas for circuit designs. Figure L1.2 The underneath of the breadboard ● The breadboard contains an array of holes where the leads of components and jumper wires can be inserted. 56 kO.7 kO Digital multimeter Background information Breadboard guide ● ● ● The Universal Solderless Breadboard. 3. or usually known as the breadboard.2 kO. Figure L1. and the .3(a) is a simple circuit. 15 kO. It offers an easy way to change components or wire connections on the breadboard without soldering. is one type of circuit board. 4.1 is a photograph of a small breadboard. 2.2. which is laid out as shown in Figure L1.1 MO. 470 O. The bottom of the board has many strips of metal. The breadboard is a good training tool and is usually used in the lab to perform experiments on electric or electronic circuits.2 is what the underneath of the breadboard looks like.9 kO.7 kO (2). Resistor colour code guide (four band) ● Most resistors are very small and it is hard to print the values on them. 18 O. and Figure L1. 8.3(b) shows how to build this circuit on the breadboard. Figure L1.1 A breadboard Figure L1. Usually the small resistors have different colour bands on them. These strips connect the holes on top of the board. The top and bottom rows will be used to connect the power supply. hold the resistor so that the colour bands are closest to the left end as shown in Figure L1. The resistor colour code is shown in Table L1.com/Discussion:Remember-Electrical-ResistorColor-Codes .3 Building a simple circuit on the breadboard standard resistor colour code can be used to interpret the values of different resistors.1 Resistor colour code Colour Black Brown Red Orange Yellow Green Blue Violet Grey White Digit 0 1 2 3 4 5 6 7 8 9 Note: Memory aid: Better Be Right Or Your Great Big Venture Goes West Source: http://www. and the fourth band represents the tolerance of the resistance. ● Figure L1. the third band represents the number of zeros to add to the integers (multiplier). To determine the value of a resistor from the colour band markings.4.Basic concepts of electric circuits + 27 E R (a) (b) Figure L1. Table L1. and the tolerance of the resistance is shown in Table L1.4 Resistor colour bands ● ● The first two colour bands on the left side of the resistor represent two digits (0–9).wikihow.1.2. its colour bands should be yellow (4). Multimeter guide ● ● A multimeter or VOM (volt–ohm–millimetre) is an electrical and electronic measuring instrument that combines functions of voltmeter. ● Turn the central selector switch pointing to the ohms range (with O sign). Method for measuring resistance with a DMM (ohmmeter function): ● Turn off the power supply if the resistor has been connected in the circuit. it indicates that its resistance lies between 900 and 1100 O: R ¼ 1 000 O Æ 10% ¼ 1 000 Æ 100 ¼ 900 to 1100 O Example: If one needs to find a resistor with the value of 470 O. Better insert the resistors into the holes of the breadboard to measure them. and to where the maximum range of the estimated resistor value is closed.2 Tolerance of resistance Colour Gold Silver Tolerance (%) +5 +10 Example: If a resistor has colour bands of brown (1). . adjust the multimeter range from the maximum to the lower range until suitable resistance is read. since you will add your own resistance (in parallel) to the resistor. black (0). ● Make the measurement by connecting the resistor in parallel with the two leads of the multimeter (connect or touch one lead from the multimeter to one end of the resistor.5. ● Turn off the multimeter. and connect or touch the other lead of the multimeter to the other end of the resistor).28 Understandable electric circuits Table L1. Note: To get more accurate measurement result. DMM is a very commonly used instrument. red (add two zeros) and silver (tolerance is 10%) from left to right side respectively. ohmmeter. ammeter. be sure not to grab the resistor’s leads with your hands when you are measuring resistors. violet (7) and brown (add one zero). ● Insert the multimeter’s leads into the sockets labelled COM and V/O as shown in Figure L1. etc. and turn on the multimeter. since it is easier to use and has a higher level of accuracy. There are two types of multimeters: digital multimeter (DMM) and analog multimeter. ● If measuring an unknown resistor. violet.1 MO Colour code Yellow. brown. gold Resistance range 446. and fill in the ‘Colour code’ and the ‘Resistance range’ columns in Table L1.1 to find the six resistors listed in Table L1. Use the resistor colour code chart in Table L1. Table L1.5 O . List the colour band identification for these six resistors.Basic concepts of electric circuits 29 Figure L1.7 kO 3.9 kO 82 kO 1. Familiarize with the resistor colour code.5 Multimeter Procedure 1.3.5 O Measured value 470.3 Resistor Example: 470 O 12 O 100 O 2.5 O7493.3 from the lab. 2 kΩ 100 Ω 2.30 Understandable electric circuits 2.7 kΩ 15 kΩ 470 Ω 18 Ω 56 kΩ 12 Ω 8.6 Construct circuits on the breadboard Conclusion The conclusion may include the following information: ● ● ● ● lab objectivities accomplished results. and measure the six resistors in Table L1. 12 Ω 100 Ω 2.7 kΩ 3.9 kΩ 4.1 MΩ Figure L1. Construct the circuits shown in Figure L1. Show your circuits to the instructor to get check-up and signature. 3.7 kΩ 1.3 with multimeter and fill in the ‘Measured value’ column in Table L1. Get the multimeter to function as an ohmmeter.6 on the breadboard with the right resistors. errors and error analysis problems encountered during the experiment and their solutions knowledge and skills obtained from the lab .3. if using for example a force of 1 N to lift an object to 1 m.1 Work You may have learned in physics that work is the result when a force acts on an object and causes it to move a certain distance.1. node. Work (W ) is the product of the force (F ) and the displacement (S) in the direction of the motion. network and loop understand the concepts of the ideal voltage source and the actual voltage source. Work W ¼ F6S where W is. 2.1 Power and energy 2. the 1 J of work done in overcoming the downward force of gravity as shown in Figure 2. Quantity Work Force Displacement Quantity Symbol W F S Unit Joule Newton Meter Unit symbol J N m .1. you will be able to: ● ● ● ● ● ● ● define energy and power calculate power know the reference directions of power analyse and calculate circuits with Kirchhoff’s voltage law (KVL) analyse and calculate circuits with Kirchhoff’s current law (KCL) define the branch.Chapter 2 Basic laws of electric circuits Objectives After completing this chapter. after you eat and sleep. The law of conservation of energy is one of most important rules in natural science.3 Power Power refers to the speed of energy conversion or consumption. W ¼ (Fcosy)S ¼ FS.1 Work Note: When the force (F) and displacement (S) do not point in the same direction. such as walking. or the amount of work done per unit of time. Work is done after the electrons or charges are moved to a certain distance in a circuit as a result of applying an electric field force from the power supply.1. For example. it will take a shorter period of time to lift it. and not electrical power.32 Understandable electric circuits 1N F 1m 1N Figure 2. 2. and if a lower power is applied to the object (a kid is lifting it). the formula to calculate work will be: W ¼ (F cos y)S ● ● where the angle y is the angle between force (F) and displacement (S). It is the same in an electric circuit. Lamp: electrical energy ! light energy. For example: Electrical generator: mechanical energy ! electrical energy. reading. etc. The hydro bill that you receive is for electrical power – the amount of electrical energy consumed in 1 or 2 months. .1. your body converts the stored energy to keep you doing daily work. Even though you can’t ever really see it. it will take a longer period of time to lift it. you use energy to do work every day. cos08 ¼ 1. writing. Battery: chemical energy ! electrical energy. when y is 0 degree. If a higher power is applied to the object (an adult is lifting it). 1 N object lifted to 1 m may have different time rates depending on the amount of power applied.2 Energy Energy is the ability to do work. but can only be converted from one form to another. it is a measure of how fast energy is transforming or being used. ‘Converted’ means ‘never disappeared’ in physics terms. 2.1. but a transfer of energy. For example in Figure 2. Our daily consumption of electricity is electrical energy. it is not work itself. It states that energy can neither be created nor destroyed. So power is defined as the rate of doing work. running. Power is the speed of energy conversion.Basic laws of electric circuits 33 Energy and power ● ● Energy is the ability to do work. Since current is the amount of charge (Q) that flows past a given point at the certain time: I ¼ Q=t and voltage is the amount of work that is required to move electrons between two points: V ¼ W=Q or Wc ¼ QV: Substituting work W into the power equation gives P ¼ W=t ¼ QV=t ¼ IV: It can also be expressed as the form of a derivative: p ¼ ðdw=dtÞ ¼ ðdw=dqÞðdq=dtÞ ¼ vi Substituting Ohm’s law into the power equation P ¼ IV obtains the other two different power equations: P ¼ VI ¼ ðIRÞI ¼ I 2 R P ¼ VI ¼ V V V ¼ R R 2 ðOhm’s law : V ¼ IRÞ V Ohm’s law : I ¼ R (Electrical) Power P P ¼ IV ¼ I 2 R ¼ V 2 =R Quantity Power ðor P ¼ IE ¼ E2 =RÞ Unit Watt Unit symbol W Quantity symbol P . and it is a measure of how fast electrons or charges are moving in a circuit. or work done per unit of time: Power ¼ Work=time or P ¼ W=t Quantity symbol W t P Or: Unit Joule Second Watt Kilowatt-hour Hour Watt Unit symbol J s W kWh h W Quantity Electrical Power: Work or Energy Time Power Electrical power is the speed of electrical energy conversion or consumption in an electric circuit. current I ¼ 1 A and resistance R ¼ 10 O. If power is given in a circuit. power is positive. so V ¼ PR Example 2.1 proved that the three power equations are equivalent since each equation leads to the same value of power at 10 W. When a component in a circuit has non-mutually related reference polarity of current and voltage. meaning the component releasing (or providing) of energy. P 4 0. the correct equation will be obtained to calculate the unknown power.2 Memory aid for power equations Example 2. Solution: P ¼ IV ¼ ð1 AÞð10 VÞ ¼ 10 W P ¼ I 2 R ¼ ð1 AÞ2 ð10 OÞ ¼ 10 W P ¼ V 2 =R ¼ ð10 VÞ2 =10 O ¼ 10 W Example 2. respectively.2: If power consumed on a 2. calculate the power in this circuit by using three power equations. P V I I 2 P R P V2 R Figure 2.1: In a circuit. using mathematical skill to manipulate the power equations and solving for current I and voltage V. section 1.e. i. By covering power in any diagram. respectively.5 O resistor is 10 W in a circuit. meaning the component absorption (or consumption) of energy. calculate the current flowing through this resistor. i.34 Understandable electric circuits The above three power equations can be illustrated in Figure 2.3. voltage V ¼ 10 V.1.2 as the memory aid for power equations.4 The reference direction of power When a component in a circuit has mutually related reference polarity of current and voltage (refer to chapter 1.6. we can express current I and voltage V as follows: pffiffiffiffiffiffiffiffiffiffi since P ¼ I2R or I2 ¼ P/R. . P 5 0. Solution: I¼ pffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi P=R ¼ 10 W=2:5 O ¼ 2 A 2. so I ¼ pffiffiffiffiffiffiffi P=R since P ¼ V2/R or V2=PR. power is negative. The concept of the reference direction of power can be illustrated in Figure 2.e.3). 4(a and b). Example 2. V1 ¼ 6 V.3: Determine the reference direction of power in Figure 2. Power for R2 (b to c): P2 ¼ V2I ¼ (14 V)(2 A) ¼ 28 W (absorption). Example 2. R2. (b) P ¼ I (7V) ¼ (2 A)(73 V) ¼ 76 W (P 5 0. If a circuit has non-mutually related reference polarity of current and voltage: P 5 0 (releasing energy). the resistor absorbs energy). Solution: Power for R1 (a to b): P1 ¼ V1I ¼ (6 V)(2 A) ¼ 12 W (absorption).4 Illustrations for Example 2.3 Solution: (a) P ¼ IV ¼ (2 A)(3 V) ¼ 6 W (P 4 0. Determine the powers dissipated on the resistors R1.5. Power for R1 and R2 (a to d): P3 ¼ (7E)I ¼ (720 V)(2 A) ¼ 740 W (releasing). and R1 and R2 in series in this figure.3 The reference direction of power The reference direction of power ● ● If a circuit has mutually related reference polarity of current and voltage: P 4 0 (absorption energy). . V2 ¼ 14 V and E ¼ 20 V in a circuit as shown in Figure 2. the resistor releases energy).4: I ¼ 2 A. P1 þ P2 þ P3 ¼ 12 W þ 28 W 7 40 W ¼ 0 (energy conservation). I V 3V 2A V I 2A 3V (a) (b) Figure 2.Basic laws of electric circuits 35 I I V V (a) P > 0 (b) P < 0 Figure 2. it is same as having a round trip. and they have the same potential positions. Kirchhoff’s laws are the most important fundamental circuit laws for analysing and calculating electric circuits after Ohm’s law. c.2. 1824–1887) at Berlin University developed the two laws that established the relationship between voltage and current in an electric circuit. Figure 2. a E1 = 10 V I V1 = 10 V R1 R2 b V2 = 10 V E2 = 10 V R3 d V3 = 10 V c Figure 2.2. 2.2 Kirchhoff’s voltage law (KVL) In 1847.1 Closed-loop circuit A closed-loop circuit is a conducting path in a circuit that has the same starting and ending points. For example.4 2.5 Circuit for Example 2. physics professor Kirchhoff (Gustav Kirchhoff. it would be a closed-loop circuit. If the current flowing through a circuit from any point returns current to the same starting point.36 Understandable electric circuits a I =2A a R1 E R2 V1 b V2 c d Figure 2. or the sum . a German physicist.6 A closed-loop circuit 2. since current I starts at point a.2 Kirchoff’s voltage law #1 KVL #1 states that the algebraic sum of the voltage or potential difference along a closed-loop circuit is always equal to zero at any moment.6 is a closed-loop circuit. d and returns to the starting point a. so the starting and ending points are the same. As current flows through a closed-loop circuit. passes through points b. if the voltage reference polarity is from negative to positive. The algebraic sum used in KVL #1 means that there are voltage polarities existing in a closed-loop circuit. and the loop direction is clockwise. SV ¼ 0. ● ● Assign a positive sign (þ) for voltage (V or E) in the equation SV ¼ 0. if the voltage reference polarity is from positive to negative and the loop direction is clockwise.5 V I E1 = 2.e. i. i. V1 = 2.7 Circuit for Example 2.5 V R1 Figure 2. if the voltage reference polarity and the loop direction are opposite. Example 2.7: V1 þ V2 þ V3 À E2 À E1 ¼ 0 ð2:5 þ 2:5 þ 2:5 À 5 À 2:5Þ V ¼ 0 KVL #1 SV ¼ 0 ● Assign a þve sign for V or E if its reference polarity (þ to 7) and loop direction (clockwise) are the same. The voltage in KVL includes voltage rising from the voltage sources (E) and voltage dropping on circuit elements or loads.5 V R2 E2 = 5 V R3 V3 = 2. Assign a negative sign (7) for voltage (V or E) in the equation SV ¼ 0. ● Assign a 7ve sign for V or E if its reference polarity (7 to þ) and loop direction (clockwise) are opposite.e. if the voltage reference polarity and the loop direction are the same. .5 V V2 = 2.5a: Verify KVL #1 for the circuit of Figure 2.Basic laws of electric circuits 37 of voltages in a closed-loop is always equal to zero.e.7.5 Solution: Applying SV ¼ 0 in Figure 2. i. It requires assigning a loop direction and it could be in either clockwise or counter-clockwise directions (usually clockwise). 7: V1 þ V2 þ V3 ¼ E2 þ E1 ð2:5 þ 2:5 þ 2:5ÞV ¼ ð2:5 þ 5ÞV 7:5 V ¼ 7:5 V KVL #2 SV ¼ SE ● ● Assign a þve sign for V if its reference polarity and loop direction are the same.2. assign a þve sign for E if its polarity and loop direction are opposite. SV ¼ SE: (10 þ 10 þ 10) V ¼ 30 V . 2. ● ● Assign a positive sign (þ) for V.7. KVL #1.e. Example 2. if its reference polarity and the loop direction are the same. SV ¼ 0: (10 þ 10 þ 10 7 30) V ¼ 0 KVL #2. Assign a negative sign (7) for the voltage source (E) in the equation. SV ¼ SE. the total voltage drops on all the resistors should be equal to the voltage for the DC power supply.e. Assign a positive sign (þ) for voltage source (E) in the equation if its reference polarity and loop direction are opposite. if its reference and the loop directions are opposite. i. if its polarity is from þve to –ve and the loop direction is clockwise. assign a negative sign (7) for V.8.e. Assign a 7ve sign for E if its reference polarity and loop direction are the same.8.4 Experimental circuit of KVL KVL can be approved by an experimental circuit in Figure 2. i.5b: Verify KVL #2 for the circuit of Figure 2. i.2. if its reference and loop directions are the same. Solution: Applying SV ¼ SE in Figure 2.38 Understandable electric circuits 2.3 KVL #2 KVL can also be expressed in another way: the sum of the voltage drops (V) around a closed loop must be equal to the sum of the voltage rises or voltage sources in a closed-loop circuit. If using a multimeter (voltmeter function) to measure voltages on all resistors and power supply in the circuit of Figure 2. assign a 7ve sign for V if its reference direction and loop direction are opposite. if its polarity is from negative to positive and the loop direction is clockwise. 625 7 1.9 Circuit for Example 2.5 Ω R2 = 5 Ω E2 = 5 V R3 = ? R4 = 4 Ω Figure 2.25 A)(5 O) ¼ 1.6: Determine resistance R3 in the circuit of Figure 2.9.25 A)(2.Basic laws of electric circuits V 39 10 V 100 Ω 30 V V 100 Ω V 10 V 100 Ω 10 V V Figure 2.625 V V2 ¼ I R2 ¼ (0.125 V Therefore.25 7 1)V ¼ 2. V3 2:125 V R3 ¼ ¼ ¼ 8:5 O I 0:25 A .8 Experimental circuit of KVL Example 2.25 A)(4 O) ¼ 1 V Solve for V3 from V1 þ V2 þ V3 þ V47E ¼ 0: V3 ¼ E 7 V1 7 V2 7 V4 ¼ (5 7 0.5 O) ¼ 0.25 A R1 = 2. V3 ¼ ? I Applying KVL #1. SV ¼ 0: V1 þ V2 þ V3 þ V4 7E ¼ 0 Therefore: V1 ¼ I R1 ¼ (0. I = 0.25 V V4 ¼ I R4 ¼ (0.6 Solution: V3 R3 . where Vab ¼ E 7 V1 7 V4 ¼ (5 7 1.5 V.10 can be calculated using KVL #2 as follows: SV ¼ SE : V þ Vab ¼ E Vab ¼ E À V ¼ ð10 À 1ÞV ¼ 9V Example 2. a V1 = 1.5 V.40 Understandable electric circuits 2.2.7: Determine the voltage across points a to b (Vab) in the circuit of Figure 2.10 KVL extension Vab in the circuit of Figure 2.5 7 1)V ¼ 2.5 KVL extension KVL can be expanded from a closed-loop circuit to any scenario loop in a circuit. because voltage or potential difference in the circuit can exist between any two points in a circuit. where Vab ¼ V2 þ V3 ¼ (2 þ 0. Method 2: SV ¼ 0: V2 þ V3 7 Vab ¼ 0.5 V V4 = 1 V b V2 = 2 V Figure 2.5)V ¼ 2. .7 Solution: Vab can be solved by two methods as follows: Method 1: SV ¼ 0: V1 þ Vab þ V4 7 E ¼ 0.5 V E=5V V3 = 0.11 Circuit for Example 2. I V=1V E = 10 V R a Vab b Figure 2.11. and it does not matter which path or branch is used to solve for voltage between these two points. Therefore. the result should be the same. For instance.1 KCL #1 KCL #1 states that the algebraic sum of the total currents entering and exiting a node or junction of the circuit is equal to zero. Assign a negative sign (7) to the current in the equation if current is exiting the node. I5 and I6 are the currents exiting the node A. 2.3.7 show that voltage across two points a and b is the same. and it has six branches.3. 2. i. SIin ¼ SIout . ● ● Assign a positive sign (þ) to the current in the equation if current is entering the node.12 Nodes and branches Applying KCL #1: I1 þ I2 þ I3 7 I4 7 I5 7 I6 ¼ 0 KCL #1 SI ¼ 0 ● Assign a þve sign for current in KCL if I is entering the node.12. SI ¼ 0. point A is a node in Figure 2.2 KCL #2 KCL can also be expressed in another way: the total current flowing into a node is equal to the total current flowing out of the node. I1. ● Assign a 7ve sign for current in KCL if I is exiting the node.e. i. I3 I2 A I1 I6 I4 I5 Figure 2. A branch is a current path between two nodes with one or more circuit components in series. the physical property of KVL is that voltage does not depend on the path.e. I2 and I3 are the currents flowing into node A. I4.6 The physical property of KVL The results from Example 2.2. A node or junction is the intersectional point of two or more current paths where current has several possible paths to flow.Basic laws of electric circuits 41 2.3 Kirchhoff’s current law (KCL) 2. 9: Determine the current I1 at node A and B in Figure 2.42 ● Understandable electric circuits Assign a positive sign (þ) to current Iin in the equation if current is entering the node. assign a negative sign (7) for Iin if current is exiting the node.13 Circuit for Example 2.14 Circuit for Example 2. assign a negative sign (7) for Iout if current is entering the node. I2 = 10 A I3 = 7 A I4 = 8 A I1 = 15 A a I5 = 10 A Figure 2. we get (15 þ 10 7 7 7 8 7 10)A ¼ 0 Example 2.14. ● Example 2.8: Verify KVL #1 and #2 for the circuit of Figure 2.13.8 Solution: KCL #2: SIin ¼ SIout: I1 þ I2 ¼ I3 þ I4 þ I5 Substituting I with its respective values. I1 I2 A I3 I4 B Figure 2.9 Solution: Node A: SI ¼ 0 : I1 À I2 À I3 À I4 ¼ 0 I1 ¼ I2 þ I3 þ I4 SIin ¼ SIout : Node B: SI ¼ 0 : I2 þ I 3 þ I 4 À I 1 ¼ 0 SIin ¼ SIout : I2 þ I3 þ I4 ¼ I1 . we get (15 þ 10)A ¼ (7 þ 8 þ 10)A KCL #1: SI ¼ 0: I1 þ I2 7 I3 7 I4 7 I5 ¼ 0 Substituting I with its respective values. Assign a positive sign (þ) to current Iout in the equation if current is exiting the node. A B C Figure 2.16 Circuit for Example 2. Water flowing into a pipe should be equal to the water flowing out of the pipe. and prove it by using another one). B and C merging together to a converging point and forms the main water flow out of the converging point to the downstream creek. Water flowing in a pipe can be analogized as current flowing in a conducting wire with KCL. if current is entering the node. I1 10 A I2 20 A I4 = ? 5A I3 Figure 2.10: Determine current I3 (you may calculate it by using one KCL.10 . assign a 7ve sign for Iin. water flows in the three upstream creeks A.15 Creeks Example 2. Assign a þve sign for Iout if current is exiting the node. if current is exiting the node.Basic laws of electric circuits 43 KCL #2 SIin ¼ SIout ● ● Assign a þve sign for Iin if current is entering the node. assign a 7ve sign for Iout.15. in Figure 2. For example. 25 A Figure 2. Another property of KCL is the continuity of current (or charges). I3 ¼ I1 þ I2 ¼ 0. and they are equal to the source branch current I3 (exiting).e. charges can neither be created nor destroyed or the amount of charges that enter the node equals the amount of charges that exit the node.3. Start from the unknown value in the problem and find the right equation that can solve this unknown. the water or current will never discontinue at any moment in a pipe or conductor.15 A A I3 = 0.3.3 Physical property of KCL The physical property of KCL is that charges cannot accumulate in a node.17 Experimental circuit for KCL Measure branch currents I1 and I2 (entering) using two multimeters (ammeter function). i.4 Procedure to solve a complicated problem It does not matter which field of natural science the problems belong to or how complicated they are. I4 ¼ I2 À I1 À I3 ¼ ð20 À 10 À 5ÞA ¼ 5 A SIin ¼ SIout : I1 þ I3 þ I4 ¼ I2 ð10 þ 5 þ 5ÞA ¼ 20A 20A ¼ 20A ðhence provedÞ The KCL can be proved by an experimental circuit in Figure 2.e.44 Understandable electric circuits Solution: SI ¼ 0 : I1 À I2 þ I3 þ I4 ¼ 0. The following steps outline the procedure: 1. the procedure for analysing and solving them are all similar. what arrives at a node is what leaves that node.25 A.1 A A A I2 = 0.17. 2. which is similar to the continuity of flowing water. This results from conservation of charges. . E = 10 V I1 = 0. 2. i. 11: I1 ¼ ? 4A C I1 I2 3A 5A 4A 1A A I3 B 2A Figure 2. solve for I3: I3 ¼ 2 A (there are no more unknown elements in this equation except for I3). the unknown in this equation is I2. Repeat until the unknown in the original problem is solved. Repeat steps 1 and 2 until there are no more unknowns in the equation. 4. At node A: I3 þ 4 A ¼ (5 þ 1)A. and solve the unknown. Determine the new unknown of the equation in step 1 and find the equation to solve this unknown. Now let’s try to use this method to solve I1 in Example 2.11 Solution: The unknown in this problem is I1.1) and solve for I1: I1 þ 4 A ¼ (5 þ 3)A. . ● At node C: I1 þ 4 A ¼ I2 þ 3 A I2 ¼ ? ð2:1Þ ● (Besides I1. so I2 ¼ 5 A.) Find the right equation to solve I2.Basic laws of electric circuits 45 2.18 Circuit for Example 2.11. the unknown in this equation is I3. 3. therefore. Substitute the solution from the last step into the previous equation. Find the right equation to solve I1. I1 ¼ 4 A.2) and solve for I2: I2 þ 3 A ¼ (4 þ 2 þ 2)A. At node B: I2 þ 3 A ¼ 4 A þ I3 þ 2 A I3 ¼ ? ð2:2Þ ● ● ● (Besides I2. Substitute I3 ¼ 2 A into (2.) Find the right equation to solve I3. Substitute I2 ¼ 5 A into (2. Example 2. I3 must be current entering node A to satisfy SIin ¼ SIout.19 Supernode Example 2. I7 ¼ I1 ¼ 5 A. I4 and I7 in the circuit of Figure 2. ● At node A: Since current entering node A is I1 ¼ 5 A.20 Circuit for Example 2.19 within the dashed circle can be treated as an extended node or supernode A. i. and this circuit can be treated as a supernode.46 Understandable electric circuits 2. I3 ¼ 1 A. and current leaving node A is I2 ¼ 6 A.12 Solution: Treat the circuit between the nodes A and D (inside of the circle) as a supernode. so I2 4 I1. . The circuit between nodes a and b in Figure 2. therefore.5 Supernode The concept of the node can be extended to a circuit that contains several nodes and branches. KCL can be applied to it: SIin ¼ SIout or I1 ¼ I2 A I1 I2 a b Figure 2. I1 = 5 A A I3 I4 C I6 = 1 A B I5 = 4 A I2 = 6 A D I7 = ? Figure 2.20. I1 þ I3 ¼ I2 or 5 A þ I3 ¼ 6 A. therefore.12: Determine the magnitudes and directions of I3.e.3. and current entering the node A should be equal to current exiting the node D. and it is a source that can provide EMF (electromotive force) and current to operate the circuit.Basic laws of electric circuits ● 47 ● At node B: Since current entering node B is I2 ¼ 6 A and currents exiting node B is I5 ¼ 4 A. B. and a loop may include several such windowpanes. A–B–D–A. Note: A mesh is always a loop. I4 must be current exiting node B to satisfy SIin ¼ SIout.13 Solution: ● ● ● ● Node: four nodes – A. CD and AD Mesh: 1. i. 2 A ¼ 1 A þ 1 A. but a loop is not necessary a mesh. BD. 2 and 3 Loop: 1. 2. I2 ¼ I4 þ I5 or 6 A ¼ I4 þ 4 A.21.13: List nodes. AC. etc. A–B–D–C–A. 2. Loop: A complete current path where current flows back to the start. so I2 4 I5.4 Voltage source and current source A power supply is a circuit device that provides electrical energy to drive the system. BC. 3. meshes and loops in Figure 2. I4 ¼ 2 A. branches. Mesh: A loop in the circuit that does not contain any other loops (nonredundant loop). . C and D Branch: six branches – AB.6 Several important circuit terminologies ● ● ● ● Node: The intersectional point of two or more current paths where current has several possible paths to flow. 2. B 1 A C 3 2 D Figure 2. therefore.e. Example 2. Prove it at node C: I4 ¼ I3 þ I6. A mesh can be analogized as a windowpane.3.21 Illustration for Example 2. The power supply can be classified into two categories: voltage source and current source. 2 A ¼ 2 A (proved). Branch: A current path between two nodes where one or more circuit components is in series. . since nothing is perfect. so it is an independent voltage source.22(b).22 Ideal voltage source Ideal voltage source ● ● It can provide a constant terminal voltage that is independent of the variations in its external circuit.23(a). DC generator or DC power supply. no matter what its load resistance RL is. and it can provide maximum current to the load. The real voltage source (or voltage source) can be represented as an ideal voltage source VS in series with an internal resistor RS as shown in Figure 2. Once a load resistor RL is connected to the voltage source (Figure 2.23(b)). will not change even if an external circuit such as a load. The terminal voltage Vab for an ideal voltage source is a constant.1. and same as the source voltage (Vab ¼ VS). will not reach a perfect constant output voltage after it is connected to an external circuit or load. a VS b (a) Ideal voltage source a I VS b RL V V VS 0 (b) Ideal voltage source with a load I or t (c) Characteristic curve Figure 2. Since the internal resistance RS is usually very small. This means that the voltage of the ideal voltage source is independent of variations in its external circuit or load. Vab ¼ VS.1 Ideal voltage source An ideal voltage source is a two-terminal circuit device that can provide a constant output voltage.4. Voltage of the ideal voltage source.1 Voltage source 2. across its terminals. The real voltage sources all have a non-zero internal resistance RS (RS 6¼ 0). The ideal voltage source has a zero internal resistance (RS ¼ 0).1. . Vab will be a little bit lower than the source voltage VS (Vab ¼ VS 7 IRS). Vab. and is shown in Figure 2. such as a battery. Its internal resistance.22(c). Its current depends on variations in its external circuit.48 Understandable electric circuits 2.2 Real voltage source Usually a real-life application of a voltage source. is connected to it as shown in Figure 2. the current in the ideal voltage source also changes since I ¼ V/RL. The characteristic curve of an ideal voltage source is shown in Figure 2.4. so when the load resistance RL changes. VS.22(a). RS ¼ 0. the terminal voltage of the source Vab will change if the load resistance RL changes. etc. Current in the ideal voltage source is dependant on variations in its external circuit.4. 2. RL. so the voltage drop on the internal resistance (IRS) is also very small.Basic laws of electric circuits a RS VS b (a) Real voltage source RS VS I RL b 0 (b) Real voltage source with a load Vab a V VS 49 ↓ ↑ VRS = IRS I or t (c) Characteristic curve Figure 2. the older battery will have a higher internal resistance RS and a lower terminal voltage Vab. and the terminal voltage Vab also changes. i. RS ( RL. and therefore.14 .e.005 Ω Figure 2. Real voltage source (voltage source) It has a series internal resistance RS.23 Real voltage source A smaller internal resistance can also provide a higher current through the external circuit of the real voltage source because I ¼ VS/(RS þ RL). a RS VS RS RL 5Ω VS 3V b (a) RS = 0. Example 2. Once the load resistance RL changes.24 Circuits for Example 2.24(a and b). This is why the terminal voltage of the real voltage source is not possible to keep at an ideal constant level (Vab 6¼ VS).14: Determine the terminal voltages of the circuit in Figure 2. The internal resistance of a real voltage source usually is much smaller than the load resistance. The terminal voltage of the real voltage source is: Vab ¼ VS 7 IRS. current I in this circuit will change. the terminal voltage of the real voltage source (Vab) is approximately stable: Vab ¼ VS À IRS % VS When a battery is used as a real voltage source. and RS ( RL.005 Ω 3V b (b) RS = 25 Ω I 50 Ω RL 5Ω a 0. can the terminal voltage of the source be kept approximately stable. I¼ VS 3V ¼ % 0:5994 A ðRS þ RL Þ ð0:005 O þ 5 OÞ Vab ¼ IRL ¼ ð0:5994 AÞð5 OÞ ¼ 2:997 V When RS ¼ 50 O. 2. Only when the internal resistance is very small. I ¼ ðVS =RS Þ: a RS VS b (a) With a load I RL Vab RS VS b (b) Open circuit I=0 Vab a RS VS b (c) Short circuit a Vab = 0 Figure 2.25(b)): Vab ¼ VS .50 Understandable electric circuits When RS ¼ 0:005 O. i. an electronic element you may have heard. Open circuit: when there is no external load RL connected to a voltage source (Figure 2. Vab ¼ 0. In this case. the terminal voltage Vab is much less than the source voltage VS. A real voltage source has three possible working conditions: ● ● ● When an external load RL is connected to a voltage source (Figure 2. Current of the ideal . But when RS ¼ 50 O .997 V % VS ¼ 3 V.e. can be approximated as an example of a current source. I ¼ 0: Short circuit: when a jump wire is connected to the two terminals of a voltage source (Figure 2.005 O.25(a)): Vab ¼ VS À I RS .275 V ( VS ¼ 3 V.2. such as when RS ¼ 0.25(c)): Vab ¼ 0.1 Ideal current source An ideal current source is a two-terminal circuit device that can provide a constant output current IS through its external circuit.4.25 Three states of a voltage source 2. I ¼ ðVS =RS þ RL Þ. A transistor. Vab ¼ 2.4. I¼ Vab VS 3V % 0:055 A ¼ ðRS þ RL Þ ð50 O þ 5 OÞ ¼ IRL ¼ ð0:055 AÞð5 OÞ ¼ 0:275 V The above example indicates that the internal resistance has a great impact on the terminal voltage and current of the voltage source.2 Current source The current source is a circuit device that can provide a stable current to the external circuit. the terminal voltage Vab is very close to the source voltage VS. IS represents the current for current source. a IS 0. in Figure 2.27. The ideal current source has an infinite internal resistance (RS ¼ ?).26(b).02 A RL Vab b Figure 2. I IS IS V or t (a) Symbol of an ideal current source (b) Characteristic curve Figure 2. respectively.15 . Its two-terminal voltage is determined by the external circuit or load. This means the current of the ideal voltage source is independent of variations in its external circuit or load. and its characteristic curve is shown in Figure 2.15: The load resistance RL is 1 000 and 50 O. Vab ¼ ISRL. so it can provide a maximum current to the load.27 Circuit for Example 2.Basic laws of electric circuits 51 current source will not change even an external circuit (load RL) is connected to it. Its voltage depends on variations in its external circuit. Determine the terminal voltage Vab for the ideal current source in the circuit. Its internal resistance RS ¼ ?. The symbol of an ideal current source is shown in Figure 2.26 Ideal current source Ideal current source ● ● ● It can provide a constant output current IS that does not depend on the variations in its external circuit. Example 2. and the direction of the arrow is the current direction of the source.26(a). so it is an independent current source. 28 Open circuit and short circuit of an ideal current source 2.28(b).29 A real current source . the current of the source will change if the load resistance RL changes.2. as shown in Figure 2. Once a load resistor RL is connected to the current source as shown in Figure 2. as shown in Figure 2. a I IS RS RL b Figure 2.28(a).4. I ¼ IS. The real current source (or current source) can be represented as an ideal current source IS in parallel with an internal resistor RS.2 Real current source Usually a real-life application of current source will not reach a perfect constant output current after it is connected to an external circuit or load. a a I ∞ IS Vab 0 I IS 0 Vab b (a) Open circuit b (b) Short circuit Figure 2. Since the internal resistance RS of the current source usually is very large. the load current I will be a little bit lower than the source current IS. as the real current sources all have a non-infinite internal resistance RS. I ¼ 0. Vab ¼ 0. Vab ¼ ?.29.52 Understandable electric circuits Solution: When RL ¼ 1 000 O. Short circuit. Vab ¼ IS RL ¼ ð0:02AÞ ð1 000 OÞ ¼ 20 V When RL ¼ 50 O. Vab ¼ IS RL ¼ ð0:02AÞ ð50 OÞ ¼ 1 V The conditions of open circuit and short circuit of an ideal current source are as follows: ● ● Open circuit. Table 2. The internal resistance of a real current source usually is much greater than the load resistance (RS ) RL). Each physical quantity has a SI unit. which is the international authority that ensures dissemination and modifications of the SI units to reflect the latest advances in science and commerce. .1 International system of units (SI) The international system of units (SI) was developed at the General Conference of the International Weight and Measures.1 SI basic units Quantity Length Mass Time Current Temperature Amount of substance Intensity of light Quantity symbol l M t I T m I Unit Metre Kilogram Second Ampere Kelvin Mole Candela Unit symbol m kg s A K mol cd SI units International system of units (SI) is the world’s most widely used modern metric system of measurement.5. There are seven base units of the SI system. A higher internal resistance RS can provide a higher current through the external circuit of the real current source.Basic laws of electric circuits 53 Once the load resistance RL changes.1. This is why the current of the real current source is not possible to keep at an ideal constant level. There are seven basic units of the SI system and they are listed in Table 2.5 International units for circuit quantities 2. Real current source (current source) ● ● It has an internal resistance RS (RS ) RL). which means the international system of units or the metric system to most people. and therefore the output current of the real current source is approximately stable. RS is in parallel with the current source. the current in the load will also changes. SI originates from the French ‘Le Syste`me International d’Unite´s’. SI system is the world’s most widely used modern metric system of measurement. 2. and multiply the positive exponent of 10 (moving the decimal point three places each time). In general science. Some derived SI units for circuit quantities are given in Table 2.16(a). In circuit analysis. A metric prefix is a modifier on the root unit that is in multiples of 10. centi and kilo are used. . 2.2. Example 2. the most common metric prefixes.5.2 Some circuit quantities and their SI units Quantity Voltage Resistance Charge Power Energy Electromotive force Conductance Resistivity Quantity symbol V R Q P W E or VS G r Unit Volt Ohm Coulomb Watt Joule Volt Siemens Ohm Á metre Unit symbol V S C W J V S OÁm As you study the circuit theory more in-depth. Table 2. such as nano and pico are used. more metric prefixes. such as milli. move the decimal point to the left.2 Metric prefixes (SI prefixes) Some time there are very large or small numbers when doing circuit analysis and calculation. you may use and add more circuit quantities and their derived SI units in this table. A metric prefix (or SI prefix) is often used in the circuit calculation to reduce the number of zeroes.3 contains a complete list of metric prefixes.16: ðaÞ ðbÞ ðcÞ ðdÞ ðaÞ ðbÞ ðcÞ ðdÞ 47 000 O ¼ ð?Þ kO 0:0505 A ¼ ð?Þ mA 0:0005 V ¼ ð?Þ mV 15 000000 000 C ¼ ð?Þ GC 47000 O ¼ 47  103 O ¼ 47 kO 0:0505 A ¼ 50:5  10À3 A ¼ 50:5 mA 0:0005 V ¼ 500  10À6 V ¼ 500 mV 15000 000000 C ¼ 15  109 ¼ 15 GC Solution: Note: ● If a number is a whole number. Large and small numbers are made by adding SI prefixes. In Example 2. Table 2.54 Understandable electric circuits All other metric units can be derived from the seven SI basic units that are called ‘derived quantities’. 47 000O ¼ 47 6 103O (moving the decimal point three places to the left). Basic laws of electric circuits Table 2.3 Metric prefix table Prefix Yotta Zetta Exa Peta Tera Giga Mega myria kilo hecto deka deci centi milli micro nano pico femto atto zepto yocto Symbol (abbreviation) Y Z E P T G M my k h da d c m m (mu) n p f a z y Exponential (power of 10) 1024 1021 1018 1015 1012 109 106 104 103 102 10 1071 1072 1073 1076 1079 10712 10715 10718 10721 10724 Multiple value (in full) 55 1 000 000 000 000 000 000 000 000 1 000 000 000 000 000 000 000 1 000 000 000 000 000 000 1 000 000 000 000 000 1 000 000 000 000 1 000 000 000 1 000 000 10 000 1 000 100 10 0.1 0.01 0.001 0.000 001 0.000 000 001 0.000 000 000 001 0.000 000 000 000 001 0.000 000 000 000 000 001 0.000 000 000 000 000 000 001 0.000 000 000 000 000 000 000 001 Note: The most commonly used prefixes are shown in bold. ● ● If a number is a decimal number, move the decimal point to the right, and multiply the negative exponent of 10. In Example 2.16(b), 0.0505 A ¼ 50.5 6 1073 A (moving the decimal point three places to the right). If numbers have different prefixes, convert them to the same prefix first, then do the calculation. Example 2.17: Determine the result of 30mA+2000mA. Solution: 30 mA þ 2 000 mA ¼ 30  10À3 A þ 2 000  10À6 A ¼ 30  10À3 A þ 2  10À3 A ¼ ð30 þ 2ÞmA ¼ 32 mA Summary Basic concepts ● Power: the speed of energy conversion, or work done per unit of time, P ¼ W/t. 56 ● ● Understandable electric circuits Energy: the ability to do work. The reference direction of power: ● If a circuit has mutually related reference polarity of current and voltage: P 4 0 (absorption energy). ● If a circuit has non-mutually related reference polarity of current and voltage: P 5 0 (releasing energy). Branch: a current path between two nodes where one or more circuit components in series. Node: the intersectional point of two or more current paths where current has several possible paths to flow. Supernode: a part of the circuit that contains several nodes and branches. Loop: a complete current path where current flows back to the start. Mesh: a loop in the circuit that does not contain any other loops. Ideal voltage source: can provide a constant terminal voltage that does not depend on the variables in its external circuit. Its current depends on variables in its external circuit, Vab ¼ VS, RS ¼ 0. Real voltage source: with a series internal resistance RS (RS ( RL), the terminal voltage of the real voltage source is: Vab ¼ VS 7 IRS. Ideal current source: can provide a constant output current Is that does not depend on the variations in its external circuit, RS ¼ ?. Its voltage depends on variations in its external circuit. Real current source: with an internal resistance RS in parallel with the ideal current source, RS ) RL. ● ● ● ● ● ● ● ● ● Formulas ● ● ● ● ● ● Work: W ¼ FS Power: P ¼ W/t Electrical power: P ¼ IV ¼ I2R ¼ V2/R KVL #1: SV ¼ 0 ● Assign a þve sign for V or E if its reference polarity and loop direction are the same. ● Assign a 7ve sign for V or E if its reference polarity and loop direction are opposite. KVL #2: SV ¼ SE ● Assign a þve sign for V if its reference polarity and loop direction are the same; assign a 7ve sign for V if its reference direction and loop direction are opposite. ● Assign a 7ve sign for E if its reference polarity and loop direction are the same; assign a þve sign for E if its polarity and loop direction are opposite. KCL #1: SIin ¼ 0 ● Assign a þve sign for I if current is entering the node. ● Assign a 7ve sign for I if current is exiting the node. Basic laws of electric circuits ● 57 ● KCL #2: SIin ¼ Iout ● Assign a þve sign for Iin if current is entering the node; assign a 7ve sign for Iin if current is exiting the node. ● Assign a þve sign for Iout if current is exiting the node; assign a 7ve sign for Iout if current is entering the node. Some circuit quantities and their SI units Quantity Voltage Resistance Charge Power Energy Electro motive force Conductance Resistivity Quantity symbol V R Q P W E or VS G r Unit Volt Ohm Coulomb Watt Joule Volt Siemens Ohm Á metre Unit symbol V O C W J V S OÁm ● The commonly used metric prefixes Prefix pico nano micro milli kilo mega giga tera Symbol (abbreviation) p n m m K M G T Exponential (power of 10) 10712 1079 1076 1073 103 106 109 1012 Experiment 2: KVL and KCL Objectives ● ● ● ● ● Construct and analyse series and parallel circuits. Apply Ohm’s law and plot I–V characteristics. Experimentally verify KVL. Experimentally verify KCL. Analyse experimental data, circuit behaviour and performance, and compare them to theoretical equivalents. Background information ● Ohm’s law: V ¼ IR 58 ● Understandable electric circuits I–V characteristics: I R 0 V ● ● KVL: SV ¼ 0, or SV ¼ SE KCL: SI ¼ 0, or SIin ¼ SIout Equipment and components ● ● ● ● ● ● ● Digital multimeter (DMM) Breadboard DC power supply Switch Resistors: 240 O, 2.4 kO, 91 O, 2.7 kO, 3.9 kO and 910 O Some alligator clips Some wires and leads with banana-plug ends Notes: (apply these notes to all experiments in this book) ● ● ● ● ● The ammeter (function) of the multimeter should be connected to the circuit after the power supply has been turned off. The voltmeter (function) of the multimeter should be connected in parallel with the component to measure voltage, and ammeter (function) should be connected in series with the component to measure current. Turn off the power supply before doing any circuit rearrangement, otherwise it will damage or harm experimental devices and components. Connect the negative terminal of the power supply to the ground using the black wire, and connect the positive terminal of the power supply to the component using the red wire. Connect other circuit components using different colour wires other than red and black. Use the actual resistance values to do the calculation. Multimeter guide ● ● Recall: A multimeter is an electrical and electronic measuring instrument that combines functions of ammeter, voltmeter, ohmmeter, etc. Method for measuring voltage with a digital multimeter (voltmeter function): ● Turn on the power supply after components have been connected in the circuit. Basic laws of electric circuits ● 59 Insert the multimeter’s leads into the sockets labelled COM and V/O and turn on the multimeter (Figure L2.1). Figure L2.1 Multimeter ● ● ● ● ● Turn the central selector switch pointing to the voltage ranges with the DCV sign (DCV is for measuring DC voltage, and ACV is for measuring AC voltage), and where the estimated voltage value should be less than the maximum range. Make the measurement by connecting the component in parallel with the two leads of the multimeter. Connect or touch the red lead from the multimeter to terminal of the component, which is expected to have the more positive voltage, and connect or touch the black lead to the other terminal of the component. Read the displayed voltage value on the scale. Turn off the multimeter after the measurement. Method for measuring current with a digital multimeter (ammeter function): ● Insert the multimeter’s leads into the sockets labelled COM and A. ● Connect the multimeter in series with the resistor branch that is going to make a measurement. 60 ● Understandable electric circuits Turn the central selector switch pointing to the current ranges (with mA or 10 A sign) where the estimated current value is closed to the maximum range. Turn on the power supply. Read the displayed current value on the scale. Turn off the multimeter after the measurement. ● ● ● Procedure Part I: Kirchhoff’s voltage law (KVL) 1. Use the resistor colour code to choose three resistors with resistor values listed in Table L2.1. Measure each resistor using the multimeter (ohmmeter function). Record the values in Table L2.1. Table L2.1 Resistance Colour code resistor value Measured value R1 240 O R2 2.4 kO R3 91 O 2. Construct the series circuit shown in Figure L2.2 on the breadboard. R1 = 240 E 9V R3 = 91 Ω R2 = 2.4 kΩ Figure L2.2 A series circuit 3. Calculate circuit current and voltages cross each resistor in Figure L2.2 (assuming the switch is turned on). Record the values in Table L2.2. Table L2.2 I Formula for calculations Calculated value Measured value VR1 VR2 VR 3 VT Basic laws of electric circuits 61 4. Set the power supply to 9 V, then turn on the switch, connect the multimeter (voltmeter function) in parallel with each resistor and power supply, and measure voltages across each resistor and power supply. Record the values in Table L2.2. 5. Use the direct method or indirect method to measure the circuit current. Record the value in Table L2.2. ● Direct method: Connect the multimeter (ammeter function) in series with the circuit components, then turn on the switch and measure circuit current directly. ● Indirect method: Apply Ohm’s law to calculate the current with measured voltage and resistance. 6. Use measured values to plot I–V characteristics for 240 O resistor. 7. Substitute the measured voltage values from Table L2.2 into KVL equations to verify SV ¼ 0 and SV ¼ SE. Part II: Kirchhoff’s current law (KCL) 1. Construct a parallel circuit as shown in Figure L2.3 to the breadboard. 10 V 2.7 kΩ 3.9 kΩ 910 Ω Figure L2.3 A parallel circuit 2. Calculate each branch current and total current in the circuit of Figure L2.3 (assuming the switch is turned on). Record the values in Table L2.3. Table L2.3 Current Formula for calculation Calculated values Measured values I1 I2 I3 IT 3. Set the power supply to 10 V, turn on the switch, measure each branch current and total current in the circuit by using direct or indirect methods (get the multimeter to function as an ammeter). Record the values in Table L2.3. 4. Substitute the measured current values from Table L2.3 into KCL equations to verify SI ¼ 0 and Iin ¼ SIout. 62 Understandable electric circuits Conclusion Write your conclusions below: . parallel circuits and series–parallel circuits know how to determine the equivalent resistance for series. It is very important to construct electric circuits in different ways as to make practical use of them. It has all its elements connected in one loop of wire. There is only one pathway by which charges can travel in a series circuit. i. parallel and series–parallel resistive circuits calculate the resistance. The water flows through the pipe from tank to tank. The same amount of charges will flow in each component of the circuit. the current flow is the same throughout the circuit. parallel and series–parallel resistive circuits are very often used electrical or electronic circuits.e. voltage.1. The same amount of water will flow in each tank. such as a light bulb. parallel and series–parallel resistive circuits understand and apply the voltage-divider (VDR) and current-divider (CDR) rules identify the wye (Y) and delta (D) circuits know the method of wye (Y) and delta (D) conversions apply the method of D–Y conversions to simplify bridge circuits understand the method for measuring the unknown resistance of a balanced bridge circuit 3. The same is true of an electrical circuit.Chapter 3 Series–parallel resistive circuits Objectives After completing this chapter.1 Series resistive circuits A series circuit is the simplest circuit. so it has just one current in the series circuit. 3.1 Series resistive circuits and voltage-divider rule Series. you will be able to: ● ● ● ● ● ● ● ● identify series circuits. It can be analogized by water flowing in a series of tanks connected by a pipe. . current and power for series. once the power turns on..1 Series circuit Many practical series circuits may not be as easily identifiable as Figures 3.1 illustrates an electrical circuit with three light bulbs (resistors) connected in series.2 (a–c) Series resistive circuits and (d) series circuit . the same current reading will be read on each ammeter.1 and 3.64 Understandable electric circuits Series circuit ● ● ● The components are connected one after the other. the ammeter will still read the same current value. Figure 3. E Figure 3.2(a). A practical example of a series circuit is a string of old Christmas lights. The current flow through each component is always the same.2(b and c) are also series circuits but drawn in different ways. There is only one current path. As long as the circuit elements are connected one after the other. Rn + Vn − (d) Figure 3. (a) R1 + (b) R2 R3 + V3 − E + - (c) V1 − + V 2 − . Figure 3. . If an ammeter is connected behind each light bulb.. For example. . and there . If we swap the position of the ammeter and the light bulb. and the terminal of the resistor connecting to the negative side of the voltage source is negative. Total series resistance (RT) or equivalent resistance (Req) RT ¼ R1 þ R2 þ Á Á Á þ Rn . It does not matter if there is a different arrangement of the elements.1. The total resistance of a series resistive circuit is always greater than any single resistance in that circuit. The total voltage VT in the circuit of Figure 3.1. 3.Series–parallel resistive circuits 65 is only one current path for the circuit.1.2 Total series resistance (or equivalent resistance) The total resistance (RT) of a series resistive circuit is the sum of all resistances in the circuit. the total voltage will be as follows: Total series voltage (VT or E) VT ¼ E ¼ V1 þ V2 þ Á Á Á þ Vn VT ¼ IR1 þ IR2 þ Á Á Á þ IRn ¼ IRT ð3:1Þ For a series resistive circuit. 3. the source voltage shared by each resistor. The equivalent resistance of a series resistive circuit is the amount of resistance that a single resistor would need to equal the overall effect of the all resistors that are present in the circuit.2(d) can be determined by Kirchhoff’s voltage law (KVL) and Ohm’s law.1) of the total voltage gives VT ¼ IðR1 þ R2 þ Á Á Á þ Rn Þ and R T ¼ R 1 þ R 2 þ Á Á Á þ Rn RT is the mathematical equation for computing the total resistance (or equivalent resistance Req) of a series resistive circuit.1 Total series voltage The voltage across the source or power supply (total voltage) is equal to the sum of the voltage that drops across each resistor in a series circuit. For n resistors connected in series. It is also called the equivalent resistance (Req) because this resistance is equivalent to the sum of all resistances when you look through the two terminals of the series resistive circuit. The terminal of the resistor connecting to the positive side of the voltage source is positive. (3.e. i.1. it is said that they are connected in series. Multiply the current I on both sides of the total voltage equation VT ¼ E ¼ V1 þ V2 þ Á Á Á þ Vn. that the current flowing through each element is always the same and that the current is always the same at any point in a series circuit.3 Series current From the definition of a series circuit we know that there is only one current path in a series circuit. such as the one in Figure 3. the total power is actually the power supplied by the source.1. The total power (PT) consumed by a series circuit is the sum of power dissipated by the individual resistor.66 Understandable electric circuits 3.4 Series power Each of the resistors in a series circuit consumes power.1. can be determined from Ohm’s law as follows: Series current (I) I¼ VT E V1 V2 Vn ¼ ¼ ¼ ¼ ÁÁÁ ¼ RT RT R1 R2 Rn 3. Since this power must come from the source.2(d). which is dissipated in the form of heat. Therefore. The current I flowing in a series resistive circuit. to get the total power PT ¼ IE ¼ IV1 þ IV2 þ Á Á Á þ IVn.1.1. the total power in a series resistive circuit can be expressed as follows: Total series power (PT) PT ¼ P1 þ P2 þ Á Á Á þ P n or PT ¼ IE ¼ I2RT ¼ (E2/RT) The power dissipated by the individual resistor in a series resistive circuit is as follows: 2 V1 R1 2 V2 P2 ¼ I 2 R2 ¼ IV2 ¼ R2 ÁÁÁ P1 ¼ I 2 R1 ¼ IV1 ¼ Pn ¼ I 2 Rn ¼ IVn ¼ 2 Vn Rn . 2 Voltage-divider rule (VDR) The VDR can be exhibited by using a pot (short for potentiometer).e. Using a voltmeter to measure the voltage across the pot.1: A series resistive circuit is shown in Figure 3. A pot is a variable resistor whose resistance across its terminals can be varied by turning a knob.3 Circuit for Example 3.1 Solution: (a) RT ¼ R1 þ R2 þ R3 ¼ (10 þ 20 þ 30)kO ¼ 60 kO (b) I ¼ (E/RT) ¼ (60 V/60 kO) ¼ 1 mA (c) V1 ¼ IR1 ¼ (1 mA)(10 kO) ¼ 10 V (d) VT ¼ IRT ¼ (1 mA)(60 kO) ¼ 60 V. VT ¼ E ¼ 60 V (checked) (e) PT ¼ IE ¼ (1 mA)(60 V) ¼ 60 mW or PT ¼ I2RT ¼ (1 mA)2(60 kO) ¼ 60 mW (checked) 3. and the voltage will decrease when the arrow moves down.Series–parallel resistive circuits 67 Example 3. the voltage relative to the negative side of the 100 V voltage source is ½ E ¼ 50 V when the arrow (knob) is at the middle of the potentiometer. The voltage will increase when the arrow moves up. as shown in Figure 3. Determine the following: (a) (b) (c) (d) (e) Total resistance RT Current I in the circuit Voltage across the resistor R1 Total voltage VT Total power PT R1 = 10 kΩ E 60 V R3 = 30 kΩ R2 = 20 kΩ Figure 3. This is the principle of the voltage divider.3.1. the voltage divider is a design technique used to create different output voltages that is proportional . i. A pot is connected to a voltage source.4(a). using the same method we can obtain the general form of the VDR as follows: I¼ E E ¼ R1 þ R2 þ Á Á Á þ Rn RT E RX RX ¼ E RT RT or VX ¼ V T RX RT VX ¼ IRX ¼ where RX and VX are the unknown resistance and voltage. . Therefore.4(a) is equivalent to (b) since R ¼ R1 þ R2 ¼ 100 kO. a potentiometer itself is an adjustable voltage divider. I E 100 V ← R1 = 50 kΩ R2 = 50 kΩ R = 100 kΩ E 100 V (a) (b) Figure 3. I¼ E R1 þ R 2 E R1 50 kO ¼ 50 V R1 ¼ E ¼ 100 V R 1 þ R2 R1 þ R 2 ð50 þ 50ÞkO E R2 50 kO R2 ¼ E ¼ 100 V ¼ 50 V R1 þ R 2 ð50 þ 50ÞkO R 1 þ R2 V1 ¼ IR1 ¼ V2 ¼ IR2 ¼ The above two equations are the VDRs for a series circuit of two resistors. and RT and VT are the total resistance and voltage in the series circuit. The circuit in Figure 3.4(b).4 Voltage divider In Figure 3. the voltage divider means that the source voltage E or total voltage VT is divided according to the value of the resistors in the series circuit. The output voltage from the divider changes when any of the resistor values change. Actually. When there are n resistors in series.68 Understandable electric circuits to the input voltage. . R1 = 10 kΩ E 60 V R2 = 20 kΩ R3 = 10 kΩ Figure 3. Example 3.5 Circuit for Example 3. þR V2 ¼ VT R1R2 2 þR Note: The numerator of the VDR is always the unknown resistance (this is worth memorizing). . The knob of the pot in the circuit will eventually let you adjust the volume of the audio equipment. nÞ ● When there are only two resistors in series: V1 ¼ VT R1R1 2 .2 Solution: Use the general form of the VDR VX ¼ E(RX/RT) R2 R2 20 kO ¼E ¼ 60 V ¼ 30 V ð10 þ 20 þ 10Þ kO RT R 1 þ R 2 þ R3 R3 R3 10 kO ¼E ¼ 60 V ¼ 15 V ð10 þ 20 þ 10Þ kO RT R 1 þ R 2 þ R3 V2 ¼ E V3 ¼ E The practical application of the voltage divider can be the volume control of audio equipment. .5.Series–parallel resistive circuits 69 VDR ● General form: VX ¼ VT RX RT or VX ¼ E RX RT ðX ¼ 1. 2. . .2: Use the VDR to determine the voltage drops across resistors R2 and R3 in the circuit of Figure 3. Example 3.1. An earth ground usually consists of a ground rod or a conductive pipe driven into the soil. such as a metal plate. the earth is an electrically neutral body. the voltage with the single-subscript notation (such as Vb) is the voltage drop from the point b with respect to ground.6 Circuit for Example 3. so that it also provides a point that has zero voltage. A chassis ground is a connection to the main chassis of a piece of electronic or electrical equipment. Vbe and Vb in the circuit of Figure 3. And the voltage with the double-subscripts notation (such as Vbc) is the voltage drop across the two points b and c (each point is represented by a subscript). a R1 = 10 kΩ b R2 = 20 kΩ E 120 V R4 = 10 kΩ c R3 = 20 kΩ d e Figure 3. and provides a reference voltage level in which all other voltages in a circuit are measured. Since an equal number of negative and positive charges are distributed throughout the earth at any given time. All the common points are electrically connected together through metal plates or wires. There are two types of circuit grounds: one is the earth ground and another is the common ground (or chassis ground).70 Understandable electric circuits 3.3 . there is a ground for each electric circuit. So the earth is always at zero potential (0 V) and measurements can be made by using earth as a reference.3: Determine Vbc. Chassis ground is also called common ground. The neutral point in the alternating circuit (AC) is an example of the common ground. The difference between these two grounds can be summarized as follows: ● Earth ground: Connecting one terminal of the voltage source to the earth.6.3 Circuit ground Usually. All chassis grounds should lead to earth ground. The symbol for it is: ● Common ground or chassis ground: The common point for all elements in the circuit. The symbol for the common point is: In a circuit. Electrical circuit grounding is important because it is always at zero potential (0 V). This is because there is only one current path in a series circuit. Thus. Old style Christmas lights were often wired in series. Parallel resistive circuits also can be analogized by flowing water. When the water has passed the islands. Vb ¼ Vbe ¼ 100 V Ground and voltage subscript notation ● ● ● ● Earth ground: connects to the earth (V ¼ 0). An obvious advantage of the parallel circuit is that burn out or removal of one bulb does not affect the other bulbs. Single-subscript notation: the voltage from the subscript with respect to ground. and the unknown resistance RX ¼ R2 þ R3 þ R4). 3. the circuit is broken. When water flowing in a river across small islands.2 Parallel resistive circuits and the current-divider rule 3. This scenario is . Double-subscript notation: the voltage across the two subscripts. This is the main disadvantage of a series circuit. so the other bulbs will stay lit. the whole string of lights went off. the one water path will be divided by the islands and split into many more water paths.1 Parallel resistive circuits If any one of the light bulbs or resistors burns out or is removed in a series resistive circuit.Series–parallel resistive circuits Solution: Vbc ¼VR2 ¼ E R2 R2 ¼E RT R1 þ R2 þ R3 þ R4 20 kO 20 kO ¼ 120 V ¼ 40 V ¼120 V ð10 þ 20 þ 20 þ 10ÞkO 60 kO R 2 þ R3 þ R 4 ð20 þ 20 þ 10ÞkO ¼ 100 V ¼ 120V RT 60 kO 71 Vbe ¼ E (Use the general form of the VDR VX ¼ E (RX/RT). Common ground (chassis ground): the common point for all components in the circuit (V ¼ 0). They continue to operate because there is still a separate. so if one light bulb burned out. There the unknown voltage VX ¼ Vbe. A parallel resistive circuit is composed of two or more series circuits connected to the same power source.2. it has more than one path for charges or current to follow. it will become a single water path again. no charges or current will move through the circuit and the entire circuit would stop operating. independent path from the source to each of the other bulbs. The total current I as illustrated in Figure 3.8 are also parallel circuits but drawn in different ways.72 Understandable electric circuits illustrated in Figure 3.7(b).7(b) leaves the positive terminal of the voltage source and flows to node a (or supernode – chapter 2).7(a). Note that the node does not have to physically be one single point.7(a).7(b). the total current I divides into three currents I1. These three currents flow through their resistors and rejoin at node b. As long as several branches are connected together. then that part of the circuit is considered to be a node. As long as the circuit elements are connected end to end and there are at least two current paths in the circuit. which is a connecting point for the four branches. I I1 R1 a I3 R3 E I2 R2 b (a) (b) Figure 3. R1 R2 R4 R1 R3 R2 R1 R3 R2 Figure 3.7 Parallel circuit In the parallel resistive circuit. I2 and I3. Many practical series circuits may not be as easily identifiable as Figure 3. The circuit in Figure 3. the source current divides among the available resistive branches in different paths. it is said that they are connected in parallel. From the parallel circuit in Figure 3. and circuits in Figure 3. we can see that circuit elements (resistors) are connected in parallel if the ends of one element are connected directly to the corresponding ends of the other. It does not matter if there are different arrangements of the elements.8 Parallel resistive circuits .7(b) is parallel resistive circuit equivalent to Figure 3. The total current then flows from b back to the negative terminal of the source. At node a. 2. If all the resistors are light bulbs and have the same resistances as in Figure 3.2 Parallel current If the parallel circuit was a river.1. 3. .Series–parallel resistive circuits 73 A parallel circuit has two main advantages when compared with series circuits. There are at least two current paths in the circuit. and the total current entering and exiting parallel resistive circuit is the same. The first is that a failure of one element does not lead to the failure of the other elements.9. IT I1 E R1 V 1 R2 V I2 V2 In … … Vn Rn Figure 3. the voltage between these two notes must be the same.7(a)).1. The other is that more elements may be added in parallel without the need for increasing voltage.9. The voltage across each component is the same. All these must be equal to the supply voltage E for the parallel resistive circuit shown in Figure 3. Parallel circuit ● ● ● The components are connected end to end. This is the same with the current in the parallel resistive circuit.1 Parallel voltage Since all resistors in a parallel resistive circuit are connected between the two notes. the total volume of water in the river would be the sum of water in each branch (Figure 3.9 V and I in a parallel circuit The voltage drop across each resistor must equal the voltage of the source in a parallel resistive circuit.2. This can be expressed in the following mathematical equation: Parallel voltage V ¼ E ¼ V1 ¼ V 2 ¼ Á Á Á ¼ V n 3. The total current is equal to the sum of currents in each resistive branch. In this case. the voltage drop across each resistor must be the same. they will glow at the same brightness as they each receive the same voltage. 1. . and the characteristics of the total current in a parallel circuit can be expressed in the following equations: Parallel current ● Branch currents: I1 ¼ Total current: IT ¼ V . The total resistance in a parallel circuit can be found by applying Ohm’s law to the equation of the total current: V V V V 1 1 1 ¼ I1 þ I2 þ Á Á Á þ In ¼ þ þ ÁÁÁ þ ¼V þ þ ÁÁÁ þ Req R1 R2 Rn R 1 R2 Rn Dividing the voltages on both sides of the equal sign in the above equation gives 1 1 1 1 ¼ þ þ ÁÁÁ þ Req R1 R2 Rn Solving for Req from the above equation will give the equivalent resistance for the parallel circuit: Req ¼ 1 ð1=R1 Þ þ ð1=R2 Þ þ Á Á Á þ ð1=Rn Þ So the total resistance of a set of resistors in a parallel resistive circuit is found by adding up the reciprocals of the resistance values and then taking the reciprocal of the total. . . therefore. Geq ¼ 1 1 1 1 ¼ þ þ ÁÁÁ þ ¼ G1 þ G2 þ Á Á Á þ Gn Req R1 R2 Rn . It will be more convenient to use the conductance than the resistance in the parallel circuits. R1 I2 ¼ V V .9? It depends on the amount of resistance in each branch.3 Equivalent parallel resistance How much current is flowing through each branch in the parallel resistive circuit in Figure 3. In ¼ R2 Rn ● V ¼ I1 þ I2 þ Á Á Á þ In Req 3. Since the conductance G ¼ 1/R. the branch currents will be different. . The branch currents can be determined by Ohm’s law.74 Understandable electric circuits If resistances are different in each branch of a parallel circuit.2. given Req ¼ 1.25 kO. R1 ¼ 20 kO. Determine (a) R2. the total power equation for the parallel circuits is obtained as follows: IT V ¼ I1 V þ I2 V þ Á Á Á þ In V ¼ PT Total parallel power PT ¼ P1 þ P2 þ Á Á Á þ Pn or 2 V PT ¼ IT V ¼ IT Req ¼ Req 2 The power consumed by each resistor in a parallel circuit is expressed as: 2 P1 ¼ I1 V ¼ I1 R1 ¼ V2 V2 V2 2 2 .10.2. R3 ¼ 2 kO and I3 ¼ 18 mA. . Pn ¼ In V ¼ In Rn ¼ R1 R2 Rn Example 3. By multiplying a voltage V to both sides of the equation for total current IT ¼ I1 þ I2 þ Á Á Á þ In. 3.Series–parallel resistive circuits When there are only two resistors in parallel: Req ¼ 1 1 R1 R2 ¼ ¼ ð1=R1 Þ þ ð1=R2 Þ ðR1 þ R2 Þ=ðR1 R2 Þ R1 þ R2 75 Usually parallel can be expressed by a symbol of ‘//’ such as: R1 // R2 // Á Á Á // Rn. (b) IT and (c) P3. .4: A parallel circuit is shown in Figure 3. . . . P2 ¼ I2 V ¼ I2 R2 ¼ .4 Total parallel power The total power is the sum of the power dissipated by the individual resistors in a parallel resistive circuit. Equivalent parallel resistance and conductance ● Req ¼ 1 ¼ R1 ==R2 == Á Á Á ==Rn ð1=R1 Þ þ ð1=R2 Þ þ Á Á Á þ ð1=Rn Þ R1 R2 ¼ R1 ==R2 R1 þ R2 ● Geq ¼ G1 þ G2 þ Á Á Á þ Gn When n ¼ 2: Req ¼ Note: The total resistance of the series resistive circuit is always greater than the individual resistance. So usually for parallel circuits the equivalent resistance is used instead of the total resistance.1. and the total resistance of the parallel resistive circuit is always less than the individual resistance. 2. . . This was shown in Figure 3. It is like a river.10 Figure for Example 3. As previously mentioned.7(a). Geq ¼ G1 þ G2 þ G3 G2 ¼ Geq À G1 À G3 ¼ ¼ 0:25 mS . the water flowing through the river will be divided by small islands and the flow is split creating more water paths. parallel circuits can be analogized by flowing water.9.76 Understandable electric circuits IT I1 E R1 R2 R3 I3 I2 Figure 3. .2 Current-divider rule (CDR) The VDR can be used for series circuits. In ¼ R2 Rn . R1 I2 ¼ E E . . In this circuit: I1 ¼ E . 1 1 ¼ ¼ 4 kO G2 0:25 mS 1 1 Req ¼ ¼ ð1=R1 Þ þ ð1=R2 Þ þ ð1=R3 Þ ð1=20 kOÞ þ ð1=4 kOÞ þ ð1=2 kOÞ ¼1:25 kO ðprovedÞ R2 ¼ E V3 I3 R3 ð18 mAÞð2 kOÞ ¼ ¼ ¼ ¼ 28:8 mA Req Req 1:25 kO Req 1 1 1 1 1 1 À À ¼ À À Req R1 R3 1:25 kO 20 kO 2 kO (b) IT ¼ (c) P3 ¼ I23R3 ¼ (18 mA)2(2 kO) ¼ 648 mW 3. The equations of the current divider can be derived by the following: A parallel resistive circuit with n resistors was shown in Figure 3. and the CDR can be used for parallel circuits.4 Solution: (a) Since R2 ¼ 1/G2. determine G2 first. . RT R1 . When there are only two resistors in parallel: Req ðR1 R2 =ðR1 þ R2 ÞÞ R2 ¼ IT ¼ IT R1 R1 R1 þ R2 I1 ¼ IT I2 ¼ IT Req ðR1 R2 =ðR1 þ R2 ÞÞ R1 ¼ IT ¼ IT R2 R1 þ R2 R2 CDR ● General form: IX ¼ IT Req X R GX or IX ¼ IT Geq ● When there are two resistors in parallel: I1 ¼ IT R1R2 2 . the numerator is the other resistance (other than the unknown resistance). . Note: The CDR is similar in form to the VDR. . þR I2 ¼ IT R1R1 2 þR There IX and RX are unknown current and resistance. . R1 þ R2 R2 R 1 þ R2 VX ¼ VT V1 ¼ VT V 2 ¼ VT . R1 I2 ¼ IT Req Req . and IT is the total current in the parallel resistive circuit. When there are two resistors in parallel.Series–parallel resistive circuits Inserting E ¼ ITReq into the above equations gives: 77 I1 ¼ IT Req . Recall the VDR: RX . The difference is that the denominator of the general form current divider is the unknown resistance. In ¼ IT R2 Rn These are the general form current-divider equations. the less the current flows through that branch.11 Circuit for Example 3. I2 and I3 in the circuit of Figure 3. IT = 60 mA E I1 R1 = 10 Ω I2 R2 = 20 Ω I3 R3 = 30 Ω Figure 3.78 Understandable electric circuits Example 3. .5 Solution: Req ¼ R1 ==R2 ==R3 ¼ ¼ 1 ð1=R1 Þ þ ð1=R2 Þ þ ð1=R3 Þ 1 % 5:455 O ð1=ð10 OÞÞ þ ð1=ð20 OÞÞ þ ð1=ð30 OÞÞ Req R1 I1 ¼ I T ¼ 60 mA 5:455 O 10 O ¼ 32:73 mA Req R2 I2 ¼ IT ¼ 60 mA 5:455 O 20 O % 16:37 mA Req R3 I3 ¼ IT ¼ 60 mA 5:455 O 30 O ¼ 10:91 mA The conclusion that can be drawn from the above example is that the greater the branch resistance. or the less the share of the total current.5: Determine the current I1.11. 6: Determine the resistance R2 for the circuit in Figure 3. I1 IT = 30 mA R1 = 10 Ω I2 = 10 mA E R2 79 Figure 3. Many circuits have various combinations of series and parallel components. The series–parallel configurations have a variety of circuit forms. or the series–parallel configurations. but combinations of series and parallel circuits. Series–parallel circuit The series–parallel circuit is a combination of series and parallel circuits.e. However.12.6 Solution: Solve R2 from the current-divider formula I2 ¼ IT ½R1 =ðR1 þ R2 Þ I2 ðR1 þ R2 Þ ¼ IT R1 I2 R2 ¼ IT R1 À I2 R1 .3 Series–parallel resistive circuits The most practical electric circuits are not simple series or parallel configurations. the same principles and rules or laws that have been introduced in the previous chapters are applied. and some of them may be very complex. i. R2 ¼ R2 ¼ R1 ðIT À I2 Þ I2 10 Oð30 À 10ÞmA 10 mA ¼ 20 O 3. The key to solving series–parallel circuits is to identify which parts of the circuit are series and which parts are parallel and then simplify them to an equivalent circuit and find an equivalent resistance.12 Circuit for Example 3.Series–parallel resistive circuits Example 3. . circuit elements are series-connected in some parts and parallel in others. This can be expressed by the equivalent circuit in Figure 3.3.8 Solution: Req ¼ [(R5 // R6 þ R4) // (R7 þ R8 ) þ R3] // R2 þ R1 . Plot the equivalent circuit if necessary. That is the equivalent resistance Req for the series–parallel circuit.13(b).13 Circuits for Example 3. In Figure 3.80 Understandable electric circuits 3. the resistor R5 is in series with R6 and in parallel with R4 and R3. R2 Req R1 R5 R6 R1 R2 (R5 + R6) // R4 // R3 R3 R4 (a) (b) Figure 3. Repeat the above steps until the resistances in the circuit can be simplified to a single equivalent resistance Req.13(a). R2 is in series with (R5 þ R6) // R4 // R3 and in parallel with R1.13. Req ¼ {[(R5 þ R6) // R4 // R3] þ R2} // R1.14 Circuits for Example 3. Note: Determine Req step by step from the far end of the circuit to the terminals of the Req.1 Equivalent resistance Method for determining the equivalent resistance of series–parallel circuits: ● ● ● ● Determine the equivalent resistance of the parallel part of the series– parallel circuits.14(a).7 Example 3. R3 E R1 R4 R7 R8 R3 R2 R5 E R1 R2 (R5 // R6 + R4) // (R7 + R8) R6 (a) (b) Figure 3. i.8: Determine the equivalent resistance Req for the circuit shown in Figure 3. Determine the equivalent resistance of the series part of the series–parallel circuits.e.7: Analysis of the series–parallel circuit in Figure 3. Example 3. IT I1 R1 E I3 R3 C B A I2 R2 I4 R4 R5 Figure 3. the total current as well as currents and voltages for each resistor can be determined by using the following steps: ● ● Apply Ohm’s law with the equivalent resistance solved from the previous section to determine the total current in the equivalent circuit IT ¼ E/Req.2 Method for analysing series–parallel circuits After determining the equivalent resistance of the series–parallel circuit. CDR.9 Solution: Req ¼ [(R3 // R4 // R5) þ R2 ] // (R6 þ R7) þ R1 3.9: Determine the Req for the circuit shown in Figure 3.Series–parallel resistive circuits Example 3.16.3. R6 R2 R7 R1 R3 R4 R5 (a) R2 81 E E R1 R6 + R7 R3 (b) R4 R5 Figure 3.10: Determine the currents and voltages for each resistor in the circuit of Figure 3. Apply the VDR. KCL and KVL to determine the unknown currents and voltages.10 .15(a).15 Circuits for Example 3.16 Circuit for Example 3. Example 3. Ohm’s law. I2 ¼ R1 R2 R2 R1 or I1 ¼ IT .5 ¼ VBC R4 þ R5 ðKCLÞ ðKVLÞ ● ● Check: IT ¼ I1 þ I2 or IT ¼ I3 þ I4.11: Determine the current IT in the circuit of Figure 3. RAB ðI5 ¼ ?. I2 ¼ IT ðthe CDRÞ R1 þ R2 R1 þ R2 VR3 ¼ VR4 þ VR5 ¼ VBC ¼ IT ½ðR4 þ R5 Þ==R3 I3 ¼ VBC . RCD RCD RCD ¼ ½ðR8 þ R9 Þ==R7 þ R6 RCD ¼ ? . I1 ¼ . I6 ¼ ?Þ RAB ¼ ? RAB ¼ ½ðR1 ==R2 þ R3 Þ==R4 þ R5 VCD VCD I6 ¼ I6 ¼ . R3 I4.17. R1 A IT R2 R3 R4 R5 I5 E B D I6 R7 C R6 R8 R9 Figure 3.17 Circuit for Example 3.11 Solution: IT ¼ I5 + I6 The method of analysis: I5 ¼ IT ¼ ? IT ¼ I 5 þ I6 VAB ¼ E. I5 ¼ VAB RAB VAB . VCD ¼ E.82 Understandable electric circuits Solution: ● Req ¼ (R1 // R2) þ [(R4 þ R5) // R3] E IT ¼ Req VAB VAB VR1 ¼ VR2 ¼ VAB ¼ IT ðR1 ==R2 Þ.5 VAB þ VBC ¼ E Example 3. They can also be used in the bridge circuit that will be discussed later. They are also referred to as tee (T) and pi (p) circuits as shown in Figure 3. the configuration of three resistors Ra. Wye (Y) and delta (D) configurations R1 Y or T configuration: D or p configuration: Rb R2 R3 Ra Rc Wye (Y) and delta (D) configurations are often used in three phase AC circuits. It is very important to know the conversion method of the two circuits and be able to convert back and forth between the wye (Y) and delta (D) configurations. Rb and Rc in the circuit of Figure 3.19.e. and the analysis method for series–parallel circuits described in previous chapters may not apply. R2 and R3 in the circuit of Figure 3.1 Wye and delta configurations Sometimes the circuit configurations will be neither in series nor in parallel. in electrical drawings. R2 and R3 in the circuit of Figure 3. R1. respectively. the problem can be easily solved. So how do we determine the equivalent resistance Req for this circuit? If we convert this to the configuration of resistors R1. For example.Series–parallel resistive circuits 83 3. Req ¼ [(R1+ Rd) // (R2 þ Re)] þ R3.4. Rb and Rc in the circuit of Figure 3. .18 Delta (D) and wye (Y) conversions The resistors of Ra.18(a) are said to be in the delta (D) configuration.4 Wye (Y) and delta (D) configurations and their equivalent conversions 3.18(b) is called the wye (Y) configuration.18(b).18(a) are neither in series nor in parallel. i. Ra Rb Req = ? Rd Rc Req = ? R1 Re Rd R3 R2 Re (a) (b) Figure 3. The delta and wye designations are from the fact that they look like a triangle and the letter Y. 84 Understandable electric circuits Δ ( Delta) Y ( Wye) R1 Ra Rb Rc R3 R2 Rc R1 R2 Rb Ra R3 π (Pi) T ( Tee) Figure 3. meaning that the resistors of the circuits between any two terminals must have the same values for both circuits as shown in Figure 3. The delta or wye conversion is used to establish equivalence for the circuits with three terminals. i:e: Rac ðYÞ ¼ Rac ðDÞ Rab ðYÞ ¼ Rab ðDÞ Rbc ðYÞ ¼ Rbc ðDÞ The following equations can be obtained from Figure 3.20: Rac ¼ R1 þ R3 ¼ Rb ==ðRa þ Rc Þ ¼ Rb ðRa þ Rc Þ Rb þ ðRa þ Rc Þ Rc ðRa þ Rb Þ Rc þ ðRa þ Rb Þ Ra ðRb þ Rc Þ Ra þ ðRb þ Rc Þ ð3:2Þ Rab ¼ R1 þ R2 ¼ Rc ==ðRa þ Rb Þ ¼ ð3:3Þ Rbc ¼ R2 þ R3 ¼ Ra ==ðRb þ Rc Þ ¼ ð3:4Þ .2 Delta to wye conversion (D!Y) There are three terminals in the delta (D) or wye (Y) configurations that can be connected to other circuits.4.19 p and T configurations 3.20. subtracting equation (3.Series–parallel resistive circuits Rc a R1 R2 b 85 Rb R3 Ra c Figure 3.4) from the sum of equations (3.4) gives Rb Ra þ Rb Rc Ra Rb þ Ra Rc þ Rb þ ðRa þ Rc Þ Ra þ ðRb þ Rc Þ À Rc Ra þ Rc Rb Rc þ ðRa þ Rb Þ ðR1 þ R3 Þ þ ðR2 þ R3 Þ À ðR1 þ R2 Þ ¼ 2R3 ¼ 2Ra Rb Ra þ Rb þ Rc R3 ¼ Ra Rb R a þ R b þ Rc ð3:5Þ Similarly.2) and (3.3) gives Rb Rc R a þ Rb þ R c R1 ¼ ð3:6Þ And subtracting equation (3.4) gives Ra Rc R a þ Rb þ R c R2 ¼ ð3:7Þ The circuit in delta configuration is converted to wye configuration as shown in Figure 3.2) and (3.21. .3) and (3.3) from the sum of equations (3.20 Delta and wye configurations Subtracting equation (3.2) from the sum of equations (3. 3 Wye to delta conversion (Y!D) The equations to convert wye to delta configuration can be derived based on (3.5).7) gives R2 R3 þ R3 R1 þ R1 R2 ¼ Ra Rc Ra Rb ðRa þ Rb þ Rc Þ þ R2 R3 þ R3 R1 þ R1 R2 ¼ 2 þ Ra Rb Rb Rc ðRa þ Rb þ Rc Þ2 Rb Rc Ra Rc ðRa þ Rb þ Rc Þ2 ðRa þ Rb þ Rc Þ 2 Ra Rb Rc ðRa þ Rb þ Rc Þ ¼ Ra Rb Rc Ra þ Rb þ Rc ð3:8Þ .5) and (3.6) and (3. (3.7). Ra þ Rb þ Rc R2 ¼ Ra Rc .6).7) and (3. Adding the products of equations (3.5–3.4.21 Delta converted to wye configuration Equations for D!Y R1 ¼ Rb Rc . and (3.86 Understandable electric circuits b b a Rb Rc Ra R3 a R1 R2 c c a b Ra R3 a Rb Rc R1 R2 b c c Figure 3. Ra þ Rb þ Rc R3 ¼ Ra Rb Ra þ Rb þ Rc 3. R1 ¼ R2 ¼ R3 ¼ RY. Ra ¼ Rb ¼ Rc ¼ RD.Series–parallel resistive circuits 87 Using (3.3.22.6). R2 For example.8) divided by (3.e.4. i. then all the resistances in the delta (D) configuration will also be the same. (3.22 Wye converted to delta configuration 3.1 If all resistors in the wye (Y) configuration have the same values.8) to divide (3. This can be obtained from Y!D or D!Y conversion equations.6) gives R2 R3 þ R3 R1 þ R1 R2 Ra Rb Rc =ðRa þ Rb þ Rc Þ ¼ Ra ¼ R1 Rb Rc =ðRa þ Rb þ Rc Þ The circuit in wye configuration is converted to delta as shown in Figure 3.5) individually will give the equations to convert wye to delta configurations as follows: Equations for Y!D R1 R2 þ R2 R3 þ R3 R1 . Use Y!D as an example. (3.e.7) and (3. RD ¼ Ra ¼ RD ¼ Rb ¼ RD ¼ Rc ¼ R1 R2 þ R2 R3 þ R3 R1 3RY RY ¼ ¼ 3RY R1 RY R1 R2 þ R2 R3 þ R3 R1 3RY RY ¼ ¼ 3RY R2 RY R1 R2 þ R2 R3 þ R3 R1 3RY RY ¼ ¼ 3RY R3 RY RY and RD . i. R1 R1 R2 þ R2 R3 þ R3 R1 Rc ¼ R3 Ra ¼ Rb ¼ R1 R2 þ R2 R3 þ R3 R1 . a R1 R2 R3 b a Rb Rc b Ra c (a) c (b) Figure 3. Rb ¼ 20 O and Rc ¼ 10 O are the same with the resistances that were given (proved). then Y to D to prove the accuracy of the equations. the delta resistance RD and wye resistance RY has the following relationship: If Ra ¼ Rb ¼ Rc ¼ RD .22. Ra ¼ Rb ¼ Rc ¼ RD ¼ 3RY In the above condition. There the delta resistances Ra ¼ 30 O. 1 RY ¼ RD 3 R1 ¼ R2 ¼ R3 ¼ RY or RD ¼ 3RY Example 3. Rb ¼ 20 O and Rc ¼ 10 O.88 Understandable electric circuits So all delta resistances (RD) have the same values. . Solution: D ! Y: R3 ¼ R2 ¼ R1 ¼ Y ! D: Ra ¼ ¼ Rb ¼ R1 R2 þ R2 R3 þ R3 R1 ½ð3:33Þð5Þ þ ð5Þð10Þ þ ð10Þð3:33Þ O2 ¼ R1 3:33 O 99:95O2 % 30 O 3:33O R1 R2 þ R2 R3 þ R3 R1 99:95 O2 ¼ % 20 O R2 5O R1 R2 þ R2 R3 þ R3 R1 99:95 O2 ¼ % 10 O R3 10 O Ra Rb ð30 OÞð20 OÞ ¼ ¼ 10 O Ra þ Rb þ Rc 30 O þ 20 O þ 10 O Ra Rc ð30 OÞð10 OÞ ¼ ¼ 5O Ra þ Rb þ Rc 30 O þ 20 O þ 10 O Rb Rc ð20 OÞð10 OÞ ¼ % 3:33 O Ra þ Rb þ Rc 30 O þ 20 O þ 10 O Rc ¼ The calculated delta resistances Ra ¼ 30 O.12: Convert D to Y in the circuit of Figure 3. Series–parallel resistive circuits 89 3. is most famous for the Wheatstone bridge circuit. The bridge was invented by Samuel Hunter Christie (1784–1865). R2 and R3 in Figure 3. the equivalent resistance Req has to be determined first. The Wheatstone bridge circuit can be used to measure unknown resistors. (a) Delta (D) and (b) wye (Y) Example 3.23(a) can be converted to Figure 3. However. To determine the equation for the total current IT of the bridge circuit. .23(b) can be determined by the equations of D ! Y conversion. He was the first person who implemented the bridge circuit when he ‘found’ the description of the device. Ra þ Rb þ Rc R3 ¼ Ra Rb R a þ Rb þ R c The equivalent resistance Req of the bridge can be determined as follows: Req ¼ R3 þ ½ðR1 þ R4 Þ==ðR2 þ R5 Þ The total current can be solved as IT ¼ E/Req. and R1. R1 ¼ Rb Rc . R a þ Rb þ R c R2 ¼ Ra Rc . A basic Wheatstone bridge circuit is illustrated in Figure 3.4.23 Wheatstone bridge circuit.4 Using D ! Y conversion to simplify bridge circuits Sir Charles Wheatstone (1802–1875).23(b) using the D ! Y equivalent conversion.13: Determine the equations to calculate the total current IT and branch current I4 for the bridge circuit in Figure 3. The normal series–parallel analysis methods cannot be used to determine this resistance. Figure 3.23(a). a British scientist.23(a). a British physicist and inventor. IT IT C Rb E A R4 Ra Rc R5 B E R1 A R4 I4 C R3 R2 R5 B I4 (a) (b) Figure 3. 24 is zero.5 Balanced bridge When the voltage across points A and B in a bridge circuit shown in Figure 3.e. the Wheatstone bridge is said to be balanced. i. the equivalent resistance Req will be Req ¼ ðRb þ R4 Þ==ðRa þ R5 Þ 3. VAB ¼ 0. A balanced Wheatstone bridge circuit can accurately measure an unknown resistor. VAB ¼ VA þ ðÀVB Þ R2 R4 ÀE R1 þ R2 R3 þ R4 VAB ¼ E ðthe VDRÞ VAB ¼ E R2 ðR3 þ R4 Þ À R4 ðR1 þ R2 Þ R2 R3 À R4 R1 ¼E ðR1 þ R2 ÞðR3 þ R4 Þ ðR1 þ R2 ÞðR3 þ R4 Þ . If the wire between A and B in the circuit of Figure 3. IT R3 R1 E A R2 R4 B Figure 3. i.e.90 Understandable electric circuits The branch current I4 ¼ IT ððR2 þ R5 Þ=½ðR1 þ R4 Þ þ ðR2 þ R5 ÞÞ (the CDR).4.24 A balanced bridge The voltage VAB is the voltage from point A to ground (VA) and then from ground to point B. First determine the voltage VAB in the points A and B.23(a) is open. (A galvanometer is a type of ammeter that can measure small current accurately.25 Measure an unknown R using a balanced bridge Adjust R2 until the current IG measured by the Galvanometer or current in the A and B branch is zero (IG ¼ 0).e. or the bridge is balanced. R2 R3 À R4 R1 ¼ 0.24. This means VAB ¼ 0. this gives: R2 R3 ¼ R4 R1 Balanced bridge When VAB ¼ 0. or when the bridge is balanced. the numerator of the above equation will be zero.Series–parallel resistive circuits 91 When VAB ¼ 0.) IT R3 R1 E A R2 IG RX B Figure 3. RX ¼ R4 at this time can be determined by the equation of the balanced bridge as follows: From: R2 R3 ¼ R4 R1 RX ¼ R4 ¼ R2 R3 R1 Solving for R4 : . R2R3 ¼ R4R1 3. using a variable (adjustable) resistor to replace R2 and connecting a Galvanometer in between terminals A and B can measure the small current IG in terminals A and B as shown in Figure 3.25.6 Measure unknown resistors using the balanced bridge The method of using the balanced bridge to measure an unknown resistor is as follows: If the unknown resistor is in the position of R4 in the circuit of Figure 3. i.4. 92 Understandable electric circuits So the unknown resistor value can be determined from the ratios of the resistances in a balanced Wheatstone bridge. .25. and the current flow through each component is always the same. Example 3. Solution: From R2R3 ¼ R4R1 solving for R4 R X ¼ R4 R2 R3 ð330 OÞð470 OÞ ¼ ¼ 1:551 kO 100 O R1 Summary Series circuits ● ● Series circuits: All components are connected one after the other. 2. Common ground or chassis ground: the common point for all components in the circuit (V ¼ 0). . . Determine the unknown resistance RX. nÞ When there are only two resistors in series: V1 ¼ V T ● ● R1 . . R2 ¼ 330 O and R3 ¼ 470 O in a balanced bridge circuit as shown in Figure 3. there is only one circuit path. Total series voltage: VT ¼ E ¼ V1 þ V2 þ Á Á Á þ Vn ¼ IRT VT ¼ IR1 þ IR2 þ Á Á Á þ IRn ¼ IRT ● ● Total series resistance (equivalent resistance Req): RT ¼ R1 þ R2 þ Á Á Á þ Rn Series current: I¼ VT E V1 V2 Vn ¼ ¼ ¼ ¼ ÁÁÁ ¼ RT RT R1 R2 Rn 2 ● ● E Total series power: PT ¼ P1 þ P2 þ Á Á Á þ Pn ¼ IE ¼ I 2 RT ¼ RT The VDR General form: VX ¼ VT RX RT or VX ¼ E RX RT ðX ¼ 1. .14: R1 ¼ 100 O. R1 þ R2 V2 ¼ V T R2 R1 þ R2 The earth ground: Connects to the earth (V ¼ 0). . there are at least two current paths in the circuit. R1 þ R2 I2 ¼ I T R1 R1 þ R2 . and the voltage across each component is the same. Parallel voltage: V ¼ E ¼ V1 ¼ V2 ¼ Á Á Á ¼ Vn Parallel currents: I1 ¼ V . R1 I2 ¼ V V . . Double-subscript notation: the voltage across the two subscripts.Series–parallel resistive circuits ● 93 ● Single-subscript notation: the voltage from the subscript with respect to ground. . . Parallel circuits ● ● ● Parallel circuits: The components are connected end to end.) When there are two resistors in parallel: I1 ¼ IT R2 . In ¼ R2 Rn IT ¼ ● V ¼ I1 þ I2 þ Á Á Á þ In Req Equivalent parallel resistance: Req ¼ 1 ¼ R1 ==R2 == Á Á Á ==Rn ð1=R1 Þ þ ð1=R2 Þ þ Á Á Á þ ð1=Rn Þ When n ¼ 2: Req ¼ ● ● R1 R2 ¼ R1 ==R2 R1 þ R2 Equivalent parallel conductance: Geq ¼ G1 þ G2 þ Á Á Á þ Gn Total parallel power: 2 PT ¼ P1 þ P2 þ Á Á Á þ Pn ¼ IT Req ¼ V2 ¼ IT V Req ● The CDR General form: IX ¼ IT Req GX or IX ¼ IT RX Geq (IX and RX are unknown current and resistance. 94 Understandable electric circuits Series–parallel circuits ● ● ● Series–parallel circuits are a combination of series and parallel circuits. Method for determining the equivalent resistance of series–parallel circuits: ● Determine the equivalent resistance of the parallel part of the series– parallel circuits. ● Determine the equivalent resistance of the series part of the series– parallel circuits. ● Plot the equivalent circuit if necessary. ● Repeat the above steps until the resistance in the circuit can be simplified to a single equivalent resistance Req. Method for analysing series–parallel circuits: ● Apply Ohm’s law to determine the total current: IT ¼ ● E Req Apply VDR, CDR, Ohm’s law, KCL and KVL to determine the unknown currents and voltages. Wye and delta configurations and their conversions R1 ● Y or T circuit: D ! Y: R1 ¼ R2 R3 , D or p circuit: Rb Rc Ra ● Rb Rc ; R a þ R b þ Rc R2 ¼ Ra Rc ; R a þ Rb þ R c R3 ¼ Ra Rb R a þ R b þ Rc ● Y ! D: R1 R2 þ R2 R3 þ R3 R1 ; R1 R1 R2 þ R2 R3 þ R3 R1 Rc ¼ R3 Ra ¼ Rb ¼ R1 R2 þ R2 R3 þ R3 R1 ; R2 ● If Ra ¼ Rb ¼ Rc ¼ RD and R1 ¼ R2 ¼ R3 ¼ RY: 1 RY RD 3 or RD ¼ 3RY ● The balanced bridge: When VAB ¼ 0, R2R3 ¼ R4R1 Series–parallel resistive circuits 95 Experiment 3: Series–parallel resistive circuits Objectives ● ● ● ● ● ● ● ● Review series and parallel resistive circuits. Construct and analyse series–parallel resistive circuits. Measure voltages and currents for series–parallel resistive circuits. Review the applications of KCL and KVL. Verify the theoretical analysis, and compare the experimental results with theory calculations. Apply the CDR to circuit analysis. Design and test a voltage divider. Measure unknown resistors using a Wheatstone bridge circuit. Background information ● ● Equivalent (or total) series resistance: Req ¼ RT ¼ R1 þ R2 þ Á Á Á þ Rn Equivalent parallel resistance: Req ¼ 1 ¼ R1 ==R2 == Á Á Á ==Rn ð1=R1 Þ þ ð1=R2 Þ þ Á Á Á þ ð1=Rn Þ ● When there are only two resistors in parallel: Req ¼ R1 R2 ¼ R1 ==R2 R1 þ R2 ● ● For a balanced Wheatstone bridge: when VAB ¼ 0, R2 R3 ¼ R4R1 CDR: IX ¼ IT Req RX When there are only two resistors in parallel: I1 ¼ IT ● R2 ; R 1 þ R2 I2 ¼ IT R1 R1 þ R2 VDR: VX ¼ VT RX RT or VX ¼ E RX RT When there are only two resistors in parallel: V1 ¼ VT R1 ; R1 þ R2 V2 ¼ V T R2 R 1 þ R2 96 Understandable electric circuits Equipment and components ● ● ● ● ● Digital multimeter Breadboard DC power supply Switch Resistors: ● four resistors with any values, ● one 10 kO variable resistor, ● 360 O, 510 O, 5.1 kO, 750 O, 1.2 kO, 2.4 kO, 5.1 kO, 910 O, 2.4 kO, 6.2 and 9.1 kO each and ● two 1.1 kO. Procedure Part I: Equivalent series and parallel resistance 1. Take four unknown resistors and record their colour code resistor values in Table L3.1. 2. Get the multimeter to function as an ohmmeter and measure the values of these four resistors. Record the values in Table L3.1. Table L3.1 Resistance Colour code resistor value Measured value R1 R2 R3 R4 3. Connect four resistors in series with the DC power supply as shown in Figure L3.1. Adjust the source voltage to the suitable value according to the value of the resistors, and then connect the DC power supply to the circuit in Figure L3.1 (consult your instructor before you turn on the switch). R1 A B R2 C R3 D R4 E VS Figure L3.1 A series circuit 4. Calculate the equivalent series resistance Req, voltage across each resistor VAB,, VBC, VCD, VDE, current I, and record in Table L3.2. Use direct method or indirect method to measure current I. Series–parallel resistive circuits Recall: 97 Direct method: Connect the multimeter (ammeter function) in series with the circuit components, turn on the switch and measure circuit current directly. ● Indirect method: Applying Ohm’s law to calculate current by using the measured voltage and resistance. 5. Turn on the switch for the circuit in Figure L3.1, get the multimeter function as an ohmmeter, voltmeter and ammeter, respectively, and measure Req, VAB,, VBC, VCD, VDE, current I. Record the values in Table L3.2. ● Table L3.2 Req Formula for calculations Calculated value Measured value VAB VBC VCD VDE I 6. Connect four resistors in parallel with the DC power supply as shown in Figure L3.2. VS R1 R2 R3 R4 Figure L3.2 A parallel circuit 7. Calculate the equivalent parallel resistance Req, currents IR1 ; IR2 ; IR3 ; IR4 and IT. Record the values in Table L3.3. 8. Turn on the switch for the circuit in Figure L3.2, measure Req, IR1 ; IR2 ; IR3 ; IR4 and IT using the multimeter (ohmmeter and ammeter functions). Record the values in Table L3.3. Table L3.3 Req Formula for calculations Calculated value Measured value IT IR1 IR2 IR3 IR4 98 Understandable electric circuits Part II: Series–parallel resistive circuit 1. Connect a series–parallel circuit as shown in Figure L3.3 on the breadboard. A 360 Ω 10 V 510 Ω 5.1 kΩ 2.4 kΩ D 750 Ω C B 1.2 kΩ Figure L3.3 A series–parallel circuit 2. Calculate the equivalent resistance Req, currents IT, I5.1 kO (current flowing through the branch of 5.1 kO resistor), and voltages VAB, VBC and VCD for the circuit in Figure L3.3. Record the values in Table L3.4. 3. Turn on the switch for the circuit in Figure L3.3, measure Req, IT, I5.1 kO, VAB, VBC and VCD. Record the values in Table L3.4. Table L3.4 Req Formula for calculations Calculated value Measured value IT VAB VBC VCD I5.1 kO Part III: Voltage divider 1. Design and construct a voltage divider as shown in Figure L3.4. When E ¼ 12 V, VA ¼ 6 V (voltage across the resistor R2), and I ¼ 6 mA, calculate resistance R1 and R2. Record the values in Table L3.5. 2. Measure resistance R1 and R2 using the multimeter (ohmmeter function) for the circuit in Figure L3.4. Record the values in Table L3.5. I = 6 mA R1 E 12 V A R2 Figure L3.4 Voltage divider circuit Series–parallel resistive circuits Table L3.5 R1 Formula for calculations Calculated value Measured value R2 VR1 99 VR2 3. Calculate voltages VR1 and VR2 for the circuit in Figure L3.4. Record the values in Table L3.5. 4. Measure voltages VR1 and VR2 using the multimeter (voltmeter function) for the circuit in Figure L3.4. Record the values in Table L3.5. Part IV: Wheatstone bridge 1. Measure the value of each resistor of RX using the multimeter (ohmmeter function) in Table L3.6. Record the values in Table L3.6. Table L3.6 Colour code value for RX Multimeter measured RX value Formula to calculate R3 Multimeter measured R3 value Bridge measured RX value 910 O 2.4 kO 6.2 kO 9.1 kO 2. Construct a bridge circuit as shown in Figure L3.5 on the breadboard, and connect the 910 O RX resistor (R4 ¼ RX) to the circuit. R1 = 1.1 kΩ E = 10 V A V R2 = 1.1 kΩ R3 B R4 = RX Figure L3.5 Bridge circuit 100 Understandable electric circuits 3. Calculate the value of the variable resistor R3 for the balanced bridge circuit (when RX ¼ 910 O, and VAB ¼ 0) in Figure L3.5. Record the value in Table L3.6. 4. Turn on the switch, and carefully adjust the variable resistor R3 when using the multimeter (voltmeter function) to measure the voltage across terminals A and B until the multimeter voltage is approximately zero. 5. Turn off the switch, use the multimeter (ohmmeter function) to measure the value of the variable resistor R3. Record the value in Table L3.6. 6. Calculate the value of RX when VAB ¼ 0 using the formula of the balanced bridge (use measured R3 value). Record the values in Table L3.6. 7. Turn off the switch for the circuit in Figure L3.5, then replace the other three RX resistors listed in Table L3.6 one by one to the circuit, and repeat steps 3 to 6. 8. Compare the measured and calculated RX values. Are there any significant differences? If so, explain the reasons. Conclusion Write your conclusions below: So the source equivalent conversion actually means that the source terminals are equivalent. current source and their equivalent conversions 4. The source equivalent conversion means that if loads are connected to both the terminals of the two sources after conversion.1). though the internal characteristics of each source circuit are not equivalent. you will be able to: ● ● ● ● ● ● convert voltage source to an equivalent current source and vice versa know the methods of voltage sources in series and parallel know the methods of current sources in series and parallel understand the branch current analysis method and apply it to circuit analysis understand the mesh analysis method and apply it to circuit analysis understand the node voltage analysis method and apply it to circuit analysis 4. The reference polarities of voltage and current of the sources should be the same before and after the conversion as shown in Figures 4. If the internal resistance RS in Figure 4.1(a) and the source current is IS ¼ E/RS in Figure 4. the load voltage VL and current IL of the two sources should be the same (Figure 4. When performing the source equivalent conversion. . we need to pay attention to the polarities of the sources.1. then the current source and voltage source can be equivalently converted.1(a and b) is equal.1 Source equivalent conversion It is sometimes easier to convert a current source to an equivalent voltage source or vice versa to analyse and calculate the circuits.1 Voltage source.Chapter 4 Methods of DC circuit analysis Objectives After completing this chapter.2 (notice the polarities of sources E and IS in the two figures).1(b).1 and 4. the source voltage is E ¼ ISRS in Figure 4. 1(a and b) are equal after connecting a load resistor RL to the two terminals of these circuits. IS ¼ E=RS E ¼ IS =RS The following procedure can verify that the load voltage VL and load current IL in two circuits of Figure 4. .2 Polarity of conversion Source equivalent conversion ● ● Voltage source ! Current source RS ¼ RS .1 Sources equivalent conversion a a RS ES IS RS b (a) (b) b Figure 4. Current source ! Voltage source RS ¼ RS .102 Understandable electric circuits a IL RS RL E a + VL IS RS IL RL + VL – b (a) (b) b – Figure 4. 1(b): RS E ¼ R S þ R L RS þ R L RS VL ¼ IL RL ¼ IS RL RS þ RL IL ¼ IS (Applying the current-divider rule and E ¼ ISRS) So the load voltages and currents in the two circuits of Figure 4.3(b): IS ¼ E 6V ¼ ¼ 3A RS 2 O RS 2O IL ¼ IS ¼ 0:5 A ¼ 3A ð2 þ 10ÞO RS þ RL . For Figure 4.3(b). RS is still 2 O in Figure 4.1(a and b) are the same.3(a) to an equivalent current source and calculate the load current IL for the circuit in Figure 4. and the source conversion equations have been proved. a IL RS = 2 Ω E=6V a IL RL = 10 Ω IS RS RL = 10 Ω b (a) (b) b Figure 4.1: Convert the voltage source in Figure 4.3(a and b).1(a): E R S þ RL RL RL VL ¼ E ¼ IS RS RS þ RL RS þ RL IL ¼ (Applying the voltage-divider rule and E ¼ ISRS) ● The current source in Figure 4.Methods of DC circuit analysis ● 103 The voltage source in Figure 4.1 Solution: The equivalent current source after the source conversion is shown in Figure 4.3(b).3 Circuit for Example 4. Example 4. 2.1. and determine the voltage source ES and internal resistance RS in Figure 4.3(a): IL ¼ E 6V ¼ ¼ 0:5 A RS þ RL ð2 þ 10ÞO Example 4.5.2 Solution: RS ¼ 10 O ES ¼ IS RS ¼ ð5 AÞð10 OÞ ¼ 50 V 4.4 Circuit for Example 4.4(a) to an equivalent voltage source.2 Sources in series and parallel 4.2: Convert the current source in Figure 4.104 Understandable electric circuits For Figure 4.4(b).1 Voltage sources in series A circuit of voltage sources in series and its equivalent circuit are shown in Figure 4. that is the equivalent internal resistance RS for series voltage sources is the sum of the individual internal resistances: RS ¼ RS1 þ RS2 þ Á Á Á þ RSn and the equivalent voltage E or VS for series voltage sources is the algebraic sum of the individual voltage sources: E ¼ E1 þ E2 þ Á Á Á þ En or VS ¼ V1 þ V2 þ Á Á Á þ Vn . R1 = 5 Ω IS = 5 A RS = 10 Ω R1 = 5 Ω RS = 10 Ω E2 = 50 V E1 = 20 V R2 = 10 Ω E1 = 20 V R2 = 10 Ω (a) (b) Figure 4.1. Voltage sources connected in series are similar with the resistors connected in series. Voltage sources in series RS ¼ RS1 þ RS2 þ Á Á Á þ RSn E ¼ E1 þ E2 þ Á Á Á þ En or VS ¼ V1 þ V2 þ Á Á Á þ Vn ● ● Assign a +ve sign if En has same polarity as E (or VS) Assign a 7ve sign if En has different polarity from E (or VS) 4.2 Voltage sources in parallel A circuit of voltage sources in parallel and its equivalent circuit are shown in Figure 4.5.2.. + En RSn En (a) (b) RS = RS1 + RS2 + . If voltage sources .Methods of DC circuit analysis 105 Assign a positive sign (+) if the individual voltage has the same polarity as the equivalent voltage E (or VS). where batteries are connected in series to increase the total equivalent voltage.. RS1 RS2 E2 E = E1 – E2 + .6.... Figure 4. assign a negative sign (7) if the individual voltage has a different polarity from the equivalent voltage E (or VS) as shown in Figure 4.. The equivalent voltage for the parallel voltage sources is the same as the voltage for each individual voltage source: E ¼ E 1 ¼ E 2 ¼ Á Á Á ¼ En or VS ¼ VS1 ¼ VS2 ¼ Á Á Á ¼ VSn and the equivalent internal resistance RS is the individual internal resistances in parallel: RS ¼ RS1 == RS2 == Á Á Á == RSn Note: Only voltage sources that have the same values and polarities can be connected in parallel by using the method mentioned above.1.5 Voltage sources in series A flashlight is an example of voltage sources in series. + RSn E1 . . which will be discussed in chapter 5 (section 5. = En Figure 4. Current sources connected in parallel can be replaced by a single equivalent resistance RS in parallel with a single equivalent current IS. // RSn E = E1 = E2 = ..+ ISn IS1 RS1 IS2 RS2 . ISn RSn RS = RS1 // RS2 // . En RS = RS1// RS2// . Voltage sources in parallel RS þ RS1 == RS2 == Á Á Á == RSn E ¼ E1 ¼ E2 ¼ Á Á Á ¼ En or VS ¼ VS1 ¼ VS2 ¼ Á Á Á ¼ VSn Only voltage sources that have the same values and polarities can be in parallel. You may have experienced using jumper cables by connecting the dead battery in parallel with a good car battery or with a booster (battery charger) to recharge the dead battery... 4. RSn (a) (b) Figure 4.2. It can provide twice the amount of current to the battery of the ‘dead’ vehicle and successfully start the engine. An example of an application for voltage sources connected in parallel is for boosting (or jump starting) a ‘dead’ vehicle. it can be solved by using Millman’s theory..4).7....3 Current sources in parallel A circuit of current sources in parallel and its equivalent circuit are shown in Figure 4.. IS = IS1 – IS2 +.106 Understandable electric circuits RS1 E1 E2 RS2 RSn .7 Current sources in parallel .6 Voltage sources in parallel having different values and polarities are connected in parallel.... It is the process of using the power from a charged battery to supplement the power of a discharged battery.1. 3A ? A 7A Figure 4.Methods of DC circuit analysis 107 The equivalent resistance RS is the individual internal resistances in parallel: RS ¼ RS1 == RS2 == Á Á Á == RSn The equivalent current IS is the algebraic sum of the individual current sources: IS ¼ IS1 þ IS2 þ Á Á Á þ ISn Assign a positive sign (+) if the individual current is in the same direction as the equivalent current IS.2. otherwise if the current entering point A does not equal the current exiting point A in Figure 4. so there must be only one current flowing through it.1.4 Current sources in series Only current sources that have the same polarities and same values can be connected in series. == RSn IS ¼ IS1 þ IS2 þ Á Á Á þ ISn ● ● Assign a +ve sign for ISn if it has the same polarity as IS Assign a 7ve sign for ISn if it has different polarity from IS 4. assign a negative sign (7) if the individual current is in a different direction from the equivalent current IS. KCL would be violated at point A.8. . . There is only one current path in a series circuit.8 KCL is violated at point A Current sources in series Only current sources that have the same polarities and values can be connected in series. . This is the same concept as Kirchhoff’s current law (KCL). Current sources in parallel RS ¼ RS1 == RS2 == . If an electric circuit or network has more than one source.9. The branch current analysis is a circuit analysis method that writes and solves a system of equations in which the unknowns are the branch currents.108 Understandable electric circuits Example 4.9 Circuit for Example 4. This method applies Kirchhoff’s laws and Ohm’s law to the circuit and solves the branch currents from simultaneous equations.3 Solution: The process of source equivalent conversion is shown in the circuit of Figure 4. Once the branch currents .9.2 Branch current analysis The methods of analysis stated in chapter 3 are limited to an electric circuit that has a single power source. 2Ω 4V a 2Ω a + 2A 4Ω 4A 4Ω 8Ω 8Ω 2V RL = 2 Ω VL = ? 2 + 4 = 6 A 2V b – + 4V 4//4 = 2 Ω RL = 2 Ω VL 8//8 = 4 Ω 2V b – 2Ω 2Ω 4V RL = 2 Ω a + VL (2 Ω) (6 A) = 12 V 4Ω 2V b – Figure 4. The branch current analysis is one of several basic methods for analysing electric circuits. it can be solved by the circuit analysis techniques that are discussed in chapters 4 and 5.3: Determine the load voltage VL in Figure 4. Determine VL by using the voltage-divider rule as follows: VL ¼ Vab ¼ IRL ¼ E RL RT ðÀ4 þ 12 À 2ÞV ¼ ð2 OÞ ¼ 1:2 V ð2 þ 2 þ 4 þ 2ÞO 4. Apply KVL to each mesh (or windowpane). ● Assign an arbitrary reference direction for each branch current. or Equation # ¼ branch # – (nodes # 71). The circuits in Figure 4. Apply KCL to numbers of independent nodes (n 7 1).10 Nodes and meshes Branch current analysis A circuit analysis method that writes and solves a system of Kirchhoff’s current law (KCL) and voltage law (KVL) equations in which the unknowns are the branch currents (it can be used for a circuit that has more than one source). branch and independent loop (or mesh).10 have three meshes (or independent loops) and different number of nodes (the dark dots).1 Procedure for applying the branch circuit analysis 1. 4. other circuit quantities such as voltages and powers can also be determined.Methods of DC circuit analysis 109 have been solved. The branch current analysis technique will use the terms node. Label the circuit. 1 3 2 1 2 3 1 2 3 Figure 4. let us review the definitions of these terms. ● Assign loop direction for each mesh (choose clockwise direction). Mesh: A loop in the circuit that does not contain any other loop (it can be analysed as a windowpane). 3.2. . ● ● ● ● Node: The intersectional point of two or more current paths where current has several possible paths to flow. where n is the number of nodes. ● Label all the nodes. Loop: A complete current path that allows current to flow back to the start. 2. Branch: A current path between two nodes where one or more circuit components is in series. and the number of KVL equations should be equal to the number of meshes. ● Assign clockwise loop direction for each mesh as shown in Figure 4. Example 4.11.11.11.4: Use the branch current analysis method to determine each branch current. Label the circuit as shown in Figure 4.11. Solve the simultaneous equations resulting from steps 2 and 3. 1. we should write two KVL equations. Or Equation # ¼ branch # À ðnodes # À 1Þ ¼ 3 À ð2 À 1Þ ¼ 2 Mesh 1: I1 R1 þ I3 R3 À E1 ¼ 0 Mesh 2: À I2 R2 À I3 R3 þ E2 ¼ 0 ð4:2Þ ð4:3Þ . using determinant or substitution methods to determine each branch current.4 Solution: This circuit contains two voltage sources. ● Assign an arbitrary reference direction for each branch current as shown in Figure 4. ● Label the nodes a and b.11 Circuit for Example 4. and n ¼ 2): I1 þ I2 ¼ I3 ð4:1Þ 3.11. and cannot be solved by using the methods we have learned in chapter 3. Apply KCL to ðn À 1Þ ¼ ð2 À 1Þ ¼ 1 number of independent nodes (there are two nodes a and b.110 Understandable electric circuits 4. power on resistor R2 and also the voltage across the resistor R1 in the circuit of Figure 4. Apply KVL to each mesh (windowpane). The procedure of applying the branch current analysis method is demonstrated in the following example. As there are two meshes in Figure 4. 2. The number of KVL equations should be equal to the number of meshes. 5. let us try to use the branch current analysis method. R1 = 2 Ω I1 E1 = 10 V 1 I3 R3 = 3 Ω 2 E2 = 5 V a I2 R2 = 2 Ω b Figure 4. Calculate the other circuit unknowns from the branch currents in the problem if necessary. Solve the simultaneous equations resulting from steps 2 and 3. otherwise assign a negative sign (7).Methods of DC circuit analysis 111 Recall KVL #1 (SV ¼ 0): Assign a positive sign (+) for E or V ¼ IR if its reference polarity and loop direction are the same. I2 and I3 (three equations can solve three unknowns). 4. ● Rewrite the above three equations in standard form: I1 þ I2 À I3 ¼ 0 I1 R1 þ 0 þ I3 R3 ¼ E1 0 À I2 R2 À I3 R3 ¼ ÀE2 ● Substitute the values into equations: I1 þ I2 À I3 ¼ 0 2I1 þ 0 þ 3I3 ¼ 10 V 0 À 2I2 À 3I3 ¼ À5 V ● Solve simultaneous equations using determinant method: . and determine branch currents I1. . 1 . . D ¼ . 2 . . 0 1 0 À2 . À1 3À3 ¼ ð1Þð0ÞðÀ3Þ þ ð2ÞðÀ2ÞðÀ1Þ þ ð0Þð1Þð3Þ À ðÀ1Þð0Þð0Þ À ð3ÞðÀ2Þð1Þ À ðÀ3Þð2Þð1Þ ¼ 4 À ðÀ6Þ À ðÀ6Þ ¼ 16 . 0 . . . 10 . . À5 1 0 À2 D . À1 . . . 3. . À3 . I1 ¼ ¼ ð10ÞðÀ2ÞðÀ1Þ þ ðÀ5Þð3Þð1Þ À ðÀ3Þð10Þð1Þ 16 % 2:19 A . . 1 . . . 2 . . 0 0 10 À5 I2 ¼ D % 0:31 A . À1 . . . 3. . À3 . ¼ ð1Þð10ÞðÀ3Þ þ ð2ÞðÀ5ÞðÀ1Þ À ðÀ3ÞðÀ5Þð1Þ 16 . 112 Understandable electric circuits . . . 1 1 0. . . . . 0 10 . . 2 . . . 0 À2 À5 . Label the nodes.5 Solution: 1. the number of loop equations can be reduced from 3 to 2: Mesh 1: À I1 R1 þ E1 þ E2 À I2 R2 ¼ 0 Mesh 2: I2 R2 À E2 À I3 R3 ¼ 0 . therefore. Apply KVL around each mesh (or windowpanes). reference direction for branch currents and loop directions in the circuit as shown in Figure 4.5: Determine current I3 in Figure 4. a I1 E1 = 2. there are three meshes in Figure 4. Apply KCL to ðn À 1Þ ¼ ð2 À 1Þ ¼ 1 number of independent nodes (there are two nodes or supernodes a and b. I3 % 1:88 A (Negative sign (7) for I2 indicates that the actual direction of I2 is opposite with its assigned reference direction.12.) 5. There is no need to write KVL for mesh 1 since mesh 1 current is already known to be equal to the source current I1 (I1 ¼ 5 A). I2 % À0:31 A.12 using the branch current analysis. so n ¼ 2): ÀI1 þ I2 þ I3 þ I4 ¼ 0 ðI4 ¼ 6 AÞ: 3.5 Ω I4 = 6 A Figure 4. Calculate the other circuit unknowns from the branch currents: 2 P2 ¼ I2 R2 ¼ ðÀ0:31Þ2 ð2Þ % 0:19 W V1 ¼ I1 R1 ¼ ð2:19Þð2Þ ¼ 4:38 V Example 4.12 Circuit for Example 4. 2.12. Àð10ÞðÀ2Þð1Þ À ðÀ5Þð2Þð1Þ I3 ¼ ¼ % 1:88A D 16 I1 % 2:19 A.5 V 1 R1 = 1 Ω E2 = 4 V I2 2 R2 = 1.5 Ω b I3 3 R3 = 0. so you should write three KVL equations. ÀI1 À 1:5I2 ¼ À2:5 À 4 ÀI1 À 1:5I2 þ 0I3 ¼ À6:5 0I1 þ 1:5I2 À 0:5I3 ¼ 4 0I1 þ 1:5I2 À 0:5I3 ¼ 4 ÀI1 þ I2 þ I3 ¼ À6 ÀI1 þ I2 þ I3 ¼ À6 Solve the above simultaneous equations using the determinant method: .Methods of DC circuit analysis 113 4. and determine the branch current I2. Solve the simultaneous equations resulting from steps 2 and 3. . . À1 À1:5 0. . . . . D¼. 0 1:5 À0:5 . . . . À1 1 1. ¼ ðÀ1Þð1:5Þð1Þ þ ðÀ1ÞðÀ0:5ÞðÀ1:5Þ À ðÀ0:5Þð1ÞðÀ1Þ ¼ À2:75 . . . À1 À6:5 0. . . . . 4 À0:5 . . 0 . . . À1 À6 1. mesh analysis is more practical and easier to use. Applying KVL to get the mesh equations and solve unknowns implies that it will have less unknown variables. the branch currents of the circuit will be easily determined. It can be used for a circuit that has more than one source. . The branch current analysis is a fundamental method for understanding mesh current analysis. Mesh current analysis A circuit analysis method that writes and solves a system of KVL equations in which the unknowns are the mesh currents (it can be used for a circuit that has more than one source). less simultaneous equations and therefore less calculation than branch current analysis. Mesh current analysis is a circuit analysis method that writes and solves a system of KVL equations in which the unknowns are the mesh currents (a current that circulates in the mesh). Mesh current analysis uses KVL and does not need to use KCL. After solving mesh currents.3 Mesh current analysis The branch current analysis in section 4.2 is a circuit analysis method that writes and solves a system of KCL and KVL equations in which the unknowns are the branch currents. I2 ¼ D ðÀ1Þð4Þð1Þ þ ðÀ1ÞðÀ0:5ÞðÀ6:5Þ À ðÀ0:5ÞðÀ6ÞðÀ1Þ % 1:55 A ¼ À2:75 I2 % 1:55 A 4. 6: Use the mesh current analysis method to determine each mesh current and branch currents IR1 .13 Circuit for Example 4. Note: ● Convert the current source to the voltage source first in the circuit. 2.114 Understandable electric circuits 4. Example 4. Identify each mesh.6 .13. ● Self-resistor: A resistor that is located in a mesh where only one mesh current flows through it. ● If the circuit has a current source. a IR1 E1 = 30 V I1 IR2 IR3 E2 = 20 V I2 R2 = 10 Ω b E3 = 10 V R3 = 20 Ω R1 = 10 Ω Figure 4. The procedure for applying the mesh current analysis method is demonstrated in the following examples. so the number of KVL equations can be reduced. 3. Apply KVL to each mesh of the circuit. and label all the nodes and reference directions for each mesh current (a current that circulates in the mesh) clockwise. ● Mutual resistor: A resistor that is located in a boundary of two meshes and has two mesh currents flowing through it. Solve the simultaneous equations resulting from step 2 using determinant or substitution methods. Label all the reference directions for each mesh current I1 and I2 (clockwise) as shown in Figure 4. 4. if there is any. the source current will be the same as the mesh current. Solution: 1.3.13. and the number of KVL equations should be equal to the number of meshes (windowpanes). IR2 and IR3 in the circuit of Figure 4. and a negative sign (7) for each mutual-resistor voltage in KVL equations. and determine each mesh current. Calculate the other circuit unknowns such as branch currents from the mesh currents in problem if necessary (choose the reference direction of branch currents first).1 Procedure for applying mesh current analysis 1. Or Equation # ¼ branch # À ðnodes # À 1Þ Assign a positive sign (+) for each self-resistor voltage. Mesh 1 : ðR1 þ R2 Þ I1 À R2 I2 ¼ ÀE1 þ E2 Mesh 2 : ÀR2 I1 þ ðR2 þ R3 Þ I2 ¼ ÀE2 À E3 Note: These equations were written by inspection of the circuit (inspection method): First column I1 Second column I2 Source E Mesh 1: Mesh 2: (Self-resistor)I1 (Mutual resistor)I1 7 + (Mutual resistor)I2 (Self-resistor)I2 = = 7E1 þ E2 7E2 7 E3 3. and a negative sign (7) for each mutual-resistor voltage in KVL (SV ¼ SE). Solve the simultaneous equations resulting from step 2.Methods of DC circuit analysis 115 2.4): 20I1 ¼ À10 þ 10I2 1 1 I1 ¼ À þ I2 2 2 ð4:6Þ ● Substitute I1 into (4. use the number of KVL: [branch # 7 (nodes # 71)]=3 7 (2 7 1) ¼ 2. Assign a positive sign (+) for each self-resistor voltage. Apply KVL around each mesh (windowpane).5) and solve for I2: 1 1 À10 À þ I2 þ 30I2 ¼ À30 2 2 I2 ¼ À1:4 A ● Substitute I2 into (4.13). and determine the mesh currents I1 and I2: ð10 þ 10ÞI1 À 10I2 ¼ À30 þ 20 20I1 À 10I2 ¼ À10 À10I1 þ ð10 þ 20ÞI2 ¼ À20 À 10 À10I1 þ 30I2 ¼ À30 Solve for I1 and I2 using the substitution method as follows: ● ð4:4Þ ð4:5Þ Solve for I1 from (4. and the number of KVL equations is equal to the number of meshes (there are two meshes in Figure 4. Alternatively.6) and solve for I1: 1 1 I1 ¼ þ ðÀ1:4Þ 2 2 I1 ¼ À0:2 A . and the number of KVL equations is equal to the number of meshes (there are three meshes in Figure 4. Label all the nodes and the reference directions for each mesh current (clockwise).4 Nodal voltage analysis The node voltage analysis is another method for analysis of an electric circuit with two or more sources.14.7: Write the mesh equations using the mesh current analysis method for the circuit in Figure 4. 1.14(b)) Mesh 1: ðR1 þ R2 þ R3 ÞI1 À R3 I2 À R2 I3 ¼ Es À E Mesh 2: ÀR3 I1 þ ðR3 þ R4 þ R5 ÞI2 À R4 I3 ¼ E Mesh 3: ÀR2 I1 À R4 I2 þ ðR2 þ R4 þ R6 ÞI3 ¼ 0 4. Apply KVL for each mesh (windowpane). IR3 ¼ ÀI2 ¼ 1:4 A Example 4. Assuming the reference direction of unknown branch current IR2 as shown in Figure 4. The node voltage analysis is a circuit analysis . 2. R6 R2 R4 R2 R1 Es I1 E R6 I3 R4 R3 I2 R5 Is R1 E R3 R5 (a) (b) Figure 4.13.14 Circuit for Example 4.7 (ES ¼ ISR1) Solution: Convert the current source to a voltage source as shown in Figure 4.116 Understandable electric circuits 4. calculate IR2 from the mesh currents by applying KCL at node a: X I ¼ 0 : ÀIR1 À IR2 þ IR3 ¼ 0 or I1 À IR2 À I2 ¼ 0 ðsince I1 ¼ ÀIR1 and I2 ¼ ÀIR3 Þ IR2 ¼ I 1 À I2 ¼ À0:2 À ðÀ1:4Þ ¼ 1:2 A IR1 ¼ ÀI1 ¼ 0:2 A.14.14(b). as shown in Figure 4. Apply KCL to all n 7 1 nodes except for the reference node (n is the number of nodes).8: Write the node voltage equations for the circuit shown in Figure 4.resistor or self-conductance voltage and a negative sign (7) for the mutual-resistor or mutual-conductor voltage. if necessary. Solution: 1. Solve the simultaneous equations and determine each nodal voltage. The procedure to apply node voltage analysis method is demonstrated in the following example. 4. assign the reference current directions for each branch as shown in Figure 4. Example 4. Assign a positive sign (+) for the self. and choose ground c to be the reference node. ● Label all the nodes and choose one of them to be the reference node. ● Assign an arbitrary reference direction for each branch current (this step can be skipped if using the inspection method). Calculate the other circuit unknowns such as branch currents from the nodal voltages in the problem. 2. ● Method 1: Write KCL equations and apply Ohm’s law to the equations.1 Procedure for applying the node voltage analysis 1. Node voltage is voltage between a node and the reference node. 2. either resistance or conductance can be used.15(a) using node voltage analysis method. Node voltage analysis A circuit analysis method that writes and solves a set of simultaneous KCL equations in which the unknowns are the node voltages (it can be used for a circuit that has more than one source). Apply KCL to n – 1 ¼ 3 – 1 ¼ 2 nodes (nodes a and b). b and c. 4. ● Method 2: Convert voltage sources to current sources and write KCL equations using the inspection method.4.Methods of DC circuit analysis 117 method that writes and solves a set of simultaneous KCL equations in which the unknowns are the node voltages. Usually ground or the node with the most branch connections should be chosen as the reference node (at which voltage is defined as zero). Label the circuit. Label nodes a. 3. .15(a). Recall that node is the intersectional point of two or more current paths. 118 Understandable electric circuits (a) I1 E1 R1 a I3 I2 R2 R3 b I4 R4 R5 I5 E2 c (b) a R3 I1 R1 R2 R4 R5 I5 b c Figure 4. Node b: I3 À I4 À I5 ¼ 0.15(b).8. E1 À Va Va Va À Vb À À ¼0 R1 R2 R3 Va À Vb Vb Vb þ E2 À À ¼0 R3 R4 R5 Or use conductance (G ¼ 1/R) ðE1 À Va ÞG1 À Va G2 À ðVa À Vb ÞG3 ¼ 0 ðVa À Vb ÞG3 À Vb G4 À ðVb þ E2 ÞG5 ¼ 0 ● Method 2: Convert two voltage sources to current sources from Figure 4. (b) Circuit for method 2 ● Method 1: Write KCL equations and apply Ohm’s law to the equations.15 (a) Circuit for Example 4. – Use conductance: First column (Va) Second column (Vb) Source IS Node a: (G1 þ G2 þ G3)Va 7 G3Vb = I1 Node b: 7G3Va + (G3 þ G4 þ G5)Vb = 7I5 . and write KCL equations by inspection. Node a: I1 À I2 À I3 ¼ 0.15(a) to Figure 4. 16(a). and a negative sign (7) for the mutual-conductor or mutual-resistor voltage and exiting node current.Methods of DC circuit analysis – Use resistance: 1 1 1 1 þ þ Va À Vb ¼ I1 R 1 R2 R3 R3 1 1 1 1 À Va þ þ þ Vb ¼ ÀI5 R3 R3 R4 R5 119 Note: The inspection method is similar with the one in mesh current analysis. and current I1 and I2 for the circuit shown in Figure 4.9 .9: Use the node voltage analysis to calculate resistances R1 and R2. Two equations can solve two unknowns. which are the node voltages Va and Vb. Example 4. (Assign a positive sign (+) for the self-resistance/conductance voltage and entering node current. (a) I1 R1 E I2 R2 Is (b) c R1 = 12 Ω I1 E = 60 V d c a I2 R2 = 24 Ω Is = 2 A b I1 = 60/12 = 5 A R1 = 12 Ω I1' d (c) a I2 a I2 Is= 2 A I1 = 5 A R1 // R2 Is = 2 A I1 = 5 A R1 = 12 Ω R2 = 24 Ω b b Figure 4. The difference is that mesh current analysis uses mesh currents in each column.16 Circuits for Example 4.) 3. and node voltage analysis uses node voltage in each column. ● ● E À Va Va À þ IS ¼ 0 R1 R2 Or use conductance: ðE À Va ÞG1 À Va G2 þ Is ¼ 0 Solve the above equation and determine the node voltage Va: E Va Va À À þ IS ¼ 0 R1 R1 R2 E 1 1 þ IS ¼ Va þ R1 R1 R2 ðE=R1 Þ þ IS ½ð60=12Þ þ 2A 7A Va ¼ ¼ ¼ 56 V ¼ ð1=R1 Þ þ ð1=R2 Þ ½ð1=12Þ þ ð1=24ÞS 0:125 S ● Calculate the branch currents from the nodal voltages: I1 ¼ E À Va ð60 À 56ÞV ¼ ¼ 0:33 A 12 O R1 Va 56 V % À2:33 A I2 ¼ À ¼ À 24 O R2 Use method 2: Convert voltage source to current source from the circuit of Figure 4. and choose b to be the reference node. and assign the reference current direction for each branch as shown in Figure 4.10: Write node voltage equations with resistances and conductances in the circuit of Figure 4.17 using the inspection method. .120 Understandable electric circuits Solution: ● Label nodes a and b.16(c): I1 ¼ E 60 ¼ ¼ 5 A R1 ==R2 ¼ 12==24 ¼ 8 O R1 12 ● Write KCL equation to node a using the inspection method: Va ¼ I1 þ IS R1 ==R2 Va ¼ ðI1 þ Is Þ ðR1 ==R2 Þ ¼ ð5 A þ 2 AÞ ð8 OÞ ¼ 56 V (Va is the same as that from method 1) Example 4. ● Apply KCL to n 7 1 ¼ 2 7 1 ¼ 1 node (node a): ● Use method 1: Write KCL equations and apply Ohm’s law to the equations: I1 À I2 þ IS ¼ 0.16(a) to the circuit of Figure 4.16(b). Methods of DC circuit analysis R5 R1 b R2 d R3 c R4 121 a Is Figure 4. Label all nodes a.10 Solution: 1. and thus it is more convenient to solve the circuit unknowns. less nodes and more branches. .5 Node voltage analysis vs. c and d (n ¼ 4) in the circuit as shown in Figure 4. with current source(s). mesh current analysis The choice between mesh current analysis and node voltage analysis is often made on the basis of the circuit structure: ● The node voltage analysis is preferable for solving a circuit that is a parallel circuit. Write KCL equations to n 7 1 ¼ 4 7 1 ¼ 3 nodes using the inspection method. b.17. (The step to assign each branch current with reference direction can be skipped since it is used for the inspection method.) 2.17 Circuit for Example 4. Vb and Vc) 4. and choose d to be the reference node. Three equations can solve three unknowns (node voltages Va. ● Use resistance: Node a: Node b: Node c: ● 1 1 1 1 þ Va À Vb À Vc ¼ IS R 1 R5 R1 R5 1 1 1 1 1 Vb À À Va þ þ þ VC ¼ 0 R1 R1 R2 R3 R3 1 1 1 1 1 þ þ VC ¼ 0 À Va À Vb þ R5 R3 R3 R 4 R5 Use conductance: Node a: ðG1 þ G5 ÞVa À G1 Vb À G5 Vc ¼ IS Node b: À G1 Va þ ðG1 þ G2 þ G3 ÞVb À G3 Vc ¼ 0 Node c: À G5 Va À G3 Vb þ ðG3 þ G4 þ G5 ÞVc ¼ 0 3. þ RS n Assign a positive sign (+) if it has the same polarity with E (or VS). E ¼ IS RS Voltage sources in series: RS ¼ RS1 þ RS2 þ . .1. Branch current analysis A circuit analysis method that writes and solves a system of KCL and KVL equations in which the unknowns are the branch currents. Nodal voltage analysis A circuit analysis method that writes and solves a set of simultaneous of KCL equations in which the unknowns are the node voltages.1. . Voltage source ! Current source: RS ¼ RS . mesh current analysis and node voltage analysis can be used for a circuit that has more than one source. with voltage sources and requires solving circuit branch currents. otherwise assign a negative sign (7).3. Summary Source equivalent conversions and sources in series and parallel ● ● ● E RS Current source ! Voltage source: RS ¼ RS . Note: The branch current analysis. IS ¼ Voltage sources in parallel: RS ¼ RS1 ==RS2 == Á Á Á ==RSn E ¼ E 1 ¼ E 2 ¼ Á Á Á En ● ● Only voltage sources that have the same values and polarities can be in parallel. The procedure for applying the nodal voltage analysis is given in Section 4. The procedure for applying the mesh current analysis is given in Section 4. . more nodes.4. Current sources in series: Only current sources that have the same polarities and values can be in series.122 ● Understandable electric circuits The mesh current analysis is preferable for solving a circuit that has fewer meshes. Mesh current analysis A circuit analysis method that writes and solves a system of KVL equations in which the unknowns are the mesh currents (it can be used for a circuit that has more than one source).1.2. The procedure for applying the branch current analysis is given in Section 4. Methods of DC circuit analysis 123 Experiment 4: Mesh current analysis and nodal voltage analysis Objectives ● ● ● ● ● Construct circuits with two voltage sources. Equipment and components ● ● ● ● ● Multimeter Breadboard Dual-output DC power supply Switches (2) Resistors: 1. circuit behaviour and performance. 9.1. a I1 E1 = 12 V R1 = 1.8 kΩ E2 = 5 V R3 = 3 kΩ b Figure L4. Experimentally verify the node voltage analysis method. It can be used for a circuit that has more than one source. and compare them to the theoretical equivalents.9 kO Background information ● ● Mesh current analysis: A circuit analysis method that writes and solves a system of KVL equations in which the unknowns are the mesh currents. 8.8 kΩ IR2 R2 = 1.2 kO.1 kO and 3.1 Circuit for mesh current analysis . Experimentally verify the mesh current analysis method. Calculate voltage VR2 and current IR2 using the mesh current analysis method (assuming the switches are turned on). Construct a circuit as shown in Figure L4. 3 kO. 2. It can be used for a circuit that has more than one source. Analyse the experimental data. Lab procedure Part I: Experimentally verify the mesh current analysis method 1.8 kO (2). Experimentally verify the methods of solving a circuit with two power supplies.1 on the breadboard. Record the values in Table L4. Nodal voltage analysis: A circuit analysis method that writes and solves a set of simultaneous KCL equations in which the unknowns are the node voltages. 124 Understandable electric circuits Table L4. Connect the multimeter (voltmeter function) in parallel to resistor R2 and measure VR2 .1. I2 and I3.2 on the breadboard. and turn on the two switches. 4. . Compare the measured values and calculated values. are there any significant differences? If so. Record the values in Table L4. respectively. Table L4. Record the value in Table L4. Set outputs of the dual-output power supply to 12 and 5 V. Record the values in Table L4. Calculate branch currents I1.1 Circuit for mesh current analysis Quantity Formula for calculations Calculated value Measured value IR2 VR2 3. Construct a circuit shown in Figure L4.2.1.2 kΩ I2 R2 = 3. Calculate the nodal voltage Va using the nodal voltage analysis method (assuming the two switches are turned on). Record the value in Table L4.2 Circuit for nodal voltage analysis 2. explain the reasons.2.1 kΩ b Figure L4. Part II: Experimentally verify the nodal voltage analysis method 1.2 Circuit for nodal voltage analysis Va Formula for calculations Calculated value Measured value I1 I2 I3 3. Use direct method or indirect method to measure current IR2 . 5.9 kΩ I3 E2 = 12 V R3 = 9. a I1 E1 = 10 V R1 = 8. and then turn on the two switches. Connect the multimeter (voltmeter function) in parallel to resistor R2 and measure Va.2. Compare the measured values and calculated values. 6. Record the values in Table L4. I2 and I3 using either the direct method or indirect method. respectively. are there any significant differences? If so.2. Set outputs of the dual-output power supply to 10 and 12 V.Methods of DC circuit analysis 125 4. explain the reasons. 5. Measure branch currents I1. Record the values in Table L4. Conclusion Write your conclusions below: . . Thevenin’s theorem. The branch current method.e. When the practical circuits are more and more complex. In electrical network analysis. These theorems include the superposition theorem. i. and the method of transferring maximum power to the load determine currents or voltages of the parallel voltage source circuits using Millman’s theorem determine currents or voltages of networks using the substitution theorem The main methods for analysing series and parallel circuits in chapter 3 are Kirchhoff’s laws. Norton’s theorem. mesh or loop analysis method and node voltage analysis method also use KCL and KVL as the main backbone. This is because you need to solve the higher-order mathematic equations when using these methods. Millman’s theorem and the substitution theorem. the fundamental rules are still Ohm’s law and Kirchhoff’s laws. the applications of the above methods solving for currents and voltages can be quite complicated. you have to use complex algebra to handle multiple circuit unknowns. The scientists working in the field of electrical engineering have developed more simplified theorems to analyse these kinds of complex circuits (the complicated circuit is also called the network). This chapter presents several theorems useful for analysing such complex circuits or networks. you will be able to: ● ● ● ● ● ● ● understand the concept of linear circuits determine currents or voltages of networks using the superposition theorem understand Thevenin’s and Norton’s theorems and know how to convert their equivalent circuits determine currents or voltages of networks using Thevenin’s and Norton’s theorems understand the maximum power transfer theorem. especially in multi-loop electric circuits.Chapter 5 The network theorems Objectives After completing this chapter. . e. Superposition theorem The unknown voltages or currents in a network are the sum of the voltages or currents of the individual contributions from each single power supply. The linear circuit can also be defined as follows: as long as the input/ output signal timing does not depend on any characteristic of the input signal. The actual unknown currents and voltages with all power sources can be determined by their algebraic sum.1 Introduction When several power sources are applied to a single circuit or network at the same time.1. The components of a linear circuit are the linear components. use series/parallel analysis to determine voltages and currents in the modified circuits. 2. Then.2 Steps to apply the superposition theorem 1. replace the voltage source with the short circuit (placing a jump wire). by setting the other inactive sources to zero. the superposition theorem can be used to separate the original network into several individual circuits for each power source working separately. An example of linear component is a linear resistor. Analyse and calculate this circuit by using the single source series–parallel analysis method. A linear circuit has an output that is directly proportional to its input. Turn off all power sources except one. . i. This method can avoid complicated mathematical calculations.1 Superposition theorem 5. 5. The voltage and current (input/output) of this linear resistor have a directly proportional (a straight line) relationship. 5. The pre-requirement of applying some of the above network theorems is that the analysed network must be a linear circuit.1. this is the meaning of the theorem’s name – ‘superimposed’. Linearity property The linearity property of a component describes a linear relationship between cause and effect. Redraw the original circuit with a single source. and replace the current source with an open circuit. it will be a linear circuit.128 Understandable electric circuits Network A network is a complicated circuit. 1 Circuit for Example 5.1(a) by using the superposition theorem.1: Determine the branch current Ic in the circuit of Figure 5. Determine Ic0 in the circuit of Figure 5.1(b). Repeat steps 1 and 2 for the other power sources in the circuit.1 2. (The result should be positive when the reference polarity of the unknown in the single source circuit is the same as the reference polarity of the unknown in the original circuit. Determine the total contribution by calculating the algebraic sum of all contributions due to single sources. Solution: 1.) Note: The superposition theorem can be applied to the linear network to determine only the unknown currents and voltages. otherwise it should be negative. Example 5.1(b): R0eq ¼ R5 ==ðR3 þ R4 Þ þ ðR1 þ R2 Þ 16  ð8 þ 8Þ ¼ þ ð8 þ 8Þ O ¼ 24 O 16 þ ð8 þ 8Þ . since power is a nonlinear variable. It cannot calculate power. Choose E1 to apply to the circuit first and use a jump wire to replace E2 as shown in Figure 5. Power can be calculated by the voltages and currents that have been determined by the superposition theorem. 4. Ia′ R1 = 8 Ω E1 = 48 V R2 = 8 Ω Ia R1 = 8 Ω E1 = 48 V R2 = 8 Ω Ib R3 = 8 Ω E2 = 24 V R4 = 8 Ω (a) Ic = R5 = 16 Ω Ib′ R3 = 8 Ω R5 = 16 Ω R4 = 8 Ω (b) Ic′ Ia′′ Ib′′ R5 = 16 Ω E2 = 24 V R4 = 8 Ω R3 = 8 Ω + R1 = 8 Ω Ic′′ R2 = 8 Ω (c) Figure 5.The network theorems 129 3. When E is applied only to the circuit (using an open circuit to replace the current source I1).2(a) by using the superposition theorem.2: Determine the branch current I2 and power P2 of the circuit in Figure 5.) Ib 00 ¼ E2 24 V ¼ ¼ 1A 00 24 O Req Ic ¼ I b R 1 þ R2 R1 þ R2 þ R5 ð8 þ 8ÞO ¼ 1A ¼ 0:5 A ð8 þ 8 þ 16ÞO 4. calculate I20 by assuming the reference direction of I20 as shown in Figure 5.1(c) and calculate Ic00 : Req 00 ¼ R5 ==ðR1 þ R2 Þ þ ðR3 þ R4 Þ 16  ð8 þ 8Þ ¼ þ ð8 þ 8Þ O ¼ 24 O 16 þ ð8 þ 8Þ (View from the E2 branch in the circuit of Figure 5.1(c) to determine Req00 .1(b) to determine Req0 . Solution: 1. Calculate the sum of currents Ic0 and Ic00 : 0 Ic ¼ Ic þ Ic ¼ ð1 þ 0:5ÞA ¼ 1:5 A Example 5. 2. When E2 is applied to the circuit.2(b).) Ia 0 ¼ E1 48 V ¼ ¼ 2A Req 0 24 O Ic 0 ¼ Ia 0 R3 þ R4 R3 þ R4 þ R5 ð8 þ 8ÞO ¼ 2A ¼ 1A ð8 þ 8 þ 16ÞO 3. replace E1 with a short circuit as shown in Figure 5. Determine I20 in the circuit of Figure 5.130 Understandable electric circuits (View from the E1 branch in the circuit of Figure 5.2(b): I2 0 ¼ ¼ E ðR2 ==R3 Þ þ R1 25 V ¼ 0:25 A ¼ 250 mA ½ðð100  100Þ=ð100 þ 100ÞÞ þ 50O . Calculate the sum of currents I20 and I200 : I2 ¼ ÀI2 0 þ I2 00 ¼ ðÀ250 þ 25ÞmA ¼ À225 mA ¼ À0:225 A I20 is negative as its reference direction in Figure 5.2(c). The negative I2 implies that the actual direction of I2 in Figure 5.) 4.2(a).2(c): I2 00 ¼ I1 R1 R1 þ R2 ==R3 50 O ¼ 25 mA 50 þ ½ð100  100Þ=ð100 þ 100ÞO ¼ 50 mA (Apply the current divider rule to the branches R1 and R2 // R3.2(b) is opposite to that of I2 in the original circuit of Figure 5. the circuit is as shown in Figure 5.The network theorems 131 I2 E = 25 V I1 = 50 mA R2 100 Ω R1 = 50 Ω (a) I2′′ = R3 = 100 Ω I2′ E = 25 V R2 R1 = 50 Ω 100 Ω R3 = 100 Ω (b) + I1 = 50 mA R1 = 50 Ω R2 100 Ω R3 = 100 Ω (c) Figure 5. .3(a) using the superposition theorem.2 3.2 Circuit for Example 5. Calculate I20 by assuming the reference direction of I200 as shown in the circuit of Figure 5.2(a) is opposite to its reference direction. When the current source I1 is applied only to the circuit (the voltage source E is replaced by a jump wire).3: Determine the branch current I3 in the circuit of Figure 5. Determine the power P2: P2 ¼ I2 2 R2 ¼ ðÀ0:225 AÞ2 ð100 OÞ % 5:06 W Example 5. 55 kΩ R4 = 0.3 2.5 V R3 = 0. I3 R1 = 0.) There ¼ ð19:84 mAÞ I1 0 ¼ E1 E1 ¼ 0 Req ðR3 þ R4 Þ==R2 þ R1 12:5 V ¼ ½ðð0:375 þ 0:45Þ Â 0:55Þ=ðð0:375 þ 0:45Þ þ 0:55Þ þ 0:3kO % 19:84 mA .132 Understandable electric circuits Solution: 1.45 kΩ E2 = 5 V I = 5 mA = (a) R1 I1′ E1 R3 R2 R4 I3′ R1 R3 I3′′ R2 R1 R2 I3′′′ R4 R3 + I2′′ E2 R4 + I (b) (c) (d) Figure 5.3(b). Use the circuit in Figure 5.3(b) to determine I30 : I 3 0 ¼ I1 0 R2 R2 þ ðR3 þ R4 Þ 0:55 kO % 7:94 mA 0:55 kO þ ð0:375 þ 0:45ÞkO (Apply the current divider rule to the branches R2 and (R3 þ R4).3 kΩ E1 = 12. Choose E1 to apply to the circuit first and use a jump wire to replace E2 and an open circuit to replace the current source I as shown in Figure 5.3 Circuit for Example 5.375 kΩ R2 = 0. I300 and I3000 : I3 ¼ I3 0 þ I3 00 À I3 000 ¼ ð7:94 þ 1:73 À 2:21ÞmA ¼ 7:46 mA I3000 is negative since its reference direction is opposite to that of I3 in the original circuit of Figure 5. . Norton in Bell Telephone laboratory published a similar theorem.3(c) to determine I300 : I3 00 ¼ I2 00 R1 R1 þ ðR3 þ R4 Þ 0:3 kO % 1:73 mA ½0:3 þ ð0:375 þ 0:45ÞkO 133 ¼ ð6:49 mAÞ (Apply the current divider rule to the branches R1 and (R3 þ R4). French telegraph engineer M. L.2. 5. but he used the current source to replace the voltage source in the equivalent circuit.) There I2 00 ¼ E2 E2 ¼ Req 00 ðR3 þ R4 Þ==R1 þ R2 5V ¼ ½ðð0:375 þ 0:45Þð0:3ÞÞ=ðð0:375 þ 0:45Þ þ 0:3Þ þ 0:55kO % 6:49 mA . L.3(a). Use the circuit in Figure 5.1 Introduction Thevenin’s and Norton’s theorems are two of the most widely used theorems to simplify the linear circuit for ease of network analysis. American engineer E.3(d) to determine I3000 : I3 000 ¼ I R4 ðR1 ==R2 þ R3 Þ þ R4 0:45 kO % 2:21 mA ½ðð0:3  0:55Þ=ð0:3 þ 0:55Þ þ 0:375Þ þ 0:45kO ¼ ð5 mAÞ (Apply the current divider rule to the branches R4 and (R1 // R2 þ R3).) 4.The network theorems 3. These two theorems state that any complicated linear two-terminal network with power supplies can be .2 Thevenin’s and Norton’s theorems 5. Calculate the sum of currents I30 . Thevenin published his theorem of network analysis method. Forty-three years later. Use the circuit in Figure 5. In 1883. 134 Understandable electric circuits simplified to an equivalent circuit that includes an actual voltage source (Thevenin’s theorem) or an actual current source (Norton’s theorem). Thevenin’s and Norton’s theorems allow for analysis of the performance of a circuit from its terminal properties only. a RTH a VTH RL b (a) IN RN b (c) RL RL b (b) a Figure 5.4 Thevenin’s and Norton’s theorems. The equivalent means that any load resistor branch (or unknown current or voltage branch) connected between the terminals of Thevenin’s or Norton’s equivalent circuit will have the same current and voltage as if it were connected to the terminals of the original circuit. Here.4(a).4(b or c). ‘the linear two-terminal network with power supplies’ means: ● ● ● ● network: the relatively complicated circuit linear network: the circuits in the network are the linear circuits two-terminal network: the network with two terminals that can be connected to the external circuits network with the power supplies: network includes the power supplies No matter how complex the inside construction of any two-terminal network with power supplies is. they can all be illustrated in Figure 5. (a) Linear two-terminal network with the power supply. . we can draw the following conclusion: any linear two-terminal network with power supplies can be replaced by an equivalent circuit as shown in Figure 5. (c) Norton’s theorem According to Thevenin’s and Norton’s theorems. (b) Thevenin’s theorem. 2. But if Thevenin’s and Norton’s theorems are used.2 Steps to apply Thevenin’s and Norton’s theorems 1. It equals the equivalent resistance. and replaced by a single current source and a single parallel resistor for Norton’s theorem. and a current source should be replaced by an open circuit. In short we can conclude that any combination of power supplies and resistors with two terminals can be replaced by a single voltage source and a single series resistor for Thevenin’s theorem. and mark the letters a and b on the two terminals. Thevenin’s and Norton’s equivalent circuits will not be changed except for their external load branches. which has a single power source and a single resistor.The network theorems 135 Any linear two-terminal network with power supplies can be replaced by a simple equivalent circuit. the wall plug can connect to 60 or 100 W lamps). Once the load is changed. Note: The ‘TH’ in VTH and RTH means Thevenin. Norton’s theorem: Norton’s equivalent circuit is a current source – with an equivalent resistance RN in parallel with an equivalent current source IN. Open and remove the load branch (or any unknown current or voltage branch) in the network. and the ‘N’ in IN and RN means Norton. The variation of the load can be determined more conveniently by using Thevenin’s or Norton’s equivalent circuits. The key to applying these two theorems is to determine the equivalent resistance RTH and the equivalent voltage VTH for Thevenin’s equivalent circuit. the equivalent resistance RN and the equivalent current IN for Norton’s equivalent circuit. 5.2. looking at it from the a and b terminals when all sources are turned off or equal to zero in the network. Determine the equivalent resistance RTH or RN. These two theorems are used very often to calculate the load (or a branch) current or voltage in practical applications. The value of RN in Norton’s equivalent circuit is the same as RTH of Thevenin’s equivalent circuit.) That is RTH ¼ RN ¼ Rab . (A voltage source should be replaced by a short circuit. ● ● Thevenin’s theorem: Thevenin’s equivalent circuit is a voltage source – with an equivalent resistance RTH in series with an equivalent voltage source VTH. the whole circuit has to be re-analysed or re-calculated. The load resistor can be varied sometimes (for instance. It equals the short-circuit current from the original linear two-terminal network of a and b.136 Understandable electric circuits 3.5.6(a) Circuit for Example 5.e. The above procedure for analysing circuits by using Thevenin’s and Norton’s theorems is illustrated in the circuits of Figure 5. Determine Norton’s equivalent current IN. E = 48 V R1 = 12 Ω IL RL = 8 Ω R2 = 12 Ω R = 12 Ω 3 Figure 5.e. It equals the open-circuit voltage from the original linear two-terminal network of a and b. Plot Thevenin’s or Norton’s equivalent circuit. a a VTH = Vab a b b a RL b E =0 RTH VTH RL Is = 0 RTH = RN = Rab b a IN = Isc b a IN RN RL b Figure 5.5 The procedure for applying Thevenin’s and Norton’s theorems Example 5.4: Determine the load current IL in the circuit of Figure 5. IN ¼ Isc ðwhere ‘sc’ means the short circuitÞ 5. Then the load (or unknown) voltage or current can be determined.6(a) by using Thevenin’s and Norton’s theorems. Determine Thevenin’s equivalent voltage VTH.4 . i. VTH ¼ Vab 4. and connect the load branch (or unknown current or voltage branch) to a and b terminals of the equivalent circuit. i. Determine Thevenin’s equivalent voltage VTH: Use the circuit in Figure 5.6(c) RTH ¼ RN ¼ Rab ¼ R1 ==R2 ==R3 ¼ ð12==12==12ÞO ¼ 4 O 3. Open and remove the load branch RL. Determine Norton’s equivalent current IN: Use the circuit in Figure 5.6(d) to calculate the open-circuit voltage across terminals a and b.6(e) to calculate the short-circuit current in terminals a and b. and mark a and b on the terminals of the load branch as shown in the circuit of Figure 5.) .The network theorems 137 Solution: 1. . a R2 = 12 Ω R1 = 12 Ω b Figure 5.6(b).6(d) VTH ¼ Vab ¼ E R2 ==R3 R1 þ R2 ==R3 ð12==12ÞO ¼ 48 V ¼ 16 V ð12 þ 12==12ÞO (Apply the voltage divider rule to the resistors R2//R3 and R1.6(b) Circuit for Example 5. a E = 48 V R1 = 12 Ω R2 = 12 Ω R3 = 12 Ω b Figure 5. a E = 48 V R1 = 12 Ω R2 = 12 Ω R3 = 12 Ω b Figure 5.4 2.6(c). . Determine Thevenin’s and Norton’s equivalent resistances RTH and RN (the voltage source is replaced by a short circuit) in the circuit of Figure 5. a IL IN = 4 A RN = 4 Ω RL = 8 Ω b Figure 5. Connect the load RL to a and b terminals of the equivalent circuits and determine the load current IL. Plot Thevenin’s and Norton’s equivalent circuits as shown in Figure 5. in this case IN ¼ E=R1 . ● Use Thevenin’s equivalent circuit in Figure 5.6 (f and g).6(f) IL ¼ VTH 16 V ¼ % 1:33 A RTH þ RL ð4 þ 8ÞO ● Use Norton’s equivalent circuit in Figure 5.6(g) .6(f) to determine IL. a RTH = 4 Ω VTH = 16 V IL RL = 8 Ω b Figure 5.6(e) IN ¼ Isc ¼ E 48 V ¼ ¼ 4A R1 12 O Since the current in the branch E and R1 will go through a short cut without resistance – through the branches a and b – and will not go through the branches R2 and R3 that have resistances. 4.138 Understandable electric circuits E = 48 V R1 = 12 Ω a R2 = 12 Ω R3 = 12 Ω b IN = I sc Figure 5.6(g) to calculate IL. The network theorems IL ¼ IN RN RN þ RL 4O % 1:33 A ¼ 4A ð4 þ 8ÞO 139 (current divider rule) 5.2. The opening two terminals of the branch are the viewpoints for Thevenin’s and Norton’s equivalent circuits. 2.7(a) and mark the letters a and b. If we want to determine the branch current I3. we use D and C as viewpoints. Determine Thevenin’s equivalent circuit for the viewpoints D–B. A R1 E D R2 C R4 R3 B E D I2 R2 C R1 R3 A I3 B I4 R4 Figure 5. 3.7(a)).3 Viewpoints of the theorems One important way to apply Thevenin’s and Norton’s theorems for analysing any network is to determine the viewpoints of Thevenin’s and Norton’s equivalent circuits.7(a) Viewpoints for the theorem Example 5.7(a): 1. Different equivalent circuits and results will be obtained from using different viewpoints.7(b) Circuit for Example 5.5 . we use A and B as viewpoints. as shown in the circuit of Figure 5. etc. 1. The load branch (or any unknown current or voltage branch) belongs to the external circuit of the linear two-terminal network with power sources. Plot Thevenin’s equivalent circuit for calculating the current I3. if we want to determine the branch current I2. Open and remove R3 in the branch A–B of Figure 5.7(b). There could be different viewpoints for the bridge circuit as shown in Figure 5.5: For the circuit in Figure 5. Solution: (a) The viewpoints for calculating I3 should be A–B (Figure 5. R1 E R2 a b B R4 A Figure 5.7(a). Determine Norton’s equivalent circuit for the viewpoints B–C. 7(d). a VTH = Vab b R2 R4 R1 E Figure 5.7(c) RTH ¼ Rab ¼ ðR2 ==R4 Þ==R1 3.7(e) .140 Understandable electric circuits 2.7(e). as shown in the circuit of Figure 5. A RTH = Rab B R4 R4 R2 R1 RTH = Rab b a R1 R2 a b Figure 5. Connect R3 to a and b terminals of the equivalent circuit and determine the current I3: I3 ¼ VTH RTH þ R3 a RTH VTH b B A I3 R3 Figure 5.7(c). Determine RTH and Rab. Determine VTH using the circuit in Figure 5.7(d) VTH ¼ Vab ¼ E R1 R1 þ R2 ==R4 (voltage divider rule) 4. Replace the voltage source E with a short circuit. Plot Thevenin’s equivalent circuit as shown in the circuit of Figure 5. The network theorems (b) Norton’s equivalent circuit for the viewpoints B–C: 141 1.7(f) 2. Replace the voltage source with a short circuit as shown in the circuit of Figure 5. Determine IN using the circuit in Figure 5.7(h) Since the current will go through the short cut without resistance – the branch a and b – and will not go through the branch with resistance R2. R1 R2 a b R3 B RTH = Rab C Figure 5. IN ¼ IT: IN ¼ I T ¼ E R1 ==R3 . R1 E R2 a b R3 B C Figure 5. Open and remove R4 in the branch B–C of Figure 5. IT R1 E R2 a b R3 IN = Isc Figure 5.7(f).7(g). Determine RN. and mark the letters a and b on the two terminals as shown in the circuit of Figure 5.7(h).7(a).7(g) 3. 7(k).7(j) 2. R1 E D R2 a b R3 B R4 Figure 5. Determine RTH.142 Understandable electric circuits 4. Replace the voltage source with a short circuit as shown in the circuits of Figure 5.7(a)) and mark the letters a and b on the two terminals as shown in the circuit of Figure 5. Open branch D–B (Figure 5.7(j).7(k) RTH ¼ Rab ¼ ðR1 ==R2 Þ þ ðR3 ==R4 Þ . a R1 D R2 b R3 a b B R1 R2 R3 R4 R4 Figure 5. Plot Norton’s equivalent circuit as shown in the circuit of Figure 5. a B Is RN R4 b C Figure 5.7(i).7(i) (c) Thevenin’s equivalent circuit for the viewpoint D–B: 1. 7(l).6 Solution: 1. Replace the current source with an open circuit as shown in the circuit of Figure 5. IL R1 = 200 Ω R2 = 200 Ω R3 = 400 Ω I = 1 mA R4 = 500 Ω RL=100 Ω Figure 5.8(b)) and mark the letters a and b on its two terminals. D RTH VTH B b a Figure 5. .The network theorems 3.8(b).8(a) Circuit for Example 5. Determine RN. as shown in the circuit of Figure 5. a R1 = 200 Ω R2 = 200 Ω R3 = 400 Ω I = 1mA R4 = 500 Ω b Figure 5. Determine VTH using the circuit in Figure 5.8(c).7(l) Example 5.8(b) 2. Plot Thevenin’s equivalent circuit as shown in the circuit of Figure 5.7(j): VTH ¼ Vab ¼ Va þ ðÀVb Þ ¼ E R2 R4 ÀE R1 þ R2 R3 þ R4 143 4.8(a) by using Norton’s theorem. Open and remove RL in the load branch (Figure 5.6: Determine current IL in the circuit of Figure 5. Plot Norton’s equivalent circuit as shown in the circuit of Figure 5. Connect RL to the a and b terminals of the equivalent circuit. 4. a R1 = 200 Ω R2 = 200 Ω R3 = 400 Ω I = 1 mA R4 = 500 Ω IN= Isc b Figure 5.8(c) RN ¼ Rab ¼ ðR1 ==R2 þ R3 Þ==R4 ¼ ½ð200==200 þ 400Þ==500O ¼ 250 O 3. IN ¼ I. a IL IN = 1 mA RN = 250 Ω RL = 100 Ω b Figure 5. Calculate IN using the circuit of Figure 5.8(d) IN ¼ I ¼ 1 mA Since the current I will flow through the short cut without resistance – the branch a and b – and will not go through the branch with resistance.8(d).8(e).144 Understandable electric circuits a R1 = 200 Ω R2 = 200 Ω R3 = 400 Ω R4 = 500 Ω RN = Rab b Figure 5. and calculate the current IL.8(e) I L ¼ IN RN R L þ RN 250 O ¼ 1 mA % 0:71 mA ð250 þ 100ÞO . The network theorems 145 When applying Thevenin’s and Norton’s theorems to analyse networks. as shown in the circuit of Figure 5.9(c). Replace the voltage source with a short circuit. it is often necessary to combine theorems that we have learned in the previous chapters. a R1 = 30 Ω E=6V I1 = 45 mA R2 = 70 Ω I2 = 140 mA IL RL = 42 Ω b Figure 5. R1 = 30 Ω R2 = 70 Ω b a E=6V I1 = 45 mA Figure 5.7 Solution: 1.9(a)).9(a) Circuit for Example 5. Determine RN. Open and remove the current source part on the right side of the circuit from the terminals a and b (Figure 5.9(c) . and the current source with an open circuit. Example 5.9(b) 2.7: Determine Norton’s equivalent circuit for the left part of the terminals a and b in the circuit of Figure 5.9(a) and determine the current IL.9(b). as shown in the circuit of Figure 5. This is explained in the following example. R1 = 30 Ω R2 = 70 Ω a RN = Rab b Figure 5. 146 Understandable electric circuits RN ¼ Rab ¼ R1 ==R2 ¼ ð30  70ÞO ¼ 21 O ð30 þ 70ÞO 3. Determine IN using the circuit in Figure 5.9(d). Since there are two power supplies in this circuit, it is necessary to apply the network analysing method for this complex circuit. Let us try to use the superposition theorem to determine IN. R1 = 30 Ω a E=6V I1 = 45 mA R2 = 70 Ω b IN Figure 5.9(d) ● When the single voltage source E is applied only to the circuit, the circuit is shown in Figure 5.9(e). R1 = 30 Ω a E=6V R2 = 70 Ω b IN′ Figure 5.9(e) Since R2 is short circuited by the IN0 (recall that current always goes through the short cut without resistance): ; IN 0 ¼ ● E 6V ¼ 0:2 A ¼ 200 mA ¼ R1 30 O When the single current source I1 is applied only to the circuit, the circuit is shown in Figure 5.9(f). R1 = 30 Ω a I1 = 45 mA R2 = 70 Ω b I N′′ Figure 5.9(f) The network theorems Since R1 and R2 are short circuited by IN00 : IN 00 ¼ I1 ¼ 45 mA ● 147 Determine IN: IN ¼ IN 0 ÀIN 00 ¼ ð200 À 45Þ mA ¼ 155 mA 4. Plot Norton’s equivalent circuit. Connect the right side of the a and b terminals of the current source (Figure 5.9(a)) to the a and b terminals of Norton’s equivalent circuit, as shown in the circuits of Figure 5.9(g). Determine the current IL. a IL IN = 155 mA RN = 21 Ω I2 = 140 mA RL = 42 Ω a IL IN + I2 RN = 21 Ω RL = 42 Ω b b Figure 5.9(g) IL ¼ ðIN þ I2 Þ RN RL þ RN ¼ ð155 þ 140ÞmA 21 O % 98:33 mA ð42 þ 21ÞO 5.3 Maximum power transfer Practical circuits are usually designed to provide power to the load. When working in electrical or electronic engineering fields you are sometimes asked to design a circuit that will transfer the maximum power from a given source to a load. The maximum power transfer theorem can be used to solve this kind of problem. The maximum power transfer theorem states that when the load resistance is equal to the source’s internal resistance, the maximum power will be transferred to the load. From the last section, we have learned that any linear two-terminal network with power supply can be equally substituted by Thevenin’s or Norton’s equivalent circuits. Therefore, the maximum power transfer theorem implies that when the load resistance (RL) of a circuit is equal to the internal resistance (RS) of the source or the equivalent resistance of Thevenin’s or Norton’s equivalent circuits (RTH or RN), maximum power will be dissipated in the load. This concept is illustrated in the circuits of Figure 5.10. 148 Understandable electric circuits a RTH a VTH RL b IN RN b (c) RL = RTH RL b (b) a RL = RN RL VS b (d) Rs RL = RS a IL RL (a) Figure 5.10 The maximum power transfer The maximum power transfer theorem When the load resistance is equal to the internal resistance of the source (RL ¼ RS); or when the load resistance is equal to the Thevenin’s/ Norton’s equivalent resistance of the network (RL ¼ RTH ¼ RN), the maximum power can be transferred to the load. The maximum power transfer theorem is used very often in radios, recorders, stereos, CDs, etc. If the load component is a speaker and the circuit that drives the speaker is a power amplifier, when the resistance of the speaker RL is equal to the internal resistance RS of the amplifier equivalent circuit, the amplifier can transfer the maximum power to the speaker, i.e. the maximum volume can be delivered by the speaker. Using the equivalent circuit in Figure 5.10(d) to calculate the power consumed by the load resistor RL gives PL ¼ IL 2 RL ¼ VS RS þ RL 2 RL ð5:1Þ When RL ¼ RS, the maximum power that can be transferred to the load is PL ¼ VS 2 ð2RS Þ R ¼ 2 S VS 2 4RS The network theorems 149 The maximum load power PL ¼ VS 2 4RS If VS ¼ 10 V; RS ¼ 30O; and RL ¼ 30O Then PL ¼ VS 2 102 V % 833 mW ¼ 4RS 4ð30 OÞ The maximum power transfer theorem can be proved by using an experiment circuit as shown in Figure 5.11. When the variable resistor RL is adjusted, it will change the value of the load resistor. Replacing the load resistance RL with different values in (5.1) gives different load power PL, as shown in Table 5.1. a Rs = 30 Ω Vs = 10 V b RL Figure 5.11 The experiment circuit Table 5.1 The load power RL (O) 10 20 30 40 50 PL (W) 0.625 0.8 0.833 0.816 0.781 Such as when RL ¼ 10 O: 2 VS PL ¼ I 2 RL ¼ RL RS þ RL 2 10 V ¼ 10 O ¼ 0:625 W ð30 þ 10ÞO 150 Understandable electric circuits When RL ¼ 20 O: PL ¼ I 2 R L ¼ 2 VS RL RS þ RL 2 10 V ¼ 20 O ¼ 0:8 W ð30 þ 20ÞO The RL and PL curves can be plotted from Table 5.1 as shown in Figure 5.12. PL(W) 1 0.8 0.6 0.4 0.2 RL(Ω) RL = RS A 0 10 20 30 40 50 Figure 5.12 RL–PL curve Table 5.1 and Figure 5.12 shows that only when RL ¼ RS (30 O), the power for the resistor RL reaches the maximum point A (0.833 W). Proof of the maximum power transfer equation: The maximum power transfer equation RL ¼ RS can be derived by the method of getting the maximum value in calculus (skip this part if you haven’t learned calculus yet). Take the derivative for RL in (5.1), and let its derivative equal to zero, giving the following: h i d ðVS =ðRS þ RL ÞÞ2 RL dP ¼ ¼0 dRL dRL VS 2 ðRS þ RL Þ2 À 2RL ðRS þ RL Þ ðRS þ RL Þ4 ¼0 VS 2 RS 2 þ 2RS RL þ RL 2 À 2RL RS À 2RL 2 ðRS þ RL Þ4 ¼0 The network theorems 151 VS 2 RS 2 À RL 2 ðRS þ RL Þ4 ¼0 VS 2 ðRS þ RL ÞðRS À RL Þ ðRS þ RL Þ4 RS À RL ðRS þ RL Þ3 ¼0 ¼0 VS 2 i:e: RS À RL ¼ 0 or RL ¼ RS hence proved: 5.4 Millman’s and substitution theorems 5.4.1 Millman’s theorem Millman’s theorem is named after the Russian electrical engineering professor Jacob Millman (1911–1991) who proved this theorem. A similar method, known as Tank’s method, had already been used before Millman’s proof. The method using series–parallel power sources was stated in chapter 4. However, the series–parallel method can only be used in power sources that have the same polarities and values. Millman’s theorem in this chapter can be used to analyse circuits of parallel voltage sources that have different polarities and values. This can be shown in the circuits of Figure 5.13. a R1 E1 Rm Vm b b a R2 E2 Rn … En Figure 5.13 Millman’s theorem Millman’s theorem states that for a circuit of parallel branches, with each branch consisting of a resistor or a voltage/current source, this circuit can be replaced by a single voltage source with voltage Vm in series with a resistor Rm as shown in Figure 5.13. Millman’s theorem, therefore, can determine the voltage across the parallel branches of a circuit. 152 Understandable electric circuits Millman’s theorem When several voltage sources or branches consisting of a resistor are in parallel, they can be replaced by a single voltage source. E1 E2 ::: En þ þ þ Rm ¼ R1 ==R2 == . . . ==Rn Vm ¼ Rm Im ¼ Rm R1 R2 Rn Note: Vm is the algebraic sum for all the individual terms in the equation. It will be positive if En and Vm have the same polarities, otherwise it will be negative. The letter m in Vm and Rm means Millman. Example 5.8: Determine the load voltage VL in the circuit of Figure 5.14 using Millman’s theorem. a a E1 = 40 V R1 = 100 Ω E2 = 30 V R2 = 100 Ω E3 = 20 V R3 = 100 Ω E4 = 10 V R4 = 100 Ω RL = 30 Ω Rm = 25 Ω Vm = 10 V RL = 30 Ω V L b b Figure 5.14 Circuit for Example 5.8 Solution: Rm ¼ R1 ==R2 ==R3 4 ¼ ð100==100==100==100ÞO ¼ 25 O ==R E1 E2 E3 E4 þ À À Vm ¼ Rm Im ¼ Rm R1 R2 R3 R4 40 V 30 V 20 V 10 V ¼ ð25 OÞ þ À À ¼ 10 V 100 O 100 O 100 O 100 O V L ¼ Vm RL 30 O ¼ ð10 VÞ % 5:455 V RL þ Rm ð30 þ 25ÞO 5.4.2 Substitution theorem Substitution theorem A branch in a network that consists of any component can be replaced by an equivalent branch that consists of any combination of components, as long as the currents and voltages on that branch do not change after the substitution. 15 and 5.5 V I R2 = 5 Ω R3 = 7. any branch in the circuit of Figure 5. since their voltages and currents are the same as the voltages and currents in branch a–b in the circuit of Figure 5.15.17(a).15 Circuit 1 of the substitution theorem The current and voltage of branch a–b in the circuit of Figure 5.5 Ω b Figure 5.16 can replace the a–b branch in the circuit of Figure 5.16 Circuit 2 of the substitution theorem According to the definition of the substitution theorem.9: Use a current source with a 30 O internal resistor to replace the a–b branch in the circuit of Figure 5. Example 5.9 .15 can be determined as follows: R2 6kO The voltage across branch a–b: V2 ¼ E ¼ 20 V ¼ 15 V ð2 þ 6ÞkO R1 þ R2 The current in branch a–b: a I = 2.5 mA R = 6 kΩ a I¼ E 20V ¼ 2:5 mA ¼ R1 þ R2 ð2 þ 6ÞkO a a I = 2.5 mA I = 2.5 mA + 15 V 2.17(a) Circuit for Example 5.5 mA V2 = 15 V Figure 5.15.16. a R1 = 2 Ω E = 2.The network theorems 153 This theorem can be illustrated in the circuits of Figures 5.5 mA E = 15 V + 15 V R = 2 kΩ + 5V - E = 10 V - Figure 5. a R1 = 2 kΩ E = 20 V R2 = 6 kΩ b I = 2. 17(b) shows the resultant circuit after the current source with a 30 O internal resistor replaced the a–b branch in the circuit of Figure 5.17(c) I3 0 ¼ Vab 1:5 V ¼ ¼ 50 mA R3 0 30 O Therefore. the current I30 in the R30 branch should be 50 mA. Vab ¼ E R2 ==R3 ðð5  7:5Þ=ð5 þ 7:5ÞÞO ¼ 2:5 V ¼ 1:5 V ð2 þ ð5  7:5Þ=ð5 þ 7:5ÞÞO R1 þ R2 ==R3 ðvoltage divider ruleÞ Vab 1:5 V I¼ ¼ ¼ 0:2 A ¼ 200 mA R3 7:5 O ● Determine the currents in the substituted branch and the current source branch using the circuit in Figure 5. a R1 = 2 Ω E = 2.17(c). to maintain the terminal voltage Vab ¼ 1.5 V I ′ = 50 mA I3′ = 50 mA R3′ = 30 Ω - b Figure 5.5 V in the original branch.17(b) ● Determine the voltage and current in the a–b branch of the circuit in Figure 5. I = 200 mA a + Vab = 1.17(a).17(a).154 Understandable electric circuits Solution: ● Figure 5. Using KCL we can get the current I0 in the current source branch: I 0 ¼ IÀI3 0 ¼ ð200 À 50ÞmA ¼ 150 mA .5 V R2 = 5 Ω b I′ R3′ = 30 Ω Figure 5. 3. otherwise it should be negative. Thevenin’s theorem: The equivalent circuit is a voltage source (with an equivalent resistance RTH in series with an equivalent voltage source VTH). and repeat steps 1 and 2 for the other power sources in the circuit. Norton’s theorem: The equivalent circuit is a current source (with an equivalent resistance RN in parallel with an equivalent current source IN). It equals the equivalent resistance. by setting the other inactive sources to zero. Analyse and calculate this circuit by using the single source series– parallel analysis method. Redraw the original circuit with a single source. The result should be positive when the reference polarity of the unknown in the single source circuit is the same as the reference polarity of the unknown in the original circuit. i. Open and remove the load branch (or any unknown current or voltage branch) in the network. The linear two-terminal network with the sources It is a linear complex circuit that has power sources and two terminals. and replace the current source with an open circuit. (A voltage source should be replaced by a short circuit. replace the voltage source with the short circuit. Determine the equivalent resistance RTH or RN. Superposition theorem ● ● The unknown voltages or currents in any linear network are the sum of the voltages or currents of the individual contributions from each single power supply. Steps to apply superposition theorem: 1. and mark the letter a and b on the two terminals. Steps to apply Thevenin’s and Norton’s theorems: 1. 2.The network theorems 155 Summary Linear circuit: includes the linear components (such as resistors). Turn off all power sources except one. 2. Determine the total contribution by calculating the algebraic sum of all contributions due to single sources. and a current source should be replaced by an open circuit.) That is RTH ¼ RN ¼ Rab . looking at it from the a and b terminals when all sources are turned off or equal to zero in the network. Thevenin’s and Norton’s theorems ● ● ● ● Any linear two-terminal network with power supplies can be replaced by a simple equivalent circuit that has a single power source and a single resistor.e. maximum power will be transferred to the load. It equals the opencircuit voltage from the original linear two-terminal network of a and b. circuit behaviour and performance. Equipment and components ● ● ● ● Breadboard Multimeter Dual-output DC power supply Resistors: 5. or when the load resistance is equal to the Thevenin’s/Norton’s equivalent resistance of the network (RL=RTH ¼ RN). Determine Norton’s equivalent current IN. and connect the load branch (or unknown current or voltage branch) to a and b terminals of the equivalent circuit. Maximum power transfer theorem When the load resistance is equal to the internal resistance of the source (RL ¼ RS). they can be replaced by a branch with a voltage source. i. 5. Determine Thevenin’s equivalent voltage VTH. Then the load (or unknown) voltage or current can be determined. as long as the currents and voltages on that branch do not change after the substitution. Plot Thevenin’s or Norton’s equivalent circuit. Vm ¼ Rm Im ¼ Rm E1 E2 En þ þ ÁÁÁ þ . Rm ¼ R1 == R2 == ::: == Rn R1 R2 Rn Substitution theorem A branch in a network that consists of any component can be replaced by an equivalent branch that consists of any combination of components.e. Analyse experimental data. and compare them to theoretical equivalents. VTH ¼ Vab 4. and 11 kO .5. Construct a circuit with two voltage sources. and collect and evaluate experimental data to verify applications of the superposition theorem. 7.156 Understandable electric circuits 3. i. Experiment 5A: Superposition theorem Objectives ● ● ● Understand the superposition theorem through experiment. It equals to the short-circuit current from the original linear two-terminal network of a and b.e.1. IN ¼ Isc. Millman’s theorem When several voltage sources or branches consisting of a resistor are in parallel. Reconstruct a circuit as shown in Figure L5.1(b).The network theorems 157 Background information ● ● Superposition theorem: The unknown voltages or currents in any linear network are the sum of the voltages or currents of the individual contributions from each single power supply.) Procedure 1. Record the values in Table L5.2. 5. Calculate the equivalent resistance Req00 from the terminals c and d in the circuit of Figure L5.1(c). i.5 kO R3 11 kO 2. otherwise it should be negative.1(b).2.1 Resistance Nominal value Measured value R1 5.1.1(b). Record the values in Table L5.2. Redraw the original circuit with a single source. replace the voltage source with a short circuit (by placing a jump wire). 3. 8. Calculate currents I10 and I30 in the circuit of Figure L5. Table L5. and replace the current source with an open circuit.1 kO R2 7. Record the values in Table L5. Calculate currents I100 and I300 in the circuit of Figure L5. Steps to apply superposition theorem: 1. and repeat steps 1 and 2. Record the values in Table L5. 3. 2.1(b). Analyse and calculate this circuit using the single source series–parallel analysis method.1(c). 4. Construct a circuit on the breadboard as shown in Figure L5. Turn off all power sources except one.1(c) on the breadboard. 7. by setting the other inactive sources to zero. Determine the total contribution by calculating the algebraic sum of all contributions due to single sources. (The result should be positive when the reference polarity of the unknown in the single source circuit is the same with the polarity of the unknown in the original circuit. .2. 6.e.1 using a multimeter (ohmmeter function) and record in Table L5. Calculate the equivalent resistance Req0 from the terminals a and b in the circuit of Figure L5. Record the value in Table L5. Measure the equivalent resistance Req0 and currents I10 and I30 using the multimeter (ohmmeter and ammeter functions) in Figure L5. Measure the values of resistors listed in Table L5.2. 2. explain the reasons.1 Superposition theorem Table L5. Record the value in Table L5. 11.1(c) using a multimeter (ohmmeter and ammeter functions).1 kΩ R3 = 11 kΩ d b d (b) (c) Figure L5. 12. Record the value in Table L5.1 kΩ I1 I3 I2 c E2 = 12 V R2 = 7.1(a) on the breadboard and measure current I3 in the circuit of Figure L5.1 kΩ R3 = 11 kΩ + R2 = 7. Construct a circuit as shown in Figure L5.5 kΩ R1 = 5.1(a). Conclusion Write your conclusions below: . are there any significant differences? If so.5 kΩ R3 = 11 kΩ b I1′ d (a) c I3′ I2′ a a I1′′ I3′′ I2′′ c E2 = 12 V R2 = 7. Measure the equivalent resistance Req00 and currents I100 and I300 in Figure L5.5 kΩ = b E1 = 6 V R1 = 5. Record the values in Table L5.158 Understandable electric circuits a E1 = 6 V R1 = 5. 10.2 using the calculated values.2 Resistance Formula for calculations Calculated value Measured value Req0 I 10 I30 Req00 I100 I300 I3 9.2.2. Compare the measured values and calculated values. Calculate I3 ¼ I30 þ I300 from Table L5. and collect and evaluate experimental data to verify the applications of Thevenin’s and Norton’s theorems. Procedure Part I: Thevenin’s and Norton’s theorems 1. 750 O. Analyse experimental data. maximum power will be dissipated in the load.The network theorems 159 Experiment 5B: Thevenin’s and Norton’s theorems Objectives ● ● ● ● Understand Thevenin’s and Norton’s theorems and the maximum power transfer theorem through this experiment. Construct electric circuits. 1 kO. or when the load resistance is equal to the Thevenin/Norton equivalent resistance of the circuit (RL ¼ RTH ¼ RN). Measure the values of the resistors listed in Table L5. 620 O.3 using a multimeter (ohmmeter function) and record in Table L5. Equipment and components ● ● ● ● Breadboard Multimeter DC power supply Resistors: 300 O. Table L5. 180 O. circuit behaviour and performance.3. and collect and evaluate experimental data to verify the applications of the maximum power transfer theorem. and compare them to the theoretical equivalents. Construct electric circuits.3 Resistor Colour code value Measured Value R1 300 O R2 620 O R3 750 O RL 180 O . and one 10 kO variable resistor Background information ● ● ● ● Any linear two-terminal network (complex circuit) with power supplies can be replaced by a simple equivalent circuit that has a single power source and a single resistor. Norton’s theorem: Norton’s equivalent circuit is an actual current source that has an equivalent resistance RN in parallel with an equivalent current source IN. Thevenin’s theorem: Thevenin’s equivalent circuit is an actual voltage source that has an equivalent resistance RTH in series with an equivalent voltage source VTH. The maximum power transfer theorem: When the load resistance is equal to the internal resistance of the source (RL ¼ RS). 5.2(b)) IL (Figure L5.2(a).3(a) using a multimeter (voltmeter function). 4.2(b and c). Record the value in Table L5.4 RTH VTH (Vab) RN IN (ISC) IL (Figure L5. a E1 = 12V R1 = 300 Ω R3 = 750 Ω R2 = 620 Ω RTH a IL IN RL = 180 Ω b RN a IL RL = 180 Ω ETH RL = 180 Ω b b (a) (b) (c) Figure L5. a E1 = 12 V R1 = 300 Ω R 3 = 750 Ω R1 = 300 Ω R2 = 620 Ω b (a) (b) R2 = 620 Ω b R3 = 750 Ω a Figure L5. Measure the open-circuit voltage Vab (Vab ¼ VTH) on the two terminals a and b in the circuit of Figure L5. .4.2 Thevenin’s and Norton’s equivalent circuits Table L5. Calculate RN and IN of Norton’s equivalent circuit in Figure L5.3 Thevenin’s and Norton’s circuits 6.2(a).2(c)) Formula for calculations Calculated value Measured value 3. Calculate the load current IL in Thevenin’s and Norton’s equivalent circuits of Figure L5. Record the values in Table L5.4.4. Record the values in Table L5.3(a) on the breadboard. Record the values in Table L5. Construct a circuit as shown in Figure L5.4. Calculate RTH and VTH of Thevenin’s equivalent circuit in Figure L5.160 Understandable electric circuits 2. 3(b).4. Record the value in Table L5.3(a). 10. Table L5. and measure the load current IL using a multimeter (ammeter function). RS represents the internal resistance for power supply or the Thevenin’s or Norton’s equivalent resistances. 8.4.4 (without connecting the load RL) using a multimeter (voltmeter function).The network theorems 161 7. Measure the equivalent resistance at terminals of a to b (Rab ¼ RTH ¼ RN) in Figure L5. Compare the measured values and the calculated values. Record the value in Table L5.4 on the breadboard. Construct a circuit as shown in Figure L5. Part II: Maximum power transfer 1.5 Load Resistor RL RL1 RL2 RL3 RL4 RL5 RL6 RL7 RL8 RL9 Open-circuit voltage Vab = Measured RL Value Measured VRL Value Calculated PL Value 1 kO . Measure the open-circuit voltage Vab in the circuit of Figure L5.2(b) (with measured VTH and RTH) on the breadboard.4. Record the value in Table L5. a Rs = 1 kΩ Vs = 10 V RL b Figure L5. state the reasons.4 Maximum power transfer circuit 2. Disconnect the power supply E in Figure L5. 9. Are there any significant differences? If so. Construct a circuit as shown in Figure L5.3(a) using a multimeter (ammeter function). Measure short-circuit current Isc (Isc ¼ IN) from a to b in Figure L5. and use a jump wire connecting the two terminals of E as shown in Figure L5.3 (b) using a multimeter (ohmmeter function). Record the value in Table L5.5. 5. Calculate power dissipated in each load resistor and record the values in Table L5. 4. When the load resistance is equal to the internal resistance of the source (RL ¼ RS ¼ 1 kO).4 and change the value of the variable resistor nine times from lower to higher values (one of the values should be 1 kO). Measure each RL and VRL using a multimeter (ohmmeter and voltmeter functions). 6. 5.5. Record the values in Table L5. Connect the 10-kO variable resistor RL to the circuit in Figure L5.162 Understandable electric circuits 3. is PL at the maximum point on the curve? Why? Conclusion Write your conclusions below: . Sketch the RL–PL curve (use PL as vertical axis and RL as horizontal axis). This is different with a resistor that consumes or dissipates electric energy. parallel and series–parallel configurations There are three important fundamental circuit elements: the resistor. Both of these electric elements can store energy that has been absorbed from the power supply. . and an inductor can store energy in the magnetic field. A capacitor can store energy in the electric field. you will be able to: ● ● ● ● ● ● ● ● describe the basic structure of the capacitor and inductor explain the charging and discharging behaviours of a capacitor understand the storing and releasing energy of an inductor define capacitance and inductance list the factors affecting capacitance and inductance understand the relationship between voltage and current in capacitive and inductive circuits calculate energy stored in capacitors and inductors determine the equivalent capacitance and inductance in series. capacitor and inductor. The resistor (R) has appeared in circuit analysis in the previous chapters. Practical electric circuits usually combine the above three basic elements and possibilities along with other devices. and release it to the circuit. Three basic circuit components ● ● ● Resistor (R) Capacitor (C) Inductor (L) A circuit containing only resistors has limited applications.Chapter 6 Capacitors and inductors Objectives After completing this chapter. The other two elements – the capacitor (C) and inductor (L) will be introduced in this chapter. but no matter how differently their shapes and sizes. Similar to resistors. and it extends from households to industry and the business world. ceramics. glass. oil. + – + – (a) (b) Figure 6.1 The construction of a capacitor A capacitor has applications in many areas of electrical and electronic circuits.1 Capacitor 6. etc. such as paper. variable and fixed.2 (a and b). etc.1. electronic engineering. and their schematic symbols are shown in Figure 6. air. there are two basic types of capacitors. A capacitor has two parallel conductive metal plates separated by an isolating material (the dielectric). (a) Fixed: unpolarized and polarized and (b) variable . computers. it is used in flash lamps (for flash camera). surge protections). There are many different types of capacitors. The dielectric can be of insulating material. plastic film. power systems (power supply smoothing. they all have the same basic construction. The basic construction of a capacitor is shown in Figure 6.164 Understandable electric circuits 6.2 Symbols of capacitor. Plate Plate Dielectric Figure 6. For instance. communications.1 The basic construction of a capacitor A capacitor can be represented by a capacitor schematic symbol as its circuit model. vacuum.1. mica. For a fixed polarized capacitor. plate B loses positive charges and thus shows negative. Electrolytic capacitors can have higher working voltages and store more charges than non-electrolytic capacitors.. Thus. A fixed capacitor is a capacitor that possesses a fixed value and cannot be adjusted.Capacitors and inductors 165 A variable capacitor is a capacitor that possesses a value that may be changed manually or automatically. and non-electrolytic capacitors are unpolarized. 6..3(a).3(b). With the switch at position 0. connect its positive (þ) lead to the higher voltage point in the circuit. and the negative pole of the voltage source will attract positive charges from the negative plate of the capacitor. For an unpolarized capacitor. Capacitor C An energy storage element that has two parallel conductive metal plates separated by an isolating material (the dielectric).Vc B C Vc = 0 E I (a) (b) Figure 6. Plate A loses electrons and shows positive. the circuit is open. and the potential difference between the two metal plates of the capacitor is zero (VC ¼ 0). the DC voltage source is connected to the two leads of the capacitor. this causes current I to flow in the circuit. Two plates of the capacitor have the same size and are made by the same conducting material. and the potential difference (VC) appears on the . so they should have the same number of charges at the initial condition.. 1 0 E 2 1 2 A + + + + + + .3 Charging a capacitor Once the three-position switch is turned on to position 1 as shown in Figure 6. a threeposition switch. and a DC (direct current) voltage source (E) is shown in Figure 6. the electric field is built up between the two metal plates of the capacitor.. From the rule ‘opposites attract and likes repel’. and negative (7) lead to the lower voltage point..1. we know that the positive pole of the voltage source will attract electrons from the positive plate of the capacitor. Electrolytic capacitors are usually polarized. it does not matter which lead connects to where.2 Charging a capacitor A purely capacitive circuit with an uncharged capacitor (VC ¼ 0). this may eventually slowly dissipate the charges. This is why a capacitor is called an energy storage element. as voltage across the capacitor VC will cause the current to flow in the circuit. 1 2 E C E 1 2 + + − + − A B I − VC (a) (b) Figure 6.e.1. the capacitor is equivalent to a voltage source. Since the insulating material will not be perfect and a small leakage current may flow through the dielectric.4. and it can maintain the potential difference across it. and the process of charging the capacitor is completed.3(b). If the voltage across the capacitor VC is measured at this time using a multimeter (voltmeter function).4 Discharging a capacitor When the switch is closed to position 2 as shown in the circuit of Figure 6.3 Energy storage element When the switch is turned off to position 0 in the circuit shown in Figure 6. After the capacitor has released all its stored energy. 6. the voltage across the capacitor will be zero (VC ¼ 0). At this time. as shown in Figure 6. So the energy storage element capacitor will keep its charged voltage VC for a long time (duration will depend on the quality and type of the capacitor). as it can store charges absorbed from the power supply and store electric energy obtained from charging.1. Charges will not be able to cross the insulating material from one plate to another.166 Understandable electric circuits capacitor with positive (þ) on plate A and negative (7) on plate B. the charging current cesses to flow (I ¼ 0). This is known as discharging a capacitor. an electric field is built up between the two plates of the capacitor. 6. the capacitor and power supply will disconnect. This is the process of charging a capacitor. and a high current causes the capacitor to release its charges or stored energy in a short time. i. Once voltage across the capacitor VC has reached the source voltage VS.4 Discharging a capacitor . Once a capacitor has transferred some charges through charging. VC ¼ E. the current in the circuit ceases to flow (I ¼ 0) and the discharge process is completed. The isolating material (dielectric) between the two metal plates isolates the charges between the two plates. Since there is no resistor in this circuit. there is no more potential difference between the source and capacitor. it is a short circuit.3(a). the capacitor and wires in the circuit forms a closed path. VC should still be the same with the source voltage (VC ¼ E) even without a power supply connected to it. gradually increase (charging) or decrease (discharge). The voltage and charge (V–Q) characteristic of a capacitor is shown in Figure 6. There is an important characteristic that implies in the charge and discharge of a capacitor. Q ¼ CV or C ¼ Q V Figure 6.5 Capacitance As previously mentioned. the capacitor starts to store energy or charges. This can be expressed by the following formula: Q V This is analogous to a pump pumping water to a reservoir. Discharging: The process of releasing energy. A passive component is a component that absorbs (but not produce) energy.Capacitors and inductors 167 The capacitor cannot release energy that is more than it has absorbed and stored. Charging/Discharging a capacitor An electric element that can store and release charges that it absorbed from the power supply. therefore it is a passive component. it will always take time. The charges (Q) that are stored are proportional to the voltage (V) across it. The higher the pressure. i. The concept of a capacitor may be analogous to a small reservoir. The process of charging a capacitor from the power supply is similar to a reservoir storing water.5 Q–V characteristic of a capacitor . That is. demonstrating that the capacitor voltage is proportional to the amount of charges a capacitor can store.e. once the source voltage is applied to two leads of a capacitor. It acts as a reservoir that stores and releases water.5.1. the voltage on the capacitor won’t be able to change instantly. ● ● Charging: The process of storing energy. The higher the voltage. the more charges a capacitor can store. the more water will be pumped into the reservoir. The process of discharging a capacitor is similar to a reservoir releasing water. 6. Just as a resistor is a component and resistance is the value of a resistor. and inversely proportional to the voltage (V) across it. C¼ Quantity Capacitance Charge Voltage Q V Quantity symbol C Q V Unit Farad Coulomb Volt Unit symbol F C V A capacitor can store 1 C charge when 1 V of voltage is applied to it. V ¼? Q 50 mC 50  10À6 C ¼ ¼ ¼ 0:05  106 V ¼ 50 KV C 1000 pF 1000  10À12 F . which is the value of the capacitor and describes the amount of charges stored in the capacitor. is directly proportional to its stored charges. Resistor is symbolized by R while resistance is R: capacitor is symbolized by C while capacitance is C. Capacitance C C. Example 6. V¼ C ¼ 1000 pF. Solution: Q ¼ 50 mC. Microfarad (mF) or picofarad (pF) are more commonly used units for capacitors. the value of the capacitor. capacitor is a component and capacitance is the value of a capacitor. 1F ¼ 1C 1V Farad is a very large unit of measure for most practical capacitors.1: If a 50 mC charge is stored on the plates of a capacitor.168 Understandable electric circuits C is the capacitance. Recall 1 mF ¼ 10À6 F and 1 pF ¼ 10À12 F Note: m is a Greek letter called ‘mu’ (see Appendix A for a list of Greek letters). That is. determine the voltage across the capacitor if the capacitance of the capacitor is 1000 pF. . and they are determined by the construction of a capacitor as shown below: ● ● ● The area of plates (A): A is directly proportional to the charge Q. Therefore.6. the larger the plate area. the more electric charges that can be stored.Capacitors and inductors 169 6.6 Factors affecting capacitance There are three basic factors affecting the capacitance of a capacitor. The factors affecting the capacitance of a capacitor are illustrated in Figure 6. A d k Figure 6. The dielectric constant (k): Different insulating materials (dielectrics) will have a different impact on the capacitance. the distance (d) between the two plates is inversely proportional to the capacitance (C).1. the stronger the produced electric field that will increase the ability to store charges. The dielectric constant (k) is directly proportional to the capacitance (C).1. The distance between the two plates (d): The shorter the distance between two plates.6 Factors affecting capacitance Factors affecting capacitance C ¼ 8:85  10À12 kA d Quantity Plates area Distance Dielectric constant Capacitance Quantity symbol A d k C Unit Square meter Meter No unit Farad Unit symbol m2 m F Dielectric constants for some commonly used capacitor materials are listed in Table 6. the more water will be pumped into the tank.8 Breakdown voltage As mentioned earlier.e. Once voltage is applied across the capacitor.2: Determine the capacitance if the area of plates for a capacitor is 0.) Leakage current A very small current through the dielectric.7 Leakage current The dielectric between two plates of the capacitor is insulating material. Solution: A ¼ 0:004 m2 . If the tank is full and still continues to increase pressure.6 2.004 m2. 100 per cent of the insulation). it is always there.1. and practically no insulating material is perfect (i.0006 2. d ¼ 0:006 m and k ¼ 5 C ¼ 8:85  10À12 kA 5  0:004m2 ¼ 8:85  10À12 ¼ 29:5 pF d 0:006m 6. and this is called the leakage current in the capacitor. That is why the charges or the energy stored on the capacitor plates will eventually leak off. the distance between the plates is 0.5 5 4 2.1 Example 6. Although the leakage current is very small.170 Understandable electric circuits Table 6. a capacitor charging acts as a pump pumping water into a reservoir. or a water tank.1. there may be a very small current through the dielectric. .1 Dielectric constants of some insulating materials Material Vacuum Air Paper (dry) Glass (photographic) Mica Oil Polystyrene Teflon Dielectric constant 1 1. The higher the pressure.5 7.006 m and the dielectric for this capacitor is mica. the tank may break down or become damaged by such high pressure. (Electrolytic capacitors have higher leakage current. 6. But the leakage current is so small that it can be ignored for the application. Differentiating the equation q ¼ Cv yields dq dv ¼C dt dt Recall that current is the rate of movement of charges. which is the maximum voltage a capacitor can have. Breakdown voltage The voltage that causes a capacitor’s dielectric to become electrically conductive. and skip the following mathematic derivation process.9 Relationship between the current and voltage of a capacitor The relationship between the current and voltage for a resistor is Ohm’s law for a resistor. As a result. when using a capacitor. which is the quantity at a specific time. Usually the lowercase letters symbolize instantaneous quantities. where Dv and Dt or dv and dt are very small changes in voltage and time. Substitute i ¼ dq/dt into the equation of dq/dt ¼ C(dv/dt) yields i¼C dv dt or i¼C Dv Dt This is Ohm’s law for a capacitor. this may explode or permanently damage the capacitor. pay attention to the maximum working voltage. and the uppercase letters symbolize the constants or average quantities. Note: If you haven’t learned calculus.Capacitors and inductors 171 This is similar to a capacitor. the charges (q) stored on the plates of the capacitor will also change. and has the i ¼ dq/dt notation in calculus. It can be obtained mathematically as follows. the capacitor’s dielectric will break down. The equation Q ¼ CV in terms of instantaneous quantity is q ¼ Cv. causing current to flow through it. The relationship between voltage and current of a capacitor can be expressed by Figure 6.7(b). This will cause current to flow in . Therefore. just keep in mind that i ¼ C(Dv/Dt) or iC ¼ C(dvC/dt) is Ohm’s law for a capacitor. The relationship between the current and voltage for a capacitor is Ohm’s law for a capacitor. A quantity that varies with time (such as a capacitor that takes time to charge/discharge) is called instantaneous quantity. If the voltage across a capacitor is too high and exceeds the capacitor’s working or breakdown voltage. It may explode or permanently damage the capacitor. 6. The relationship of voltage and current for a capacitor shows that when the applied voltage at two leads of the capacitor changes.1. The applied voltage of the capacitor can never exceed the capacitor’s breakdown voltage. 4. the greater the amount of capacitive current flows through the circuit.172 Understandable electric circuits i C + dv (b) V i C 0 (a) dt Figure 6.3 and 6. the reference polarities of voltage and current of a capacitor should be consistent. as shown in Figure 6. Note: Although there is a DC voltage source applied to the capacitive circuit in Figures 6. Voltage that does not change with time is DC voltage. A capacitor can block DC current. Ohm’s law for a capacitor The current of a capacitor iC is directly proportional to the ratio of capacitor voltage dvC/dt (or DvC/Dt) and capacitance C. The relationship of voltage and current in a capacitive circuit shows that the faster the voltage changes with time.7 Relationship between v and i of a capacitor the capacitor circuit. Similarly.7(a). the smaller the amount of current. meaning that current is zero when DC voltage is applied to a capacitor. the capacitor charging/discharging happened at the moment . iC ¼ C dvC dt or i¼C DvC Dt where dvC and dt or DvC. and Dt are very small changes in voltage and time. and if voltage does not change with time. That is. the capacitor may play an important role for blocking the DC current. This is a very important characteristic of a capacitor. Zero current means that the capacitor acts like an open circuit for DC voltage at this time. Current and the rate of change of voltage are directly proportional to each other. Therefore. The reference polarities of capacitor voltage and current should be mutually related. DC blocking Current through a capacitor is zero when DC voltage applied to it (opencircuit equivalent). the slower the voltage changes with time. the current will be zero. Substituting this into the capacitor’s current i ¼ C(dv/dt) yields p ¼ Cv dv dt Since the relationship between power and work is P ¼ W/t (energy is the ability to do work). It can store energy that it absorbed from charging and maintain voltage across it. the capacitor is equivalent to an open circuit for that circuit. i. substituting it into p ¼ Cv(dv/dt) yields dw dv ¼ Cv dt dt Integrating the above expression: ðt ðv dw dv dt ¼ C v dt dt 0 0 dt gives 1 W ¼ Cv2 2 Note: If you haven’t learned calculus. just keep in mind that W ¼ ½(Cn2). and. and skip the above mathematic derivation process. instantaneous power for this expression is p ¼ dw/dt.Capacitors and inductors 173 when the switch turned to different locations. 6.1. When the capacitor charging/dischaging has finished. Energy stored by a capacitor 1 WC ¼ Cv2 2 Quantity Energy Capacitor Voltage Quantity symbol W C V Unit Joule Farad Volt Unit symbol J F V . Energy stored by a capacitor in the electric field can be derived as follows. when the voltage across the capacitor changes within a moment. The instantaneous electric power of a capacitor is given by p ¼ ni. a capacitor is an energy storage element.e.10 Energy stored by a capacitor As mentioned earlier. The total or equivalent capacitance has the opposite form with the total or equivalent resistance Req. The total or equivalent capacitance Ceq will decrease for a series capacitive circuit and it will increase for a parallel capacitive circuit.1 Capacitors in series A circuit of n capacitors is connected in series as shown in Figure 6.3: A 15 V voltage is applied to a 2.2 mF capacitor.2. 6. Determine the energy this capacitor has stored.2 Capacitors in series and parallel Same as resistors. Solution: 1 WC ¼ Cv2 2 1 ¼ ð2:2 mFÞ Â ð15 VÞ2 2 ¼ 247:5 mJ 6.8 n capacitors in series Applying Kirchhoff’s voltage law (KVL) to the above circuit gives E ¼ V1 þ V2 þ Á Á Á þ Vn and since V ¼ Q/C. Example 6. substituting it into the above expression yields Qeq Q1 Q2 Qn ¼ þ þ ÁÁÁ þ Ceq C 1 C2 Cn .174 Understandable electric circuits The expression for energy stored by a capacitor demonstrates that the capacitor’s energy depends on the values of the capacitor and voltage across the capacitor. capacitors may also be connected in series or parallel to obtain a suitable resultant value that may be either higher or lower than a single capacitor value. Q1 Q2 E C1 V1 C2 V2 … Qn Cn Vn Figure 6.8. Since only one current flows in a series circuit. Equivalent (total) series capacitance ● n capacitors in series: Ceq ¼ 1 ð1=C1 Þþð1=C2 ÞþÁÁÁþð1=Cn Þ ● C2 Two capacitors in series: Ceq ¼ CC1þC2 1 Example 6. it also has the same form with the formula for calculating two resistors in parallel. This formula has the same form with the formula for calculating equivalent parallel resistance (1/Req) ¼ (1/R1) þ (1/R2) þ . þ (1/Rn). Qeq is the equivalent (or total) charges and Ceq is the equivalent (or total) capacitance for a series capacitive circuit respectively. i. Qeq ¼ Q1 ¼ Q2 ¼ . .9. . C1 E = 25 V C2 C3 C4 100 μF 100 μF 100 μF 100 μF Figure 6. ¼ Qn ¼ Q therefore.Capacitors and inductors 175 where E ¼ Qeq/Ceq. .9 Circuit for Example 6. each capacitor will store the same amount of charges.4: Determine the charges Q stored by each capacitor in the circuit of Figure 6. When there are two capacitors in series. Q Q Q Q ¼ þ þ ÁÁÁ þ Ceq C 1 C2 Cn Dividing by Q on both sides of the above expression gives 1 1 1 1 ¼ þ þ ÁÁÁ þ Ceq C1 C2 Cn or Ceq ¼ 1 ð1=C1 Þ þ ð1=C2 Þ þ Á Á Á þ ð1=Cn Þ This is the equation for calculating the series equivalent (total) capacitance. . i.e. .4 Solution: Since Q ¼ CV.e. solve for Ceq first. or Q ¼ CeqE. Ceq ¼ C1C2/(C1 þ C2) and Req ¼ R1R2/ (R1 þ R2). 85 6 10712kA/d) shows that if the distance between the plates of a capacitor (d) increases.10.176 Understandable electric circuits Ceq ¼ ¼ 1 ð1=C1 Þ þ ð1=C2 Þ þ ð1=C3 Þ þ ð1=C4 Þ 1 ¼ 25 mF ½ð1=100Þ þ ð1=100Þ þ ð1=100Þ þ ð1=100ÞmF Therefore. The charge stored on the individual capacitor in this circuit is Q1 ¼ C1 V .2. the total or equivalent capacitance Ceq (25 mF) is less than any one of the individual capacitances (100 mF).4 shows that when capacitors are connected in series. Ceq Figure 6. Q ¼ Ceq E ¼ ð25 mFÞ ð25 VÞ ¼ 625 mC Example 6. the capacitance (C) will decrease. This is shown in Figure 6. The formula for factors affecting the capacitance (C ¼ 8.2 Capacitors in parallel A circuit of n capacitors connected in parallel is shown in Figure 6.11.10 The physical characteristic of series Ceq 6.11 n capacitors in parallel . ðwhere V ¼ EÞ Q1 E C1 Q2 C2 … Qn Cn Figure 6. The physical characteristic of the series equivalent capacitance is that the single series equivalent capacitance Ceq has the total dielectric (or total distance between the plates) of all the individual capacitors. Q2 ¼ C2 V . Qn ¼ Cn V . As you may have noticed. the total or equivalent capacitance Ceq (450 mF) is greater than any one of the . Ceq V ¼ C1 V þ C2 V þ Á Á Á þ Cn V dividing both sides by V yields Ceq ¼ C1 þ C2 þ Á Á Á þ Cn This is the equation for calculating the parallel equivalent (total) capacitance.þ Rn). i.5 Solution: Since Q ¼ CV. . Qeq ¼ Q1 þ Q2 þ Á Á Á þ Qn therefore. i.Capacitors and inductors 177 The total charge Qeq in this circuit should be the sum of all stored charges on the individual capacitor.12 Circuit for Example 6. we can see that when capacitors are connected in parallel.5. . and capacitors in parallel result in series form as resistances.5: Determine the total charge in all the capacitors in the circuit of Figure 6. this equation has the same form with the equation for calculating series resistances (Req ¼ R1 þ R2 þ. Qeq ¼ CeqE and Ceq ¼ C1 þ C2 þ Á Á Á þ Cn ¼ ð100 þ 10 þ 20 þ 320ÞmF ¼ 450 mF therefore.e. Qeq ¼ Ceq E ¼ ð450 mFÞð60 VÞ ¼ 27 000 mC From Example 6. Equivalent (total) parallel capacitance Ceq ¼ C1 þ C2 þ . þ Cn Equation for calculating capacitance is exactly opposite the equations for calculating resistance. Capacitors in series result in parallel form as resistances.e. E = 60 V C1 = 100 μF C2 = 10 μF C3 = 20 μF C4 = 320 μF Figure 6. Example 6. . .12. 6 μF C5 = 2 μF Figure 6.4.85 6 10712kA/d).178 Understandable electric circuits individual capacitances (C1 ¼ 100 mF.3 Capacitors in series–parallel Similar to resistors. Ceq Figure 6. When serial and parallel capacitors are combined together. This is shown in Figure 6.14. the capacitance will increase.6 Solution: C4. C3 ¼ 20 mF and C4 ¼ 320 mF). From the formula of factors affecting the capacitance (C ¼ 8.5 μF C3 = 0.4.5 ð6mFÞð3mFÞ Ceq ¼ ¼ ¼ 2 mF C 1 þ C2.14 Circuit for Example 6.3.13 The physical characteristic of parallel Ceq 6. a C1 = 6 μF C2 = 2 μF b C4 = 0. capacitors may also be connected in various combinations. we can see that if the area of plates (A) of a capacitor increases.5 ¼ C2 þ C3 þ C4.2. series–parallel capacitor circuits result and an example is shown in the following. Example 6. The physical characteristic of the equation for calculating the parallel equivalent capacitance is that a single parallel equivalent capacitor Ceq has the total area of plates of the individual capacitors.5 ¼ C4 C5 ð0:5mFÞð2mFÞ ¼ ¼ 0:4 mF ð0:5 þ 2ÞmF C 4 þ C5 C2.5 ¼ ð2 þ 0:6 þ 0:4ÞmF ¼ 3 mF C1 C2.13.3.5 ð6 þ 3ÞmF .6: Determine the equivalent capacitance through two terminals a and b in the circuit of Figure 6. C2 ¼ 10 mF.3.4. When the charge changes its velocity of motion (or when the charge is accelerated). As the changing current flows through the conductor. and the movement of a charge will produce a magnetic field. whenever a changing current flows through a conductor.Capacitors and inductors 179 6. The area shows that the magnetic characteristics are called the magnetic field. the area surrounding the conductor will produce an electromagnetic field. radios.3. The difference between the two is that a capacitor stores transferred energy in the electric field. The direction of these lines of force can be determined by the right-hand spiral rule. as shown in Figure 6.1. as it is produced by the changing current-carrying conductor. both capacitors and inductors are energy storage elements. an electromagnetic field is generated. and an inductor stores transferred energy in the magnetic field.3 Inductor We have learned about two of the three important fundamental passive circuit elements (components that absorb but not produce energy). The electromagnetic field can be visualized by inserting a current-carrying conductor (wire) through a hole in a cardboard and sprinkling some iron filings on it.3. radars. Therefore. TVs. it is also called the electromagnetic field. Four fingers = the direction of magnetic lines of force or direction of the flux (the total magnetic lines of force). Inductors have many applications in electrical and electronic devices. . Since inductors are based on the theory of electromagnetism induction.1 Electromagnetism induction 6. 6. let us review some concepts of electromagnetism induction you may have learned in physics that will be used in the following section. As previously mentioned. including electrical generators. The third element is the inductor (or coil). transformers. etc.1 Electromagnetic field All stationary electrical charges are surrounded by electric fields.15. I → → E Figure 6. and therefore. the iron fillings will align themselves with the circles surrounding the conductor. these are magnetic lines of force. This is the principle of electricity producing magnetism. the resistor and the capacitor.15 Electricity produces magnetism ● Right-hand spiral rule: Thumb = the direction of current. motors. eL) that has an opposite polarity with vL will be induced. In other words. 6. and this will result in an induced current in the coil. or the more the turns the coil has. For example. and is expressed mathematically as vL ¼ N(df/dt). hence an electromagnetic field is generated. in Figure 6. the changing magnetic flux will induce an electromagnetic field and produce an induced voltage (vL). British physicist and chemist Michael Faraday discovered how an electromagnetic field can be induced by a changing magnetic flux.2 Faraday’s law In 1831. the magnetic lines of flux will be cut and a voltage vL across the coil will be induced (vL can be measured by using a voltmeter. if a magnet bar is moved back and forth in a coil of wire (conductor). . This is the principle of a magnet producing electricity. or if the coil is moved back and forth close to the magnet and through the magnetic field.3. When there is a relative movement between a conductor and a magnetic field (or a changing current through the conductor). Faraday’s law ● ● When there is a relative movement between a conductor and magnetic field. V Figure 6.16 Magnet produces electricity Faraday observed that the induced voltage (vL) is directly proportional to the rate of change of flux (df/dt) and also the number of turns (N) in the coil. vL is directly proportional to the rate of change of flux (df/dt) and the number of turns (N) in the coil. the faster the relative movement between the conductor and magnetic fields. vL ¼ N(df/dt). the higher the voltage will be produced.180 Understandable electric circuits Electromagnetic field The surrounding area of a conductor with a changing current can generate an electromagnetic field.) Or. it will induce a changing magnetic flux F (the total number of magnetic lines of force) surrounding the conductor.16. an electromotive force (emf. This electromagnetic field will produce an induced voltage and current.1. The induced voltage (vL) and induced emf (eL) have opposite polarities (E ¼ 7V). This is similar to the concept of the mutually related reference polarity of voltage and current. the current (cause) in the circuit will increase. then eL ¼ ÀL . and the minus sign for eL is to remind us that the induced emf always acts to oppose the change in magnetic flux that generates the emf and current. an induced voltage (vL) or induced emf (eL) and also an induced current (i) will be produced. > 0. When there is a relative movement between a conductor and a magnetic field (or a changing current through the conductor). Lenz’s law ● When there is a changing current through the conductor. Russian physicist Heinrich Lenz developed a companion result with the Faraday’s law. but the polarity of induced emf (eL) changes and will try to stop it from decreasing. However. or vL ¼ L dt dt dt where di/dt is the rate of change of current. The polarity of the induced emf is always opposite to the change of the original current. This is because an induced current in the circuit flows in a direction that can create a magnetic field that will counteract the change in the original magnetic flux. this emf is also called the counter emf.17. an induced voltage (vL) or induced emf (eL) and also an induced current (i) will be produced. the induced voltage (vL) has the same polarity with the direction of induced current (i).Capacitors and inductors 181 6. Lenz’s law can be expressed as follows: di di di If i > 0. When the switch is turned on in the circuit of Figure 6.17 Lenz’s law Mathematically.3. the current i will decrease. Lenz defined the polarity of induced effect and stated that an induced effect is always opposed to the cause producing it. When the switch is turned off. but the induced emf (effect) will try to stop it from increasing.1. i E L eL + + vL E R + eL vL + Figure 6. .3 Lenz’s law In 1834. According to the principle of electromagnetic induction. The schematic symbol for an air-core inductor looks like a coil of wire as shown in Figure 6.2 Inductor An inductor (L) is made by winding a given length of wire into a loop or coil around a core (centre of the coil).3 Self-inductance When current flows through an inductor (coil) that is the same as a currentcarrying conductor. Inductors may be classified as air-core inductors or iron-core inductors. eL ¼ ÀL dt or vL ¼ L dt ● The letter L in the above equation is called inductance (or self-inductance).18(b). L Air-core (a) L Variable air-core L Iron-core (b) L Variable iron-core Figure 6. the inductor can be also classified as fixed and variable. which is discussed below. 6. when there is a relative movement between an inductor and . Faraday’s law and Lenz’s law. But this coil turns out to be a very important electric/electronic element because of its magnetic properties. Similar to resistors and capacitors.3. a magnetic field will be induced around the inductor. 6.18 Schematic symbols for inductors Inductor L An inductor is an energy storage element that is made by winding a given length of wire into a loop or coil around a core. An air-core inductor is simply a coil of wire. The schematic symbol for an iron-core inductor is shown in Figure 6. Iron-core provides a better path for the magnetic lines of force and a stronger magnetic field for the iron-core inductor as compared to the air-core inductor.18(a).182 Understandable electric circuits The polarity of the induced emf (eL) is always opposite to the change di di of the original current.3. 3. The measurement of the changing current in an inductor that is able to generate induced voltage is called inductance. the inductance (L) and the current rate of change (di/dt) determine the induced voltage (nL). The inductor is symbolized by L while inductance is symbolized by L. capacitance and inductance are the value or capacity of these components. and the resistance.4 Relationship between inductor voltage and current Lenz’s law vL ¼ L(di/dt) shows the relationship between current and voltage for an inductor. The induced voltage nL is directly proportional to the inductance L and the current rate of change di/dt. and it is Ohm’s law for an inductor. So inductance is the capacity to store energy in the magnetic field of an inductor. The resistor. There. This relationship can be illustrated as in Figure 6. or induced emf (eL). the changing magnetic flux will induce an electromagnetic field resulting in an induced voltage (vL).19. Inductance L (or self-inductance) The measurement of the changing current in an inductor that is able to generate induced voltage is called inductance that is measured in henries (H).19 Characteristics of an inductor’s voltage and current Ohm’s law for an inductor An inductor’s voltage nL is directly proportional to the inductance L and the rate of change current di/dt: nL ¼ L(di/dt). i + L vL L vL = L di dt di dt Figure 6.Capacitors and inductors 183 magnetic field or when current changes in the inductor. capacitor and inductor are circuit components. Ohm’s law for an inductor nL ¼ L(di/dt) has a similar form as Ohm’s law for a capacitor iC ¼ C(dnC/dt). Quantity Inductance (or Self-Inductance) Quantity symbol L Unit Henry Unit symbol H 6. and the unit of inductance is henry (H). and also an induced current (i). . These two are very important formulas that will be used in future circuits. e. These parameters are determined by the construction of an inductor as shown in the following (if all other factors are equal): ● ● ● ● The number of turns (N) for the coil: More turns for a coil will produce a stronger magnetic field resulting in a higher induced voltage and inductance.20. An inductor can pass DC. Different materials have different degrees of permeability. and therefore it can produce a stronger magnetic field resulting in a higher inductance. i. Passing DC ● ● Voltage across an inductor is zero when a DC current flows through it (short-circuit equivalent). The permeability of the material of the core (m): A core material with higher permeability will produce a stronger magnetic field resulting in a higher inductance. di/dt ¼ 0. or the greater the change of current.5 Factors affecting inductance There are some basic factors affecting the inductance of an inductor (ironcore). 6.184 Understandable electric circuits The larger the inductance. the higher the induced voltage in the coil. This is a very important characteristic of an inductor and is opposite to that of a capacitor.3. Recall that a capacitor can block DC and acts like an open circuit for DC. Zero voltage means that an inductor acts like a short circuit for DC current. the inductor may play an important role for passing the DC current. Factors affecting inductance L¼ N 2 Am l . (Permeability of the material of the core determines the ability of material to produce a magnetic field. The cross-section area of the core (A): A larger core area requires more wire to construct a coil. and therefore producing a weaker magnetic field resulting in a smaller inductance. The length of the core (l): A longer core will make a loosely spaced coil and a longer distance between each turn. the inductor voltage (vL) is also zero.) Factors affecting the inductance of an inductor are illustrated in Figure 6. Therefore. When the current does not change with time (DC current). number of turns is symbolized by N. and the instantaneous power for this expression is p ¼ dw/dt.Capacitors and inductors 185 L μ A N Figure 6. or when core material with higher permeability is chosen. the inductance of an inductor will increase.6 The energy stored by an inductor Same as a capacitor. or when the length of core is reduced.3. The energy stored by an inductor can be derived as follows: The instantaneous electric power of an inductor is given by p ¼ ivL Since the relationship between power and work is P ¼ W =t (energy is the ability to do work).20 Factors affecting inductance Where inductance is symbolized by L. Substituting p ¼ dw/dt and vL ¼ L(di/dt) into the instantaneous power expression p ¼ inL gives dw di ¼ Li dt dt . When voltage is applied to two leads of an inductor. and this energy is then absorbed by the inductor and stored in the magnetic field as electromagnetic field builds up. From the expression of the factor affecting inductance. area of the core is symbolized by A. measured in m2. measured in henries (H). we can see that either when the number of turns of a coil increases. the current flows through the inductor and will generate energy. or when the cross-section area of the core increases. 6. permeability is symbolized by m. an inductor is also an energy storage element. 0 i:e: 1 w ¼ Li2 2 Note: If you haven’t learned calculus. determine the energy stored by the inductor and induced voltage nL. When current decreases. an inductor will not be able to release more energy than it has stored. and skip the above mathematic derivation process. an inductor releases the stored energy to the circuit. an inductor absorbs energy and stores it in the magnetic field of the inductor. This equation has a similar form with the energy equation of a capacitor (WC ¼ (½)Cv2). The equation for energy stored by an inductor shows that the inductor’s energy depends on the inductance and the inductor’s current. skip the vL part. just keep in mind that WL ¼ (½)Li2. energy W is measured in J and current i is measured in A. so it is also called a passive element. 6. Solution: WL ¼ 1 2 1 1 Li ¼ ð0:01 HÞð5eÀ2t AÞ2 ¼ ð0:01 HÞð25eÀ4t AÞ ¼ 0:125eÀ4t J 2 2 2 di d À2t vL ¼ L ¼ 0:01 ð5e Þ ¼ 0:01 HðÀ2Þð5ÞðeÀ2t AÞ ¼ À0:1eÀ2t V dt dt Note: If you haven’t learned calculus. Example 6. This is because . an inductor is formed.7: Current in a 0.7 Winding resistor of an inductor When winding a given length of wire into a loop or coil around a core.3.01 H inductor is i(t) ¼ 5e72tA. Same as a capacitor. Energy stored by an inductor 1 WL ¼ Li2 2 where inductance L is measured in H. When current increases. A coil or inductor always has resistance.186 Understandable electric circuits Integrating both sides: t ð t ð dw di dt ¼ Li dt dt dt 0 0 Therefore t ð w ¼ L idi. Example 6.8 Solution: E ¼ 20 V. Rw ¼ 5 O.8: The winding resistance for an inductor in the circuit of Figure 6.22 is 5 O. the more turns of coils there are. This is called the winding resistance of a coil (Rw).22 Circuit for Example 6. and the longer the wire.21. When the current approaches a steady state (does not change any more). and thus the wire will have a significantly higher internal resistance.Capacitors and inductors 187 there is always a certain internal resistance distributed in the wire. and WL ¼ 4 J: E 20 V I¼ ¼ ¼ 0:4 A R þ Rw ð45 þ 5ÞO L¼? . Rw L Figure 6. What is the inductance of the inductor? I R = 45 Ω Rw = 5 Ω E = 20 V L Figure 6. An inductor circuit with winding resistance is shown in Figure 6.21 Winding resistance Winding resistance Rw The internal resistance in the wire of an inductor. R ¼ 45 O. the energy stored by the inductor is 4 J. .24 Inductors in parallel .4. .23.23 Inductors in series Equivalent series inductance Leq ¼ L1 þ L2 þ .4 Inductors in series and parallel Similar to resistors and capacitors. As you may have noticed. The equivalent inductance will increase if inductors are in series. The equivalent (total) series or parallel inductance has the same form as the equivalent (total) series or parallel resistance.24. 6. L1 Leq L2 Ln … Figure 6. . þ Ln This is the equation for calculating the equivalent (total) series inductance. Leq L1 L2 … Ln Figure 6.2 Inductors in parallel A circuit of n inductors connected in parallel is shown in Figure 6. þ Rn). . inductors may also be connected in series or in parallel to obtain a suitable resultant value that may be either higher or lower than a single inductor value.4.188 Understandable electric circuits From WL = ½(Li2) solving for L: 2WL 2  4J ¼ ¼ 50 H 2 I ð0:4 AÞ2 (i ¼ I since the current approaches to steady state) L¼ 6. this formula has the same form as the formula for calculating series resistances (Req ¼ R1 þ R2 þ .1 Inductors in series A circuit of n inductors connected in series is shown in Figure 6. 6. and the equivalent (total) inductance will decrease if inductors are in parallel. þ (1/Rn). these equations have the same forms as the equations for calculating parallel resistance Req ¼ 1/(1/R1) þ (1/R2) þ . .25 Circuit for Example 6. If Leq ¼ 70 H.25. . inductors may also be connected in various combinations of series and parallel.10: There are three inductors in a series–parallel inductive circuit: 40. Example 6. As you may have noticed.3 Inductors in series–parallel Similar to resistors and capacitors. 6.9: Determine the equivalent inductance for the series–parallel inductive circuit shown in Figure 6. An example of a series–parallel inductive circuit is shown in the following. L2 = 1H L4 = 1H Leq L1 = 3H L6 = 1H L3 = 2H L5 = 1H Figure 6. also Req ¼ R1R2/(R1 þ R2).9 Solution: Leq ¼ ½ðL4 þ L5 Þ==L3 þ ðL2 þ L6 Þ==L1 ½ðð1 þ 1Þ Â 2Þ=ðð1 þ 1Þ þ 2Þ þ 1 þ 1  3 Leq ¼ H ¼ 1:5 H ½ðð1 þ 1Þ Â 2Þ=ðð1 þ 1Þ þ 2Þ þ 1 þ 1 þ 3 Example 6. how are these inductors connected? .Capacitors and inductors 189 Equivalent parallel inductance ● n inductors in parallel: Leq ¼ 1 ð1=L1 Þþð1=L2 ÞþÁÁÁþð1=Ln Þ ● Two inductors in parallel: Leq ¼ LL1 L2 2 1 þL These are the equations for calculating the equivalent parallel inductance.4. 40 and 50 H. Capacitor discharging: Capacitor releases energy to the circuit. it can explode or permanently damage the capacitor. Inductance (L): The measurement of the changing current in an inductor that produces the ability to generate induced voltage is called inductance. Factors affecting capacitance: C ¼ 8:85  10À12 kA d ● ● ● Leakage current: A very small current through the dielectric. Faraday’s law: vL ¼ N df dt ● Lenz’s law : eL ¼ ÀL di dt or vL ¼ L di dt ● ● Inductor (L): An energy storage element that is made by winding a given length of wire into a loop or coil around a core. Breakdown voltage: The voltage that causes a capacitor’s dielectric to become electrically conductive. C ¼ Q/U. . Capacitance (C): The value of the capacitor. and then in series with a 50 H inductor. Summary Capacitor ● ● ● ● ● Capacitor (C): An energy storage element that has two conductive plates separated by an isolating material (the dielectric).190 Understandable electric circuits Solution: Leq ¼ 50 H þ ð40 HÞ==ð40 HÞ ¼ 70 H 40  40 or Leq ¼ 50 H þ H ¼ 70 H 40 þ 40 So two 40 H inductors are in parallel. Inductor ● ● Electromagnetic field: The surrounding area of a conductor with a changing current can generate an electromagnetic field. Capacitor charging: Capacitor stores absorbed energy. Blocking DC: A capacitor can block DC current (open-circuit equivalent). and compare them to the theoretical equivalents. þ (1/Rn)] Two resistors: Req ¼ R1R2/ (R1 þ R2) Leq ¼ 1/[(1/L1) þ (1/L2) þ . . .Capacitors and inductors ● 191 Factors affecting inductance: L¼ N 2 Am l ● Winding resistance (Rw): The internal resistance in the wire of an inductor. . þ (1/Ln)] Two inductors: Req ¼ L1L2/ (L1 þ L2) Short-circuit equivalent Elements in DC Open-circuit equivalent Experiment 6: Capacitors Objectives ● ● ● ● Understand the characteristics of a capacitor through experiment. circuit behaviour and performance. . Verify the equations of capacitors in series and parallel through experiment. Analyse experimental data. . þ Rn Capacitor iC ¼ C(dnC/dt) WC ¼ ½(Cn ) 2 Inductor nL ¼ L(di/dt) WL ¼ ½(Li2) Leq ¼ L1 þ L2 þ . . þ Cn Parallel Req ¼ 1/[(1/R1) þ (1/R2) þ . Apply the voltage divider rule in a capacitive circuit. þ Ln Ceq ¼ 1/[(1/C1) þ (1/C2) þ . . Background information ● Parallel equivalent capacitance: Ceq ¼ C1 þ C2 þ Á Á Á þ Cn Series equivalent capacitance: 1 Ceq ¼ ð1=C1 Þ þ ð1=C2 Þ þ Á Á Á þ ð1=Cn Þ ● . capacitor and inductor Characteristic Resistor Ohm’s law Energy Series V ¼ IR W ¼ pt or dw ¼ pdt Req ¼ R1 þ R2 þ . . . . The characteristics of the resistor. . þ (1/Cn)] Two capacitors: Ceq ¼ C1C2/ (C1 þ C2) Ceq ¼ C1 þ C2 þ . . Record their nominal values in Table L6.) Table L6. Capacitors: ● 30 and 470 mF electrolytic capacitors each ● Four non-electrolytic capacitors with any values Note: Electrolytic capacitors are polarized.192 Understandable electric circuits When n ¼ 2: Ceq ¼ C1 C2 C 1 þ C2 ● Voltage divider rule for capacitive circuit: Ceq Ceq VC 1 ¼ E . Connect the positive lead of the capacitor to the positive terminal of the DC power supply. (A capacitor can hold its stored charges for days or weeks and can shock you even when it is not connected to a circuit. short circuit each capacitor with a bit of wire to release or discharge the stored charges on the capacitor.1 Capacitor Nominal value Measured value C1 C2 C3 C4 . Non-electrolytic capacitors are non-polarized. VC 2 ¼ E C1 C2 Recall the voltage divider rule for resistive circuit: R1 R2 VR2 ¼ E VR 1 ¼ E Req Req Equipment and components ● ● ● ● ● Multimeter Breadboard DC power supply Z meter or LCZ meter (or any other measuring instruments that can be used to measure capacitance): ● Z meter: A measuring instrument that can be used to measure the values of capacitors.1. Procedure Part I: Capacitors in series and parallel 1. and negative lead to the negative terminal of the DC power supply. Take four non-electrolytic capacitors. so they can be connected either way in a circuit. ● LCZ meter: A measuring instrument that can be used to measure the values of capacitors and inductors. 2. Record the values in Table L6.1. explain the reasons. Compare the measured values and calculated values. Measure the value of each capacitor using a Z meter or LCZ meter and record in Table L6. .3.2 Equivalent capacitance Formula for calculations Calculated value Measured value Ceq for Figure L6. Calculate the equivalent capacitance for each circuit.3 Capacitor Nominal value Measured value C1 30 mF C2 470 mF 2.2 on the breadboard. Take two capacitors with the value shown in Table L6.2. Part II: Apply voltage divider rule in the capacitive circuit 1.1(b) Ceq for Figure L6. short circuit each capacitor with a bit of wire to release or discharge the stored charges on the capacitor.1(a) Ceq for Figure L6.1(c) 4. 5. Connect each circuit as shown in Figure L6.1 Capacitors in series and parallel Table L6. b and c) on the breadboard.Capacitors and inductors 193 2. Connect a series capacitive circuit as shown in Figure L6. Measure the equivalent capacitance for each circuit in Figure L6. C1 C2 C3 C4 C1 C2 C3 C4 C1 C2 C3 C4 (a) (b) (c) Figure L6. Record the values in Table L6. 3.1(a. Table L6.1. are there any significant differences? If so. 2 using the multimeter (voltmeter function).4. Note: Take down the measurement quickly. explain the reasons.194 Understandable electric circuits E=6V C1 = 30 μF C2 = 470 μF VC1 VC2 Figure L6.4. otherwise the capacitor will discharge gradually. are there any significant differences? If so.2 using the voltage divider rule. Conclusion Write your conclusions below: . Record the values in Table L6. Record the values in Table L6. Compare the measured values and calculated values. Calculate VC 1 and VC2 in the circuit of Figure L6. Table L6.2 Capacitor in series 3. 5. Measure the voltage across each capacitor from Figure L6.4 VC1 Formula for calculations Calculated value Measured value VC2 4. . capacitor C and inductor L. These circuits exhibit the important behaviours that are fundamental to much of analogue electronics.Chapter 7 Transient analysis of circuits Objectives After completing this chapter. The circuits in this chapter will combine the resistor(s) R with an energy storage element capacitor C or an inductor L to form an RL (resistor–inductor) or RC (resistor–capacitor) circuit.1 Transient response 7. and they are used very often in electric and electronic circuits. Analysis RL or RC circuits still use Kirchhoff’s current law (KCL) and Kirchhoff’s voltage law (KVL). the resistor R. differential equations (the equations with the derivative) will be used to describe RC and RL circuits.1 The first-order circuit and its transient response There are three basic elements in an electric circuit. The main difference between these types of circuits and pure resistor circuits is that the pure resistor circuits can be analysed by algebraic methods.1. Since the relationship of voltages and currents in the capacitor and inductor circuits is expressed by the derivative. you will be able to: ● ● ● ● ● ● ● ● ● understand the first-order circuits and concepts of the step response and source-free response of the circuits understand the initial conditions in the switching circuit understand the concepts of the transient and steady states of RL and RC circuits determine the charging/discharging process in an RC circuit determine the energy storing/releasing process in an RL circuit understand the concepts of time constants for RL and RC circuits plot the voltages and currents verse time curves for RL and RC circuits understand the relationship between the time constant and the charging/ discharging in an RC circuit understand the relationship between the time constant and the energy storing/releasing in an RL circuit 7. and a single energy storage element (L or C). That means the dynamic state of the circuit has been changed. and the energy storing/releasing of an inductor is in the magnetic field. etc. And the step response for an RC or RL circuit is the circuit responses . it is the result of the operation of a switch in the circuit after a certain time interval. and the other is the source-free response. and only one single energy storage element (inductor or capacitor). First-order circuit ● ● The circuit that contains resistor(s).2 Circuit responses A response is the effect of an output resulting from an input. The steady-state is an equilibrium condition that occurs in a circuit when all transients have finished. the transient and steady states. We have discussed the concept of charging/discharging behaviour of the energy storage element capacitor C in chapter 6. It is the stable-circuit state when all the physical quantities in the circuit have stopped changing. or the circuits that include resistor(s). current. The transient state is the dynamic state that occurs by a sudden change of voltage. The step response for a general system states that the time behaviour of the outputs when its inputs change from zero to unity value (1) in a very short time. 7.1. such as the process of charging/discharging a capacitor or energy storing/releasing for an inductor as the result of the operation of a switch.196 Understandable electric circuits RL or RC circuits that are described by the first-order differential equations. Steady state: An equilibrium condition that occurs when all physical quantities have stopped changing and all transients have finished. For the process of charging/discharging a capacitor or energy storing/releasing for an inductor. The difference is that charging/discharging of a capacitor is in the electric filed. RL or RC circuits that are described by the first-order differential equations. one is called the step response. in a circuit. The first-order RL or RC circuit has two responses. There are two types of circuit states in RL or RC circuit. are called the first-order circuits. Transient state: The dynamic state that occurs when the physical quantities have been changed suddenly. Another energy storage element inductor L also has the similarly energy storing/releasing behaviour. Following are some of the basic terms for a step response: ● ● ● The initial state: the state when an energy storage element hasn’t stored energy yet. The source-free response or natural response is opposite to the step response. Source-free (or natural) response The circuit response when the input (DC power source) is zero. Output (response): the resultant current and voltage. at the moment of time before the power turns on. And the response v(t) or output voltage for this condition is obviously also zero. and the initial condition of the capacitor or inductor is not zero (the energy has been stored to the capacitor or inductor). Input (excitation): the power supply. Step response The circuit response when the initial condition of the energy store elements (L or C) is zero. The source-free response can also be analogized as the process to release water in a reservoir or a water bottle. It is when a DC source voltage is instantly applied to the circuit. and the initial condition of the energy store elements (L or C) is not zero. since it ‘steps’ from zero to a unit constant value (1). the response v(t) will . So the step response can be also called the unit-step response. i. It is the discharging or energy releasing process of the capacitor or inductor in an RC or RL circuit.e. i. and the input (DC power source) is not zero in a very short time. The unit-step response is defined as follows: All initial conditions of the circuit are zero at time less than zero (t 5 0). When an RC or RL circuit that is initially at ‘rest’ with zero initial condition and a DC voltage source is switched on to this circuit instantly. the discharging/releasing process of the C or L.Transient analysis of circuits 197 (outputs) when the initial state of the energy store elements L or C is zero and the input (DC power source) is not zero in a very short time.e. this DC voltage source can be analogized as a unit-step function. It is the circuit response when the input is zero. The step response can be analogized as a process to fill up water in a reservoir or a water bottle. the charging/storing process of the C or L. After the power turns on (t 4 0).e. the energy store elements L or C hasn’t stored energy yet and the output current or voltage generated in this first-order circuit. i. Or the charging process of the energy storing process of the capacitor or inductor. t > 0 1 2 E=1V 3 C 1 t (b) R 0 (a) Figure 7. . vC (0þ) is the capacitor voltage at the instant time after the switch is closed.1 Step function This unit-step function can be expressed as the switch in the circuit of Figure 7. the non-zero initial capacitor voltage and inductor current can be expressed as follows: vC ð0þ Þ ¼ vC ð0À Þ and And ● iL ð0þÞ ¼ iL ð0ÀÞ ● ● ● vC (07) is the capacitor voltage at the instant time before the switch is closed. the capacitor voltage and inductor current will not change instantly. the switch turns to position 1. t ¼ 0þ is the instant time interval after switching the circuit (turn on the switch).1.1(b). The instantly turned on or turned off the switch.1(a). iL (0þ) is the inductor current at the instant time after the switch is closed. ● At this switching moment.3 The initial condition of the dynamic circuit The process of charging and discharging of a capacitor needs a switch to connect or disconnect to the DC source in the RC circuit. as shown in the circuit of Figure 7. 7. or the source input that is switched ‘on’ or ‘off’ in an RC or RL circuit is called the switching circuit. a DC power source is connected to the RC circuit. t < 0 vðtÞ ¼ 1.1(a). vL (07) is the inductor current at the instant time before the switch is closed. this concept can be described as t ¼ 07 and t ¼ 0þ. when t ¼ 0. and produces a unit-step response to the circuit.198 Understandable electric circuits be a constant unit value 1. ● t ¼ 07 is the instant time interval before switching the circuit (turn off the switch). 0. as shown in the following mathematical expression and also can be illustrated in Figure 7. At the moment when the circuit is suddenly switched. All important concepts of step response (charging) or source-free response (discharging) and transient and steady state of an RC circuit can be analysed by this simple circuit. vC and iL do not change instantly: vC ð0þÞ ¼ vC ð0ÀÞ and iL ð0þÞ ¼ iL ð0ÀÞ: 7. At the instant time before/after the switch is closed.2 The step response of an RC circuit Chapter 6 has discussed the charging and discharging process of a capacitor. t ¼ 0þ: the instant time interval after the switch is closed. 7.3? . and the DC power source (E) is connected to the RC series circuit as shown in the circuit of Figure 7.2. t ¼ 07: the instant time interval before the switch is closed.Transient analysis of circuits 199 Initial conditions ● ● ● ● Switching circuit: the instantly turned on or turned off switch in the circuit. or release the stored electric charges instantly. Such a circuit is generally referred to as an RC circuit.1 The charging process of an RC circuit Assuming the capacitor has not been charged yet in the circuit of Figure 7. But there is always a small amount of resistance in the practical capacitive circuits. a pure capacitive circuit will fill with electric charges instantly.2 An RC circuit What will happen when the switch is turned to position 1. When there are no resistors in the circuit. 1 2 E 3 R C Figure 7. Sometimes a resistor will be connected to a capacitive circuit that is used very often in the different applications of the electronic circuits.2. Figure 7.2 is a resistor–capacitor series circuit that has a switch connecting to the DC power supply. the switch is in position 2 (middle). and at this time. The capacitor voltage v is zero at the beginning.4. For this . This RC circuit is similar to a reservoir (or a water bottle) filling with water to capacity. the current and voltage in the circuit will not change any more. After this charging time interval or the transient state of the RC circuit. The voltage across the capacitor vC is not instantaneously equal to the source voltage E when the switch is closed to 1. Since there is a resistor in this circuit. i. the process of the capacitor’s charging will not finish instantly. this is shown Figure 7.4(b) indicates that capacitor voltage vC increases exponentially from zero to its final value (E). It needs time to overcome the resistance R of the circuit to gradually charge to the source voltage E.4(a). the reservoir will need more time to fill up with the water (or the water bottle will need more time to fill up with water if the tab didn’t fully open). vC ¼ E. The circuit current is zero.4 The charging process of an RC circuit Figure 7. 1 3 E R E vC= E 0 t vC (a) (b) Figure 7. the capacitor can be fully charged.e.e. And the capacitor will reach a state of dynamic equilibrium (steady state). If the door of the reservoir opened only to a certain width. the capacitor will gradually store the electric charges. Once they reach the steady state. the circuit current stops flowing.200 Understandable electric circuits 1 i + vR R C − + vC − E 3 Figure 7. and the capacitor is equivalent to an open circuit as shown in the circuit of Figure 7. i. The voltage across the capacitor will be increased until it reaches the source voltage (E). at that time no more charges will flow onto the plates of the capacitor.3 RC charging circuit The energy storing element capacitor C will start charging. the capacitor voltage is equal to the source voltage. The state of the circuit voltage or current after charging is called the steady state. section 1. Applying KVL to this circuit will result in vR þ vC ¼ E ð7:1Þ The voltage drop across the resistor is Ri (Ohm’s law) while the current through this circuit is i ¼ CðdvC =dtÞ (from chapter 6.1) yields RC ● dvC þ vC ¼ E dt ð7:3Þ Determine the capacitor voltage vC Note: If you haven’t learned calculus. The phenomena of the capacitor voltage vC increases dexponentially from zero to its final value E (or a charging process) in an RC circuit can be also analysed by the quantity analysis method as follows. there is no voltage drop across the resistor. and skip the following mathematical derivation process. vR ¼ RC dvC dt ð7:2Þ Substituting (7.3.4) is the equation for the capacitor voltage vC during the discharging process in an RC circuit. the current stops to flow.2 Quantity analysis for the charging process of the RC circuit The polarities of the capacitor and resistor voltages of the RC circuit are shown in Figure 7.2. dvC vR ¼ Ri i ¼ C dt Therefore. then just keep in mind that (7. Therefore. i. The first-order differentia (7.9). 7.e.Transient analysis of circuits 201 open circuit.2) into (7.3) can be rearranged as vC À E ¼ ÀRC dvC dt Divide both sides by 7RC À 1 dvC ðvC À EÞ ¼ RC dt rearrange À dt dvC ¼ RC vC À E . 202 Understandable electric circuits Integrating the above equation on both sides yields vC t ð ð 1 dvC À dt ¼ vC À E RC 0 0 . t . t . v . ¼ ln. vC À Ej0C RC 0 rearrange À t ¼ lnjvC À Ej À lnjÀEj RC . . . vC À E. . . ¼À t ln. ÀE . ● Determine the resistor voltage vR Applying KVL in the circuit of Figure 7.5) yields vR ¼ E À Eð1 À eÀt=RC Þ Therefore.3 v R þ vC ¼ E rearrange v R ¼ E À vC ð7:5Þ Substituting the capacitor voltage vC ¼ Eð1 À eÀt=RC Þ into (7. RC À Taking the natural exponent (e) on both sides results in elnjðvc ÀEÞ=ðÀEÞj ¼ eÀt=RC vC À E ¼ eÀt=RC ÀE Solve for vC vC ¼ Eð1 À eÀt=RC Þ ð7:4Þ The above equation is the capacitor voltage during the charging process in an RC circuit. the resistor voltage is vR ¼ EeÀt=RC . 5 mF) ¼ 6. Solution: RC ¼ (2. vR and i versus time can be plotted as in Figure 7. the curves of vC.1: For the circuit shown in Figure 7. the resistor voltage and the charging current decay exponentially from initial value E and E/R (or Imax) to zero.5 vC. Determine the resistor voltage vR and capacitor voltage vC. respectively. the charging time t ¼ 37.5 mF.Transient analysis of circuits ● 203 Determine the charging current i Dividing both sides of the equation vR ¼ EeÀt=RC by R yields vR E Àt=RC ¼ e R R Applying Ohm’s law to the left side of the above equation will result in the charging current i i¼ E Àt=RC e R Charging equations for an RC circuit ● ● ● Capacitor voltage: vC ¼ Eð1 À eÀt=RC Þ Resistor voltage: vR ¼ EeÀt=RC Charging current: i ¼ E eÀt=RC R Mathematically. vC E E vR E R i 0 t 0 t 0 t Figure 7. vR and i versus t Example 7. And t is the charging time in the equations.5. these three equations indicate that capacitor voltage increases exponentially from initial value zero to the final value E. R ¼2. According to the above mathematical equations.5 kO and C ¼ 2. if E ¼ 25 V.3.25 ms ● vR ¼ EeÀt=RC ¼ ð25 VÞðeðÀ37:5=6:25Þms Þ ¼ ð25 VÞðeÀ6 Þ % 0:062 V .5 ms.5 kO)(2. so it will need some time to release all the water. 7. the capacitor voltage vC will be 0.3 The source-free response of the RC circuit 7. It needs some time to overcome the resistance and eventually release all the charges from the capacitor. the discharging curve is shown in Figure 7. The voltage across the capacitor is vC ¼ E. the capacitor will start discharging.7. Once the capacitor has finished the discharge.3.4(a). This is similar to a reservoir that has an opened door to release the water (or the water bottle has an opened lid to pour water).204 ● Understandable electric circuits vC ¼ Eð1 À eÀt=RC Þ ¼ ð25 VÞð1 À eðÀ37:5=6:25Þms Þ ¼ ð25 VÞð1 À eÀ6 Þ ¼ 24:938 V These results can be checked by using KVL: vR þ vC ¼ E.6.6 Discharging process of the RC circuit Once the switch turns to position 3 as shown in the circuit of Figure 7. The discharging time will increase since there is a resistor in the circuit. but now it will be different than a pure capacitive circuit that can discharge instantly.1 The discharging process of the RC circuit Consider a capacitor C that has initially charged to a certain voltage value v0 (such as the DC source voltage E) through the charging process of the last section in the circuit of Figure 7.6(b). the sum of the capacitor voltage and resistor voltage must be equal to the source voltage in the RC circuit. . 1 2 E 3 C R 1 3 − vR + + i vC = E E vC = E − (a) (b) Figure 7. Substituting the values into KVL yields v R þ vC ¼ E ð0:062 þ 24:938ÞV ¼ 25 V ðcheckedÞ Thus. But the releasing door of a reservoir is not open wide enough. whose function will be the same as a voltage source in the right loop of the RC circuit in Figure 7. 7 Discharge curve of the RC circuit 7.6) dvC ÀRC ¼ vC dt Divide both sides of the above equation by 7RC dvC 1 vC ¼À RC dt ● ð7:8Þ Determine the capacitor voltage vC Note: If you haven’t learned calculus.3. and skip the following mathematical derivation process. Substitute (7. resistor voltage vR and discharging current i of the capacitor discharging circuit can be determined by the following mathematical analysis method. then just keep in mind that (7.6(b) will result in vR À vC ¼ 0 or Since vR ¼ iR and i ¼ ÀC dvC dt vR ¼ v C ð7:6Þ Substitute i into the equation of vR vR ¼ ÀRC dvC dt ð7:7Þ The negative sign (7) in the above equation is because the current i and voltage vC in the circuit of Figure 7.2 Quantity analysis of the RC discharging process The equations used to calculate the capacitor voltage vC. Integrating (7.8) on both sides yields ð ð dvC 1 dt ¼À RC vC .Transient analysis of circuits vC E 205 0 t Figure 7.6(b) have opposite polarities. Applying KVL to the circuit in Figure 7.10) is the equation for the capacitor voltage vC during the charging process in an RC circuit.7) into the left-hand side of (7. 206 Understandable electric circuits ln jvC j ¼ À 1 t þ ln A RC (ln A – the constant of the integration) or ln jvC j À ln A ¼ À 1 t RC . v . t . C. Rearrange ln . . vC ð0 þ Þ ¼ vC ð0 À Þ or vC ¼ V0 When t ¼ 0.6(b).3). ¼ À ðRCÞ A Taking the natural exponent (e) on both sides of the above equation: eln jvC =Aj ¼ eÀt=RC Therefore. Immediately before/after the switch is closed to the position 3 in the circuit of Figure 7.9) yields V0 ¼ AeÀ0=RC That is V0 ¼ A .1. such as the source voltage E.6(b). therefore. the initial condition (initial value) of the capacitor voltage should be vC ð0ÀÞ ¼ V0 v0 can be any initial voltage value for the capacitor. vC does not change instantly (from the section 7. vC ¼ eÀt=RC A or vC ¼ AeÀt=RC ð7:9Þ As the capacitor has been charged to an initial voltage value v0 before being connected to the circuit in Figure 7. substituting vC ¼ v0 in (7. 9) vC ¼ V0 eÀt=RC 207 ð7:10Þ This is the equation of the capacitor voltage for the RC discharging circuit. The curves of vC. The discharging voltage and current decay exponentially from the initial value to zero.8. t is the discharging time. and eventually the energy stored in the capacitor will be released to the circuit completely.6) vR ¼ v C Substitute (7. this means that the capacitor gradually releases the stored energy.10) into (7. and it will be received by the resistor and convert to heat energy.6) yields vR ¼ V0 eÀt=RC ð7:11Þ ● Determine the discharge current i Since i¼ vR R ðOhm s lawÞ ‘ Substitute (7. . ● Determine the resistor voltage vR According to (7. These three equations mathematically indicate that capacitor voltage vC and the resistor voltage vR decay exponentially from initial value V0 to the final value zero. and the discharging current i decays exponentially from the initial value v0/R (or Imax) to the final value zero.Transient analysis of circuits Substitute v0 ¼ A into (7.11) into the above Ohm’s law will result in i¼ V0 Àt=RC e R Discharging equations for an RC circuit ● ● ● Capacitor voltage: vC ¼ V0 eÀt=RC Resistor voltage: vR ¼ V0 eÀt=RC Discharging current: i ¼ V0 eÀt=RC R In the above equations. V0 is the initial capacitor voltage. vR and i versus time t can be illustrated as shown in Figure 7. lesser the vC variation. vR and i versus t 7.3. The time rate of this process depends on the values of circuit capacitance C and resistance R. The variation of the R and C will affect rate of the charging and discharging.3 RC time constant t In an RC circuit.e.8% of its initial value.e. The higher the R and C values (or when the time constant t increases). RC time constant t ¼ RC Quantity Resistance Capacitance Time constant Quantity symbol R C t Unit Ohm Farad Second Unit symbol O F s The time constant t represents the time the capacitor voltage reaches (increases) to 63.2% of its final value (steady state) or the time the capacitor voltage decays (decreases) below to 36.8 The curves of vC. Generally speaking.208 Understandable electric circuits vC. The product of the R and C is called the RC time constant and it can be expressed as a Greek letter t (tau). This can be shown in Figure 7.9. longer the time to reach the final or initial values. i. vR V0 i V0 R 0 t 0 t Figure 7. the time constant is the time interval required for a system or circuit to change from one state to another. . the charging and discharging is a gradual process that needs some time. t ¼ RC. the longer the charging or discharging time. the time required from the transient to the steady state or to charge or discharge in an RC circuit. i. when v0 ¼ E ¼ 100 V.4 The RC time constant and charging/discharging The capacitor charging/discharging voltages when the time is 1 t and 2 t can be determined from the equations of the capacitor voltage in the RC charging/ discharging circuit. (a) Charging and (b) discharging 7.Transient analysis of circuits vC E E τ Increases τ Increases 0 (a) t 0 (b) 209 vC t Figure 7.3. ● at t ¼ 1 t ● Capacitor charging voltage: vC ¼ Eð1 À eÀt= Þ ¼ 100 Vð1 À eÀ1 = Þ % 63:2 V ● Capacitor discharging voltage: vC ¼ V0 eÀt= ¼ 100 V eÀ1 = % 36:8 V ● at t ¼ 2 t ● Capacitor charging voltage: vC ¼ Eð1 À eÀt= Þ ¼ 100 Vð1 À eÀ2 = Þ % 86:5 V ● Capacitor discharging voltage: vC ¼ V0 eÀt= ¼ 100 V eÀ2 = % 13:5 V . That is vC ¼ Eð1 À eÀt= Þ and vC ¼ V0 eÀt= For example.9 The effect of the time constant t to vC. 5% 5% 0 1. when time has passed 4 to 5 t.5% of E 5% of E 1.10 mean that when the time constant is 1 t. If the final and initial value are 100 V.2% of the final value and discharges to 36.8% of the initial value.2 V and discharge to 36. and discharge to 0.3% of E vC E Capacitor discharging voltage: vC ¼ V0 eÀt= 36.5% of the final value. Time constant t for and charging/discharging ● When t ¼ 1 t: the capacitor charges to 63. According to this sequence.5% of the initial value.1 and Figure 7.2% of E 99.2% of E 86.1 and graphs in Figure 7.67% 1τ 2τ 3τ 4τ 5τ t 0 1τ 2τ 3τ 4τ 5τ t Figure 7.8% 0.8% of E 0. These results are summarized in Table 7.1 The capacitor charging/discharging voltages Charging/discharging time 1t 2t 3t 4t 5t vC E 63.2% Capacitor charging voltage: vC ¼ Eð1 À eÀt= Þ 63. or discharge approaching to zero. it will charge to 63.2% 86. When the time is 5 t. and discharge to 36. the capacitor will charge to 99. the capacitor will charge to 63.5% 95% 98. The data in Table 7.210 Understandable electric circuits Table 7.8% 13. if the time constant is 5 t.67% of E 99. the capacitor charging/discharging voltages can be determined when the time is 3. the circuit will reach the steady state. which means that the capacitor will charge approaching to the source voltage E.0% of E 98. and discharge to 13.10 The charging/discharging curves of the capacitor voltage Using the same method as above.67% of the initial value.10.8% of the initial value (the initial capacitor voltage). Therefore.3% of the final value. charging/discharging of the capacitor will be almost finished.8% of E 13. 4 and 5 t. the capacitor will charge to 86.2% of the final value (source voltage).8 V.5% of E 95.3% 36. the transient state of RC circuit will be finished and enter the steady state of the circuit. When the time constant is 2 t. . After 5 t. 3% of the final value and discharges to 0. Therefore. The capacitor voltage discharging to 5 V is 5% of the initial value E (100 V).11(a) is a resistor and inductor series circuit.3: In an RC circuit. the resistance is 10 kO.11. All important concepts of the magnetic storing/releasing or transient and steady state of RL circuit can be analysed by a simple circuit as shown in Figure 7. After understanding the RC circuit.005 mF) ¼ 50 ms. and an RL circuit stores the energy in the magnetic field.4 The step response of an RL circuit Figure 7. and the capacitance is 0.005 mF. C ¼ 0:005 mF. therefore. We will use the term charging/ discharging in an RC circuit. Example 7. In how much time can the capacitor voltage be discharged to 5 V after the switch is turned to position 3? Solution: E ¼ 100 V.6(a). 5 ¼ 1s or ¼ 1 ¼ 0:2 s 5 ð0:2 sÞ ¼ ¼ 40 ms R ð5k OÞ . An RC circuit stores the charges in the electric field. Solution: The transient state in the RC circuit will last 5 t. the capacitor discharging time is t ¼ 3 ¼ 3ð50 msÞ ¼ 150 ms Example 7. Table 7. R ¼ 10 kO .67% of the initial value. the source voltage is 100 V.2: In the circuit of Figure 7. ¼ RC therefore C ¼ 7. t¼? The time constant t for discharging is: t ¼ RC¼ (10 kO)(0. its method of analysis can be used to analyse the RL circuit in a similar fashion. R ¼ 5 kO. the transient state has last 1 s in this circuit. The step response (storing) and source-free response (releasing) of an RL circuit is similar to the step response and source-free response of an RC circuit. . Determine the capacitance C. Such a circuit is generally referred to as an RL circuit. and the term energy storing/releasing in an RL circuit. it runs through a switch connecting to the DC power supply.Transient analysis of circuits 211 ● When t ¼ 5 t: the capacitor charges to 99.1 and Figure 7.10 indicate that the time capacitor discharges to 5% of the initial value is 3 t. 11(a) turns to position 1. The inductor L absorbs the electric energy from the DC source and converts it to magnetic energy. iL Imax 0 5τ t Figure 7. Since there is a resistor R in the circuit of Figure 7.11(b).4. the current needs time to overcome the resistance in this RL circuit. 7.11(a) is a circuit that can be used to analyse RL step response and source-free response.11 RL circuit Figure 7. and will produce the induced voltage VL. and the DC power source is connected to the RL series circuit as shown in the circuit of Figure 7. the switch is in position 2.12. when the switch in Figure 7. the current will flow through this RL circuit.11(a). This energy storing process of the inductor in an RL circuit is similar to the electron charging process of the capacitor in an RC circuit.12 Current versus time curve in the RL circuit The phenomenon of the inductor current iL in an RL circuit increases exponentially from zero to its final value (Imax) or from the transient to the steady state can also be analysed by the quantitative analysis method below. The current iL in the RL circuit will reach the final value (maximum value) after a time interval. Therefore. it will be different as a pure inductor circuit that can store energy instantly. . the process of the inductor’s energy storing will not finish instantly. the electromagnetic field will be built up in the inductor L.212 Understandable electric circuits 1 2 E 3 L E 3 iL R 1 + VR − + VL − (a) (b) Figure 7. After the switch is turned to position 1. as shown in Figure 7. assuming the energy has not been stored in the inductor yet.11(b)? As it has been mentioned in chapter 6. What will happen when the switch is turned to position 1.1 Energy storing process of the RL circuit In the circuit of Figure 7. 3.4) and vR ¼ Ri.4.Transient analysis of circuits 213 7. where i ¼ iL. ● Determine the current iL iL ¼ E ð1 À eÀt=ðL=RÞ Þ R E ¼ ð1 À eÀt= Þ R ¼ Imax ð1 À eÀt= Þ ð7:13Þ The time constant of RL circuit is ¼ L R The final value for the current is Imax ¼ ● E R Determine the resistor voltage vR Applying Ohm’s law vR ¼ Ri ð7:14Þ Keeping in mind that i ¼ iL and substituting i by the current iL in (7. Applying KVL to this circuit will result in vL þ vR ¼ E ð7:12Þ Substituting vL ¼ L(di/dt) (from chapter 6. section 6.2 Quantitative analysis of the energy storing process in an RL circuit The polarities of the inductor and resistor voltages of an RL circuit are shown in the circuit of Figure 7.2 will yield the equation of the current in RL circuit during the process of energy storing as given in the following sections.11(b).12) yields L diL þ RiL ¼ E dt Applying a similar analysis method for the RC charging circuit in section 7.14) yields E vR ¼ R ð1 À eÀt= Þ R ¼ Eð1 À eÀt= Þ . with (7. . iL E R vR vL E E 0 t 0 t 0 t Figure 7. the curves of iL. These three equations mathematically indicate that circuit current and resistor voltage increase exponentially from initial value zero to the final value E/R and E.13 Curves of iL.12) v L þ vR ¼ E Substitute vR and solving for vL v L ¼ E À vR ¼ E À Eð1 À eÀt= Þ ¼ EeÀt= Energy storing equations for an RL circuit ● ● ● Circuit current: iL ¼ E ð1 À eÀt= Þ R Resistor voltage: vR ¼ Eð1 À eÀt= Þ Inductor voltage: vL ¼ EeÀt= In the above equations.13. According to the above mathematical equations. t is the energy storing time. respectively. and t ¼ L/R is the time constant of the RL circuit.11(b). the inductor voltage decays exponentially from initial value E to zero.4: The resistor voltage vR ¼ 10(1 7 e72t)V and circuit current iL ¼ 2(17e72t) A in an RL circuit is shown in the circuit of Figure 7. vR and vL versus time Example 7. vR and vL versus time can be illustrated in Figure 7. Determine the time constant t and inductance L in this circuit.214 Understandable electric circuits The final value for the resistor voltage is E ¼ Imax R ● Determine the inductor voltage vL According to (7. 1 Energy releasing process of an RL circuit Consider an inductor L that has initially stored energy and has the induced voltage vL through the energy storing process of the last section. R ¼ ¼ ¼ 5O R I 2A The time constant ¼ L R Solve for L L ¼ R ¼ ð5 OÞð0:5 sÞ ¼ 2:5 H 7.5 Source-free response of an RL circuit 7.14 RL circuit .Transient analysis of circuits Solution: The given resistor voltage vR ¼ Eð1 À eÀt= Þ ¼ 10ð1 À eÀ2t ÞV with E ¼ 10 V and or ¼ 1 s ¼ 0:5 s 2 E ð1 À eÀt= Þ ¼ 2ð1 À eÀ2t Þ A R À t ¼ À2t 215 The given current iL ¼ with E E 10 V ¼ 2 A E ¼ 10 V. 1 2 E 3 vR R + 1 3 L vL= E − + vR R iL − + E vL − (a) (b) Figure 7. If the switch turns to position 3 at this moment (Figure 7. the inductor voltage vL has a function just like a voltage source in the right loop of this RL circuit.14(b)).5. the current in the circuit will take time to decay from the stored initial value to zero.18) is the equation for the current in the RL circuit during the energy releasing. This means the inductor releases the energy gradually.16) diL RiL ¼À dt L .2 Quantity analysis of the energy release process of an RL circuit The equations to calculate the inductor voltage vL.216 Understandable electric circuits Without connecting the resistor R in this circuit. then just keep in mind that (7. This might produce a spark on the switch and damage the circuit components. the resistance in the circuit will increase the time required for releasing energy. Divide L on both sides in (7. But if there is a resistor R in the circuit.15) yields L ● diL ¼ ÀRiL dt ð7:16Þ Determine the circuit current Note: If you haven’t learned calculus. the inductor will release the stored energy immediately. and the resistor absorbs the energy and converts it to heat energy. resistor voltage vR and circuit current iL of the RL energy releasing circuit can be determined by the following mathematical analysis method. The current iL curve of the energy release process in the RL circuit is illustrated in Figure 7. Applying KVL to the circuit in Figure 7. at the instant when the switch turns to position 3.15 Energy release curve of the RL circuit 7. iL E R 0 t Figure 7.15.5.14(b) will result in vL þ vR ¼ 0 or vL ¼ ÀvR ð7:15Þ Substituting vL ¼ LðdiL =dtÞ and vR ¼ RiL into (7. and skip the following mathematical derivation process. 15)) results in vL ¼ ÀvR ¼ ÀRI0 eÀt= . t ¼ L/R is the time constant for the RL circuit.18) into vL þ vR ¼ 0 (as in (7. ln jiL j ¼ À t þ ln A iL L L Rearrange: R lnjiL j À ln A ¼ À t L Taking the natural exponent (e) on both sides results in elnjiL =Aj ¼ eðÀR=LÞt Solve for iL iL ¼ AeðÀR=LÞt or iL ¼ eðÀR=LÞt A 217 ð7:17Þ Since energy has been stored in the inductor before it is been connected to the circuit in Figure 7.18) vR ¼ Ri ¼ RðI0 eÀt= Þ ¼ RI0 eÀt= ● Determine the inductor voltage Substituting (7. therefore. ● Determine the resistor voltage Keep in mind that i ¼ iL and apply Ohm’s law to (7. iL does not change. iL ð0þÞ ¼ iL ð0ÀÞ or iL ¼ I0 When t ¼ 0. such as I0 ¼ E/R) Since immediately before/after the switch is closed to position 3.Transient analysis of circuits Integrating the above equation on both sides yields ð ð diL R R ¼À dt. its initial condition or value should be iL ð0ÀÞ ¼ I0 (I0 can be any initial current. substitute iL ¼ I0 into (7.14(b). iL ¼ I0 eÀt= ð7:18Þ i:e: I0 ¼ A In the above equation.17) yields I0 ¼ AeðÀR=LÞÂ0 Therefore. I0 ¼ E/R is the initial current for the inductor and t ¼ L/R is the time constant for the RL circuit. the storing and releasing of energy is a gradual process that needs time. to the final value zero.3 RL time constant t In an RL circuit. I0R and 7I0R. These three equations mathematically indicate that inductor current. The quotient of L and R is called the RL time constant t ¼ L/R. resistor voltage and inductor voltage decay exponentially from initial value I0. The curves of iL.218 Understandable electric circuits Energy releasing equations for an RL circuit ● ● ● Circuit current: iL ¼ I0 eÀt= Resistor voltage: vR ¼ I0 ReÀt= Inductor voltage: vL ¼ ÀI0 ReÀt= In the above equations. RL time constant ¼ L R Quantity symbol R L t Unit Ohm Henry Second Unit symbol O H s Quantity Resistance Inductance Time constant . The RL time constant is the time interval required from the transient to the steady state or the energy storing/releasing time in an RL circuit. vR and vL versus time can be illustrated in Figure 7. t is the energy releasing time.16 Curves of iL.16. The time rate of this process depends on the values of the circuit inductance L and resistance R.5. iL I0 I0 R vR 0 vL t 0 t 0 t I0 R Figure 7. vR and vL versus time 7. The variation of R and L will affect the rate of the energy storing and releasing. respectively. .5% of E/R 95. The higher the value of L. the time of the inductor current decays (decreases) below to 36. 5 t. E R iL I0 iL 0 (τ ↑ ) t 0 (τ ↑ ) t Figure 7.14.8% of the its initial value.67% of I0 63. Solution: The time of a transient state usually lasts 5 .2 and Figure 7.2 Relationship between the time constant and the inductor current RL energy storing/releasing time 1t 2t 3t 4t 5t Increasing the inductor current (storing): Decreasing the inductor current (releasing): iL ¼ E ð1 À eÀt= Þ R iL ¼ I0 eÀt= 36. .17 Effect of the time constant t on iL (L" or R#) 7.8% of I0 0. the lower the R (or when the time constant t increases).0% of E/R 98. according to the equations of iL ¼ ðE=RÞð1 À eÀt= Þ and iL ¼ I0 eÀt= .3% of E/R Example 7.2% of E/R 86. . The time constant ¼ L=R. . Determine the inductance L. the circuit current for an RL circuit can be determined when the time constant is 1 t. the resistance R is 100 O and the transient state has lasted 25 ms.8% of I0 13. ¼ ð25 ms=5Þ ¼ 5 ms. These results are summarized in Table 7..4 The RL time constant and the energy storing and releasing Similar to an RC circuit. This can be shown in Figure 7.17.5: In the RL circuit of Figure 7.2% of E/R 99.5% of I0 5% of I0 1.Transient analysis of circuits 219 The time constant t represents the time the inductor current reaches (increases) to 63.5. and this transient state is 5 ¼ 25 ms.2% of its final value (steady state). the longer the storing or releasing time. the lesser the iL variation and the longer the time to reach the final or initial values. respectively. . .18. 2 t. L ¼ R ¼ ð100 OÞð5 msÞ ¼ 500 mH. Table 7. 18 Relationship of inductor current and time constant Example 7. and a single energy storage element (L or C). Solution: L 40 H ¼ ¼ 20 ms R 2 kO E 1 V ðÀ0:2=20Þms % 0:5 mA e iL ¼ I0 eÀt= ¼ eÀt= ¼ R 2 kO ¼ Summary ● ● ● ● ● First-order circuit: ● The circuit that contains resistor(s).8% 13.5% 5% 0 1τ 2τ 3τ 4τ 5τ t 0 1τ 2τ 3τ 1. the discharging/releasing process of the C or L.14(b). Transient state: The dynamic state that occurs when the physical quantities have been changed suddenly. i.220 iL Understandable electric circuits iL I0 E R 63.e. E ¼ 1 V and t ¼ 0. Determine the circuit current iL in this energy releasing circuit.2 ms. i. ● RL or RC circuits that are described by the first-order differential equations.2% 86.6: In the circuit of Figure 7.8% 4τ 0.3% 36. . L ¼ 40 H.2% 99. and the input is not zero in a very short time. R ¼ 2 kO.67% 5τ t Figure 7. Step response: The circuit response when the initial condition of the L or C is zero. the charging/ storing process of the C or L. Source-free response: The circuit response when the input is zero. and the initial condition of L or C is not zero.5% 95% 98.e. Steady state: An equilibrium condition that occurs when all physical quantities have stopped changing and all transients have finished. 5% 95. vR ¼ EeÀt= ● RC discharging (the source-free response): vC ¼ V0 eÀt= .8% 0. Time vC and iL increasing (charging/storing): vC ¼ Eð1 À eÀt=RC Þ.2% 99. Construct an RC circuit. 222). Analyse the experimental data. iL ¼ I0 eÀt= 36. ¼ RC . and compare them to the theoretical equivalents.8% 13. iL ¼ E ð1 À eÀt= Þ R 63. Analyse and verify the capacitor’s charging/discharging time by experiment. vR ¼ V0 eÀt= ● RC time constant t.Transient analysis of circuits ● 221 ● ● The initial condition: ● t ¼ 07: the instant time before switching ● t ¼ 0þ: the instant time after switching ● vC ð0 þ Þ ¼ vC ð0 À Þ.2% 86.67% vC and iL decaying (discharging/releasing): 1t 2t 3t 4t 5t Experiment 7: The first-order circuit (RC circuit) Objectives ● ● ● ● Understand the capacitor charging/discharging characteristics in the RC circuit (the first-order circuit) by experiment. Background information ● RC charging (the step response): vC ¼ Eð1 À eÀt= Þ. The relationship between the time constants of RC/RL circuits and charging/storing or discharging/releasing: Summary of the first-order circuits (see p. circuit behaviour and performance.3% vC ¼ V0 eÀt=RC .5% 5% 1. vC and iL do not change instantly. collect and evaluate experimental data to verify the capacitor charging/discharging characteristics in an RC circuit.0% 98. iL ð0þÞ ¼ iL ð0ÀÞ ● Immediately before/after the switch is closed. 222 Summary of the first-order circuits Waveforms Time constant t ¼ RC vR E Circuits Equations RC charging (step response) vC E vC ¼ Eð1 À eÀt= Þ. vR ¼ V0 eÀt= .vR i V0 R i¼ t E Àt= e R 0 t 0 t Understandable electric circuits t ¼ RC RC discharging (source-free response) V0 vC ¼ V0 eÀt= . i E R vR ¼ EeÀt= . 0 t 0 i¼ iL vR E V0 Àt= e R E ð1 À eÀt= Þ R E R t RL storing (step response) iL ¼ vL E t ¼ L/R vR ¼ Eð1 À eÀt= Þ 0 iL I0 I0R t 0 vR 0 t 0 vL t t vL ¼ EeÀt= RL releasing (source-free response) Àt= 0 iL ¼ I0 eÀt= t ¼ L/R vR ¼ I0 Re vL ¼ ÀI0 ReÀt= t 0 t − I0R . 0 vC . 1 on the breadboard. etc.).Transient analysis of circuits 223 Equipment and components ● ● ● ● ● ● ● ● Multimeters (two) Breadboard DC power supply Stopwatch Z meter or LCZ meter Switch Resistors: 1 and 100 kO Capacitor: 100 mF electrolytic capacitor Procedure Part I: Charging/discharging process in an RC circuit 1. Table L7.1.2 (such as 0–10 V. 1 2 E = 10 V R = 1 kΩ C = 100 μF V V Figure L7. Wait until voltage across the capacitor reaches to steady state (does not change any more). . and observe the needles of the two multimeters (voltmeter function). Use a jump wire to short circuit the 100 mF capacitor terminals to discharge it. Construct an RC circuit as shown in Figure L7.1 An RC circuit 3. Turn on the switch to position 1 in the circuit of Figure L7.1. then measure the value of capacitor using a Z meter or LCZ meter and record in Table L7. and record the observation of capacitor voltage vC and resistor voltage vR in Table L7.1 Capacitor Nominal value Measured value C 100 mF 2. and record the measured values in Table L7. Table L7.224 Understandable electric circuits Table L7.3 using the multimeter (ohmmeter function). . Use a jump wire to short circuit the 100 mF capacitor terminals to discharge it.1 (such as 10–0 V. 3. Part II: Capacitor’s characteristics in DC circuit 1.3 Component Nominal value Measured value R 100 kO C 100 mF 2. Measure the resistors listed in Table L7. observe the needles of two multimeters (voltmeter function). Turn on the switch to position 2 in the circuit of Figure L7. etc. Record the value in Table L7.2 (use measured R and C values). Wait until voltage across the capacitor vC decreased to 0 V.2 The RC circuit for Part II 4.). Calculate the capacitor charging/discharging voltage vC and vR when t ¼ t. Record the values in Table L7.3. Calculate the time constant t for the circuit in Figure L7. record the observations of capacitor voltage vC and resistor voltage vR in Table L7.2 Switching position Turn on the switch to position 1 Turn on the switch to position 2 vC vR 4.4. 5. 1 2 R =100 kΩ E = 10 V V C = 10 μF V Figure L7.4. then measure the value of capacitor using a Z meter or LCZ meter and record in Table L7.1.3. Construct a circuit as shown in Figure L7.2 on the breadboard. but it still can approximately verify the theory. and record t. Keep the switch at position 1 and make sure it does not change.4 t Charging formula Discharging formula Calculated value for charging Calculated value for discharging Measured value for charging Measured value for discharging vC vR 7. and observe the time required for the capacitor voltage to decrease to 3. Note: Since electrolyte capacitors may conduct leakage current. Also measure vC and vR at this time and record the values in Table L7. and observe capacitor voltage vC using the multimeter (voltmeter function) until the capacitor voltage reaches and stays at 10 V (vC ¼ 10 V).Transient analysis of circuits 225 6. this is the charging time constant t.4. Then turn on the switch to position 2.4. observe time required for capacitor voltage vC charging to 6. Table L7. Conclusion Write your conclusions below: . Also measure vR at this time.6 V using both the multimeter (voltmeter function) and stopwatch (this is the discharging time constant t).3 V using both multimeter (voltmeter function) and stopwatch. the measurement and calculation may be a little different. vC and vR in Table L7. Turn on the switch to position 1. . only their magnitude changes. Before the 19th century. the resulting currents and voltages in AC circuit also periodically switch their polarities and directions.Chapter 8 Fundamentals of AC circuits Objectives After completing this chapter. Similar to DC circuits. peak. Since then. and is still commonly used in current industries. you should be able to: ● ● ● ● ● ● ● ● understand the difference between DC and AC understand the definitions of AC phase shift. phasor. This chapter will discuss the alternating current (AC) circuits. . etc. in which the voltage alternates its polarity and the current alternates its direction periodically. peak to peak. AC gradually showed its advantages and rapidly developed in the latter of the 19th century. That is.1 Introduction to alternating current (AC) 8. businesses and homes throughout the world. RMS values. the polarity of DC voltage and direction of DC current do not change. and capacitive elements in AC circuits 8. understand the relationship of period and frequency understand and define three important components of sinusoidal waveform define the phase difference between sinusoidal voltage and current convert sinusoidal time-domain quantities to phasor-domain forms. DC and AC have had constant competition. Since the AC power supply provides an alternating voltage and current. frequency.1. hence. inductive. an alternating voltage is called AC voltage and alternating current is called AC current. all resulting voltages and currents in DC circuit are constant and do not change with time. the DC power supply was the main form of electrical energy to provide electricity.1 The difference between DC and AC Previous chapters have studied DC (direct current) circuits. and vice versa analyse the sinusoidal AC circuits using phasors study the effect of resistive. The DC power supply provides a constant voltage and current. period. but its polarity or direction does not change with time (always above zero). .228 Understandable electric circuits This is because the AC power can be more cost-effective for long-distance transmission from power plants to industrial. allowing for a wide range of applications. The pulsing DC changes pulse amplitude periodically. V or I 0 t V or I 0 V or I 0 t t Figure 8. Figure 8.2 Pulsing DC waveforms Direct current (DC) ● ● The polarity of DC voltage and direction of DC current do not change.1. V or I 0 t Figure 8. but the polarity does not change. in which the amplitude of DC pulse changes periodically from zero to the maximum with time. This is why power transmission for electricity today is nearly all AC.1. so it still belongs to the DC category.2 DC and AC waveforms The DC voltage and current do not change their polarity or direction over time. It is also easy to convert from AC to DC.2 shows some pulsating DC waveforms. only their magnitude changes. A DC waveform (a graph of voltage and current versus time) is shown in Figure 8. 8. commercial or residential areas.1 DC waveform There is also a type of DC waveform known as the pulsing DC. V or I 0 + _ t V or I 0 V or I 0 t t Figure 8. i. A few examples of AC waveforms are shown in Figure 8. square wave. also belong to AC. triangle wave. although other waveforms such as square wave.) and DC quantities use uppercase letters (E. Sine AC (or AC) varies over time according to sine (or cosine) function. and is the most widely used AC.3 Period and frequency The waveform of a sinusoidal function is shown in Figure 8. AC quantities are represented by lowercase letters (e. v. and is often referred to as AC.4. saw-tooth wave.3. etc. Applying the expression of sine function to electrical quantities will obtain general expressions of AC voltage and current as follows: Sinusoidal voltage: vðtÞ ¼ Vm sinðot þ cÞ Sinusoidal current: iðtÞ ¼ Im sinðot þ cÞ 8. Sine AC voltage and current vary with sine (or we could use cosine by adding 908 to the sine wave) function.Fundamentals of AC circuits 229 AC voltage and current periodically change polarity or direction with time.). the symbol of AC source is . A sine function can be described as a mathematical expression of f ðtÞ ¼ Fm sinðot þ cÞ. The sine AC wave energy is the type of power that is generated by the utility power industries around the world.1. . This is the expression of sine function in the time domain (the quantity versus time). etc.3 AC waveforms The sinusoidal or sine AC wave is the most basic and widely used AC waveform. etc.). I. etc. Alternating current (AC) ● ● The polarity of voltage and direction of AC current periodically change with time (such as sine wave. V. in Figure 8. The frequency is measured in hertz (Hz). or the positive and negative alternations of one revolution.e. For instance.230 Understandable electric circuits f(t) Fm 0 p T 2p wt Figure 8. so it has a frequency of 2 Hz.4 Three important components of a sine function There are three important components in the expression of the sine function f ðtÞ ¼ Fm sinðot þ cÞ: peak value Fm. angular velocity O and phase shift c.4 Sinusoidal waveform ● ● Period T: It is the time to complete one full cycle of the waveform. Fm is the peak value or amplitude of the sine wave (Im for current or Vm for voltage). i. Frequency f: It is the number of cycles of waveforms within 1 s. Frequency f: Number of cycles per second. f ¼ 1=T . ● Peak value Fm: In the expression f ðtÞ ¼ Fm sinðot þ cÞ.5. f ¼ 1=T 8. It is the distance from zero of the horizontal axis to the maximum point . T is measured in seconds (s). the number of complete cycles in 1 s is 2. Period and frequency ● ● ● Period T: Time to complete one full cycle. f(t) 0 t = 1s t(s) Figure 8.5 Frequency of sine waveform ● Relationship of T and f: The frequency f of the waveform is inversely proportional to period T of the waveform.1. so the angular velocity can be determined by o¼ 2p T 1 f ¼ T The relationship between the angular velocity and frequency is 2p ¼ 2pf o¼ T Since the angular velocity is directly proportional to the frequency. It is measured in degrees or radians.6 (a)). ● If phase shift c has a negative value (c 5 0). the distance of one cycle is 2p as shown in Figure 8. it is also called the angular frequency.Fundamentals of AC circuits 231 (positive or negative) that a waveform can reach during its entire cycle (Figure 8. Angular velocity ¼ Rotating distance/Time (Same with the linear motion: Velocity ¼ Distance/Time) Since the time required for a sine wave to complete one cycle is period T. A sine wave may shift to the left or right of 08. the waveform of sine function f ðtÞ ¼ Fm ðot þ cÞ will shift to the left side of 08 as shown in Figure 8. the waveform of sine function f ðtÞ ¼ Fm sinot starts from t ¼ 0 as shown in Figure 8. ● If phase shift c ¼ 0. It is measured in volts or amperes. The range of phase shift is between 7p and þp.6 The peak value and phase of the sine wave ● Angular velocity o (the Greek letter omega): Angular velocity or angular frequency of a sine wave reflects the rate of change of the rotation of the wave. ● Phase c (the Greek letter phi): The phase or phase shift of a sine wave is an angle that represents the position of the wave shifted from a reference point at the vertical axis (08). ● If phase shift c has a positive value (c 4 0).4.6(b).6(a). It is measured in radian per second (rad/s). . f(t) wt wt f(t) Fm 0 f(t) p (a) 2p wt 0 0 ψ (b) ψ (c) Figure 8. the waveform of sine function f ðtÞ ¼ Fm ðot À cÞ will shift to the right side of 08 as shown in Figure 8.6(c). 232 Understandable electric circuits Three important components of sine function f ðtÞ ¼ Fm sinðot þ cÞ ● ● ● ● Fm: Peak value (amplitude) o: Angular velocity or angular frequency o ¼ 2p=T ¼ 2pf (p ¼ 1808) c: Phase or phase shift ● c 4 0: Waveform shifted to the left side of 08 ● c 5 0: Waveform shifted to the right side of 08 Example 8.7.1 8. waveform shifted to the right side of 08) Frequency: f ¼ 1=T Since o ¼ 2p=T and o ¼ 25 rad=s T¼ 2p 2p rad 1 1 ¼ % 0:25s f ¼ ¼ ¼ 4 Hz o 25 rad=s T 0:25 s The waveform is shown in Figure 8.1: Given a sinusoidal voltage vðtÞ ¼ 6sinð25t À 30 ÞV.7 Waveform for Example 8. determine its peak voltage.25 s 2p wt ψ = 30° Figure 8. For instance.1. given the general expressions of sinusoidal voltage and current as vðtÞ ¼ Vm sinðot þ cv Þ and iðtÞ ¼ Im sinðot þ ci Þ . and plot its waveform. phase angle and frequency. the angular displacement of their phases is called phase difference and is denoted by (lowercase Greek letter phi). v(t) Vm = 6 V 0 p T = 0. It is a phase angle by which one wave leads or lags another. Solution: Peak value: Vm ¼ 6 V Phase: c ¼ 7308 (c 5 0.5 Phase difference of the sine function For two different sine waves with the same frequency. or is a right angle (orthos means ‘straight’. or voltage lags current. . as shown in Figure 8.8(a).8(c) Current leads voltage ● If ¼ cv À ci ¼ Æp=2 (or +908). and gonia means ‘angle’).Fundamentals of AC circuits the phase difference between voltage and current is ¼ ðot þ cv Þ À ðot þ ci Þ ¼ cv À ci ● 233 If ¼ cv À ci ¼ 0.8(a) Two waveforms are in phase ● If ¼ cv À ci > 0. It is shown in Figure 8. current leads voltage. v i 0 wt Figure 8.8(b).8(b) Current lags voltage ● If ¼ cv À ci < 0.8(d). then voltage and current are orthogonal. i v 0 i v wt yi yv 0 wt f f Figure 8. voltage leads current. or current lags voltage as shown in Figure 8. the two waveforms are in phase as shown in Figure 8.8(c). v v i 0 i 0 wt yv yi f wt f Figure 8. . (b) ¼ cv À ci ¼ 60 À 20 ¼ 40 > 0 So voltage leads current 408 as shown in Figure 8.8(e).8(d) Voltage and current are orthogonal ● If ¼ cv À ci ¼ Æp (or Æ180 ). (a) vðtÞ ¼ 20 sinðot þ 30 ÞV. voltage and current are out of phase as shown in Figure 8.234 Understandable electric circuits v i wt v i wt 0 0 f= p 2 f=− p 2 Figure 8. iðtÞ ¼ 2:5 sinðot þ 20 ÞA Solution: (a) ¼ cv À ci ¼ 30 À 60 ¼ À30 < 0 So voltage lags current 308 as shown in Figure 8.8(e) Voltage and current are out of phase Phase difference ¼ cv À ci For two waves with the same frequency such as vðtÞ ¼ Vm sinðot þ cv Þ and iðtÞ ¼ Im sinðot þ ci Þ: ● ● ● ● ● If If If If If ¼ 0: v and i are in phase 4 0: v leads i 5 0: v lags i ¼ Æp=2: v and i are orthogonal ¼ Æp: v and i are out of phase Example 8. iðtÞ ¼ 12 sinðot þ 60 ÞA (b) vðtÞ ¼ 5 sinðot þ 60 ÞV.9(b).9(a).2: Determine the phase difference of the following functions and plot their waveforms. v 0 i p 2p wt Figure 8. The different expressions will provide different ways to analyse the sinusoidal AC quantity.10.2 Sinusoidal AC quantity A sinusoidal AC quantity such as AC voltage or current can be described in a number of ways.2.10 Peak and peak–peak value .2(b) 8.Fundamentals of AC circuits v wt i 30° 60° 235 20 30° 0 Figure 8. instantaneous value. The peak value is denoted by Fpk as shown in Figure 8. f(t) Fm = Fpk Fp–p 0 wt Figure 8.9(a) Figure for Example 8. the peak value of the sinusoidal waveform is one of three important components of the sine function.2(a) 5 i 0 40° 20° 60° v wt Figure 8. They can be described by their peak value. 8. peak–peak value.9(b) Figure for Example 8. and it is also because a sinusoidal wave always varies periodically and there is no one single value that can truly describe it. and is the amplitude or maximum value Fm in sine function f ðtÞ ¼ Fm sinðot þ cÞ. average value or root mean square (RMS) value.1 Peak and peak–peak value As previously mentioned. 11. or minimum to maximum peak.236 Understandable electric circuits The peak–peak value Fp–p represents the distance from negative to positive peak.12 Average value . For a sinusoidal function f ðtÞ ¼ Fm sinðot þ cÞ. etc.10. or between peak and trough of the waveform. determine the instantaneous voltage v1 (voltage at 308) and v2 (voltage at 1358) when Vm ¼ 5 V. To determine the maximum values that electrical equipment can withstand.2.3 Average value Because of the symmetry of the sinusoidal waveform.3 Solution: v1 ¼ Vm sin ot ¼ 5 sin30 ¼ 2:5 V v2 ¼ Vm sin ot ¼ 5 sin135 % 3:54 V 8. v(t) v2 v1 0 wt 30° 90° 135° Vm = 5 V Figure 8. so Fp–p ¼ 2Fpk as shown in Figure 8. f(t) Fm 0 π wt Figure 8.13.12 and 8. Example 8. as shown in Figures 8. its average value is defined as the average of its half-cycle (0 to p).2 Instantaneous value The instantaneous value of the sinusoidal waveform f(t) varies with time. current i.11 Figure for Example 8.3: Given a sinusoidal AC voltage vðtÞ ¼ Vm sin ot as shown in Figure 8. 8. Instantaneous values of the variables are denoted by lowercase letters. its average value in a complete full cycle is always zero. such as voltage v. and it is the value at any instant time t (or ot) in any particular point of a waveform.2. the peak values or peak–peak values of the AC quantities should be considered. as shown in Figure 8.13 Average value The average value of a half-cycle sinusoidal wave with a zero phase shift can be derived by using integration as follows: Note: If you haven’t learned calculus. In North America.2. the values measured and displayed on instruments and the nominal ratings of the electrical equipment are RMS values. . instantaneous value and average value For a sinusoidal waveform: ● Peak value Fpk ¼ Fm: The amplitude or maximum value ● Peak–peak value Fp–p: Fp–p ¼ 2Fpk ● Instantaneous value f(t): The value at any time at any particular point of the waveform ● Average value Favg: Favg ¼ 0:637Fm 8.Fundamentals of AC circuits f(t) Fm 0. the single-phase AC voltage 110 V from the wall outlet is an RMS value. Applications of RMS value: RMS value (also referred to as the effective value) of the sinusoidal waveform is widely used in practice.4 Root mean square (RMS) value 1.637 times the peak value. Peak value.637 0 237 π wt Figure 8. the average value of a half-cycle sinusoidal wave is 0. For example. peak–peak value. Area 1 ¼ ¼ p p Zp 0 Favg 1 f ðtÞdt ¼ p Zp Fm sinotdot 0 Favg Fm Fm ¼ ½ À cosotp ¼ À ½cosp À cos0 0 p p Fm 2Fm ðÀ1 À 1Þ ¼ % 0:637Fm ¼À p p % 0:637Fm Therefore. and skip the following mathematic derivation process.13. then just keep in mind that Favg ¼ 0:637Fm is the equation for the average value of a half-cycle sinusoidal wave. an AC source RMS value will deliver the equivalent amount of average power to a load as a DC source. ● ● The average power generated by DC voltage is Pavg ¼ I 2 R. The physical meaning of RMS value: For a sinusoidal waveform.e. whether turning on the switch 1 (connect to DC) or switch 2 (connect to AC) in Figure 8. Only the first part in the above power expression represents the average power of AC. the average AC power is equivalent to the average DC power 2 2 when the AC source is an RMS value.238 Understandable electric circuits 2. the physical meaning of the AC RMS value is the heating effect of the sine wave. i. since the average value of the second part in the 2 power expression (a cosine function) is zero. That is. If the lamp is replaced by an electric heater. For instance. PAC ¼ ðIm RÞ=2. then the heating effect delivered by 20-V DC and 20-V AC RMS will be the same.14 RMS value 3. Taking the square root of both sides of the equation gives rffiffiffiffiffi 2 Im Im I¼ ¼ pffiffiffi % 0:707Im 2 2 or Im ¼ pffiffiffi 2I % 1:414I ð8:1Þ . Quantitative analysis of RMS value: The average power generated by an AC power supply is 2 pAC ¼ i2 R ¼ ðIm sinotÞ2 R ¼ ðIm sin2 otÞR AC pAC ¼ 2 Im 1 I2 R I2 R ð1 À cos2otÞ R ¼ m À m cos2ot 2 2 2 as sin2 ot ¼ 1=2ð1 À cos2otÞ. RMS value of AC current: According to the physical meaning of RMS. I = 2A 1 2 P = 40 W R = 10 Ω 20-V DC 20-V RMS Figure 8. 20-V DC or 20-V AC RMS value will deliver the same amount of power (40 W) to the resistor (lamp).14. so ðIm RÞ=2 ¼ I 2 R or I 2 ¼ Im =2. a German-American mathematician and electrical engineer.2) is the RMS value of AC voltage. load V ¼ 0:707 Vm . developed the phasor notation in 1893. Since phasors have magnitudes and directions. and Im is the peak value or amplitude of the AC current. and Vm is the peak value or amplitude of the AC voltage.Fundamentals of AC circuits 239 The current I in (8. RMS value of a periodical function f(t): Equations (8. A phasor is a vector that contains both magnitude and direction or amplitude and phase information. they can be represented as complex numbers.e. RMS value of AC function ● ● ● RMS value or effective value of AC: An AC source with RMS value will deliver the equivalent amount ofpffiffiffi power to a pffiffiffi as a DC source. sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Z 1 T 2 f ðtÞdt ðT is the period of the functionÞ F¼ T 0 The name root mean square (RMS) is obtained from the above equation. in which the term 1/T denotes the average (mean). It can .3. It can be used to represent AC quantities. Vm V ¼ pffiffiffi ¼ 0:707 V m 2 or Vm ¼ pffiffiffi 2 V ¼ 1:414 V ð8:2Þ ● The voltage V in (8. I ¼ 0:707Im or Vm ¼ 2 V. this relation only applies to the sine wave.1) is the RMS value of the AC current. ● RMS value of AC voltage: It can be obtained by the same approach by determining the RMS value of the AC current. the following general equation can be used to determine its RMS value.2) indicate the relationship between the RMS value and the peak value. However. For a non sine wave function f(t).1 Introduction to phasor notation Charles Proteus Steinmetz. f 2(t) denotes the square pffi (square) and denotes the square root (root) value. Im ¼ 2I qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi R 1 T The general equation to calculate RMS value: F ¼ T 0 f 2 ðtÞdt 8. A phasor notation or phasor domain is a method that uses complex numbers to represent the sinusoidal quantities for analysing AC circuits. i.1) and (8.3 Phasors 8. which is pffiffiffi related by 2. 240 Understandable electric circuits represent sine waves in terms of their peak value (magnitude) and phase angle (direction). the phase angle and the angular frequency.e. Phasor notation is a method that uses complex numbers to represent the sinusoidal quantities for analysing AC circuits when all quantities have the same frequency. The key for understanding the phasor notation is to know how to use complex numbers. so the resulting voltages and currents in the circuit should also be sinusoidal values with the same frequency or angular frequency. Phasor ● ● A phasor is a vector that contains both amplitude and angle information. therefore. 8. j is called the imagery unit. in which a is called modulus of the complex number. i. The phasor notation can simplify the calculations for AC sinusoidal circuits. . In an AC circuit. the rectangular form and the polar form. they can be determined by the peak value or RMS value and the phase shift of the phasor notation. Therefore. and it can be represented as complex number. and the angle c is called argument of the complex number. the AC source voltage and the current are the sinusoidal values with the same frequency.2 Complex numbers review The complex number has two main forms. we will review some important formulas of complex numbers that you may have learned in previous mathematics courses. ● Polar form: A ¼ affc This is the abbreviated form of the exponential form A ¼ ae jc . We have learned that a sinusoidal wave can be represented by its three important components: the peak value (or RMS value). j is used to denote the imagery unit rather than i to avoid confusion.3. it is widely used in circuit analysis and calculations. Therefore. Since i has been used to represent AC current in the circuit analysis. ● Rectangular form: A ¼ x þ jy pffiffiffiffiffiffiffi ð j ¼ À1Þ where x is the real part and y is the imagery part of the complex number A. The peak value can be easily converted to the RMS value. Note that the phasor notation can be used for sinusoidal quantities only when all waveforms have the same frequency. voltages and currents in an AC circuit can be analysed by using the phasor notation. Note: The symbol i is used to represent imagery unit in mathematics. Euler’s formula can also be used for the conversion of triangular form to exponential form e jc ¼ cosc þ jsinc ● or ae jc ¼ aðcosc þ jsincÞ Operations on complex numbers: Given two complex numbers A1 ¼ x1 þ jy1 ¼ a1 ffc1 and A2 ¼ x2 þ jy2 ¼ a2 ffc2 The basic algebraic operations of these two complex numbers are given as follows: Addition: A1 þ A2 ¼ ðx1 þ x2 Þ þ jðy1 þ y2 Þ Subtraction: A1 À A2 ¼ ðx1 À x2 Þ þ jðy1 À y2 Þ Multiplication: ● ● Polar form: A1 Á A2 ¼ a1 Á a2 ffðc1 þ c2 Þ Rectangular form: A1 Á A2 ¼ ðx1 þ jy1 Þðx2 þ jy2 Þ ¼ ðx1 x2 À y1 y2 Þ þ jðx2 y1 þ x1 y2 Þ pffiffiffiffiffiffiffipffiffiffiffiffiffiffi pffiffiffiffiffiffiffi Here j2 ¼ À1 À1 ¼ ð À1Þ2 ¼ À1 is used.15 Complex number ● Convert rectangular form to polar form (refer to Figure 8.Fundamentals of AC circuits +j y a y 0 x + 241 Figure 8. Let A ¼ x þ jy ¼ affc pffiffiffiffiffiffiffiffiffiffiffiffiffiffi Applying a ¼ x2 þ y2 (Pythagorus theory) gives pffiffiffiffiffiffiffiffiffiffiffiffiffiffi A ¼ x þ jy ¼ x2 þ y2 tanÀ1 y=x ¼ affc ● Convert polar form to rectangular or triangular form. So A ¼ affc ¼ x þ jy ¼ a(cosc þ jsinc).15).15. x ¼ acosc and y ¼ asinc can be obtained from Figure 8. Division: ● Polar form: A1 a 1 ¼ ffðc1 À c2 Þ A2 a 2 Rectangular form: A1 x1 þ jy1 ðx1 þ jy1 Þðx2 À jy2 Þ x1 x2 þ y1 y2 x2 y1 À x 1 y2 ¼ ¼ ¼ þj 2 A2 x2 þ jy2 ðx2 þ jy2 Þðx2 À jy2 Þ x2 þ y2 x2 þ y 2 2 2 2 ● . which will be discussed later. a sinusoidal function is actually taking the imaginary part of the complex number f ðtÞ ¼ Jm ½Fm e jc e jot ð8:3Þ There are two terms in (8. ‘Re[ ]’ and ‘Jm[ ]’ stand for ‘real part’ and ‘imaginary part’ of complex numbers. The second term e jot is called the rotating factor that varies with time t. sine function f ðtÞ ¼ Fm sinðot þ cÞ ¼ Jm ½Fm e jðotþcÞ ¼ jc jot Jm ½Fm e e .3.3 Phasor Using the phase notation to represent the sinusoidal function is based on Euler’s formula e j ¼ cos þ jsin . For a sinusoidal function f ðtÞ ¼ Fm sinðot þ cÞ. Complex numbers ● ● ● Rectangular form: A ¼ x þ jy Polar form: A ¼ affc Conversion between rectangular and polar forms: A ¼ x þ jy ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffi y x2 þ y2 tanÀ1 ¼ affc x A ¼ affc ¼ x þ jy ¼ aðcosc þ jsincÞ ● ● Addition and subtraction: A1 Æ A2 ¼ ðx1 Æ x2 Þ þ jðy1 Æ y2 Þ Multiplication: A1 Á A2 ¼ a1 Á a2 ffðc1 þ c2 Þ ¼ ðx1 þ jy1 Þðx2 þ jy2 Þ Division: A1 a1 x1 þ jy1 ¼ ffðc1 À c2 Þ ¼ A2 a2 x2 þ jy2 ● 8. Therefore. The first term is the phasor of the sinusoidal function Fm e jc ¼ Fm ffc ¼ F . Fm e jc and e jot .3). That is. respectively. replacing with (ot þ c) in Euler’s formula gives ejðotþcÞ ¼ cosðot þ cÞ þ jsinðot þ cÞ where cosðot þ cÞ ¼ Re ½ejðotþcÞ and sinðot þ cÞ ¼ Jm ½e jðotþcÞ .242 Understandable electric circuits It will be much simpler to use the polar form on operations of multiplication and division. this is known as phasors. This diagram is called the phasor diagram. If the sinusoidal currents and voltages in an AC circuit are represented by vectors with the complex numbers.16. The angle between the rotating line and the positive horizontal axis is the phase angle c of the sinusoidal function. +j F = Fm y Fm y 0 + Figure 8. where boldface letter F represents a phasor (vector) quantity. it can be presented with a rotating line in the complex plane as shown in Figure 8.4 Phasor diagram Since a phasor is a vector that can be represented by a complex number. There is no difference between operations on phasors and complex numbers.Fundamentals of AC circuits So (8. A phasor quantity can also be repre_ sented by a little dot on the top of the letter. since both of them are vectors. such as F ¼ Fm ffc. The length of the phasor is the peak value Fm (or RMS value F).3) is F ¼ Fm ffc. similar to the boldface that indicates a vector quantity in maths and physics. the first term in (8. The sinusoidal voltage vðtÞ ¼ Vm sinðot þ cÞ and current iðtÞ ¼ Im sinðot þ cÞ in an AC circuit can be expressed in the phasor domain as: ● Peak value: _ V ¼ Vm ffcv or V ¼ Vm ffcv _ I ¼ Im ffci or I ¼ Im ffci ● RMS value: _ V ¼ V ffcv or V ¼ V ffcv _ I ¼ Iffci or I ¼ Iffci 8.3) of sine function can be written as f ðtÞ ¼ Fm sinðot þ cÞ ¼ Jm ½Fe jot 243 Therefore.3.16 Phasor diagram . (a) v ¼ À10sinð60t þ 25 ÞV (b) i ¼ 12sinð25t À 20 ÞA Solution: _ (a) V ¼ À10ff25 V _ (b) I ¼ 12ff À 20 A Example 8.17. as shown in Figure 8. it rotates counterclockwise at angular frequency o in a radius Fm of the circle. _ (a) V ¼ 120ff30 V _ (b) I ¼ 12ff0 A pffiffiffi (a) v ¼ 120 2sinðot þ 30 Þ V pffiffiffi (b) i ¼ 12 2sinot A Solution: 8. known as the rotating factor or time factor. V ¼ V ffcv _ _ Im ¼ Im ffci . The rotating factor e jot can be represented by Euler’s formula e jot ¼ cosot þ jsinot when ot ¼ Æ90 : eÆj90 ¼ cosð Æ90 Þ þ jsinð Æ90 Þ ¼ Æj.3. Therefore. +908 is also the rotating factor (Æj ¼ Æ90 ).5 Rotating factor In the sinusoidal expression of f ðtÞ ¼ Fm sinðot þ cÞ ¼ Jm ½Fm e jc e jot .5: Use the instantaneous value to express the following voltage and current in which 120 and 12 are RMS values.244 Understandable electric circuits Phasor Time domain f ðtÞ ¼ Fm sinðot þ cÞ vðtÞ ¼ Vm sinðot þ cÞ iðtÞ ¼ Im sinðot þ cÞ Phasor domain _ Fm ¼ Fm ffc or Fm ¼ Fm ffc ðPeak valueÞ _ F ¼ Fffc or F ¼ Fffc ðRMS valueÞ _ V_m ¼ Vm ffcv . the term e jot varies with time t. As time changes.4: Use the phasor notation to express the following voltage and current in which 710 and 12 are the peak values. I ¼ Iffci Example 8. . When t ¼ t1.6: In Figure 8. The geometric meaning of the sinusoidal function f ðtÞ ¼ Fm sinðot þ cÞ ¼ Im ½Fm e jc e jot represented by the rotational phasor motion can be seen from the following example.Fundamentals of AC circuits 245 Rotating factor e jot or Æj ¼ Æ90 +j Fm 0 w ψ +1 –j Figure 8. And it goes from c to 3608. Example 8.18. When t ¼ t0 ¼ 0. . The instantaneous value of the sinusoidal wave at any time is equal to the projection of its relative rotating phasor on the vertical axis ( j ) at that time.17 Rotating factor +j t = t1 Fm Fm 0 ψ w t = t0 90° 0 t = t0 t = t1 wt ψ –j Figure 8. the phasor is F ¼ Fm ffc. the phasor is F ¼ Fm ff90 .18.18 Sine wave and phasor motion A sinusoidal function can be represented by a rotating phasor that rotates 3608 in a complex plane as shown in Figure 8. e. df ðtÞ .3. and determine voltage V (or V). i. i.7: Convert the following sinusoidal time-domain expression to its _ equivalent phasor domain.6 Differentiation and integration of the phasor Note: Skip the following part and start from Example 8. Z _ F f ðtÞdt ¼ jo This is equivalent to a phasor that rotates clockwise on the complex plane by 908 (since 1=j ¼ Àj ¼ À90 ). the derivative of the sinusoidal function with respect to time can be obtained by its phasor F multiplying with jo.246 Understandable electric circuits 8. (Appendix B provides the details for how to derive the above differentiation of the sinusoidal function in phasor notation. joF dt This is equivalent to a phasor that rotates counterclockwise by 908 on the complex plane since þj ¼ þ90 .e. For a sinusoidal function f ðtÞ ¼ Fm sinðot þ cÞ. so _ V _ _ 2V À 6 j4V þ 4 ¼ 20ff30 j4 _ V ð2 À 24j À jÞ ¼ 20ff30 20ff30 20ff30 _ V¼ % ¼ 0:8ff115:43 2 À j25 25ff À 85:43 . joF or joF ðþj ¼ þ90 Þ R _ Integration: f ðtÞdt . Z dv 2v À 6 þ 4 vdt ¼ 20sinð4t þ 30 Þ dt Solution: _ V _ _ 2V À 6joV þ 4 ¼ 20ff30 jo Since o ¼ 4 in the original expression. Differentiation and integration of the sinusoidal function in phasor notation _ Differentiation: df ðtÞ=dt .8 if you haven’t learned calculus. F=jo or ð1=joÞF (1=j ¼ Àj ¼ À90 ) Example 8.) The integral of the sinusoidal function with respect to time can be obtained from its phasor divided by jo. Fundamentals of AC circuits 247 Note: The complex number of the denominate is pffiffiffiffiffiffiffiffiffiffiffiffiffiffi y Z ¼ x þ jy ¼ 2 À j25 ¼ x2 þ y2 tanÀ1 x (Since x is positive and y is negative in 2 7j25.) .) Example 8.9. _ (a) I ¼ j5eÀj30 mA _ (b) V ¼ À6 þ j8 V Solution: _ (a) I ¼ j5ff À 30 mA ¼ 5ff90 ff À 30 mA ð j ¼ 90 Þ ¼ 5ffð90 À 30 Þ mA ¼ 5ff60 mA iðtÞ ¼ 5sinðot þ 60 Þ mA qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 8 _ (b) V ¼ À6 þ j8 V ¼ ðÀ 6Þ2 þ 82 tanÀ1 ff À6 V % 10ff126:87 V (Since y is positive and x is negative. the angle should be in the fourth quadrant. it should be in the second quadrant. the algebraic operations of sinusoidal functions of the same frequency can be replaced by algebraic operations of the equivalent phasors. vðtÞ ¼ 9:85 sin ðot À 28:48 Þ V . it should be in the fourth quadrant.) vðtÞ ¼ 10sinðot þ 126:87 Þ V If the phasors are used to express sinusoidal functions.8: Convert the phasor-domain voltage and current to their equivalent sinusoidal forms (time domain). 785. Example 8.438. which is shown in Example 8.9: Calculate the sum of the following two voltages v1 ðtÞ ¼ 2 sin ðot þ 60 Þ V and v2 ðtÞ ¼ 10 sin ðot À 40 Þ V Solution: Convert the sinusoidal time-domain voltages to their equivalent phasor forms _ _ V1 ¼ 2ff60 V and V2 ¼ 10ff À40 V _ _ So V1 þ V2 ¼ 2ff60 þ 10ff À40 ¼ 2 cos 60 þ j2 sin 60 þ 10 cos ðÀ40 Þ þ j10 sin ðÀ40 Þ % 1 þ j 1:732 þ 7:66 À j6:43 ¼ 8:66 À j4:698 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi À4:698 2 þ ðÀ 4:698Þ2 tanÀ1 ¼ 8:66 8:66 % 9:85ff À 28:48 V (Since y is negative and x is positive.e. i. 19(b).4 Resistors.e. c ¼ 08. Assuming the initial phase angle is zero.e.19(c).248 Understandable electric circuits 8. i.4.19 Resistor’s AC response Where the source voltage is e ¼ Vm sinðot þ cÞ The sinusoidal current in the circuit can be obtained by applying Ohm’s law for AC circuits (v ¼ Ri). all resulting voltages and currents in the circuit are also sinusoidal and have the same frequency as AC voltage source.1 Resistor’s AC response A resistor is connected to a sinusoidal voltage source as shown in Figure 8. iR + e R vR – (a) φ (b) 0 vR iR wt Figure 8.e. inductors and capacitors in sinusoidal AC circuits Any AC circuit may contain a combination of three basic circuit elements: resistor.19(a). and voltage across the resistor is the same as the source voltage. inductors and capacitors. i. The above sinusoidal expressions of resistor voltage vR and current iR indicate that voltage and current in the circuit have the same frequency f (since o ¼ 2pf ) and the same phase angle c (or vR and iR are in phase). i. they can all be converted from the sinusoidal time-domain form f ðtÞ ¼ Fm sinðot þ cÞ to the phasor-domain form F ¼ Fm ffc. . This is also illustrated in Figure 8. Therefore. then vR iR ¼ ¼ IRm sinot R vR ¼ RiR ¼ VRm sinot This is illustrated in Figure 8. When these elements are connected to a sinusoidal AC voltage source. e ¼ vR or vR ¼ Vm sinðot þ cÞ. 8. iR ¼ e VRm ¼ sinðot þ cÞ ¼ IRm sinðot þ cÞ R R where IRm ¼ VRm =R (peak values) or I ¼ VR =R (RMS value). IR E • • +j + • vR – 0 IR y + (b) • vR • R (a) Figure 8.20(b). it will give _ _ IR ¼ GVR ðG ¼ 1=RÞ The relationship of the resistor voltage and current in an AC circuit can be presented by a phasor diagram illustrated in Figure 8.Fundamentals of AC circuits 249 vR 0 iR wt Figure 8.20 The phasor diagram of the AC resistive circuit . The peak and RMS values of the resistor voltage and the current in phasor domain also obey the Ohm’s law as follows: _ _ Peak value: IRm ¼ VRm =R or VRm ¼ IRmR _ _ RMS value: IR ¼ VR =R or VR ¼ IRR If it is expressed in terms of conductance.19(c) When c ¼ 08 Relationship of voltage and current of a resistor in an AC circuit ● Instantaneous values (time domain): vR ¼ VRm sinðot þ cÞ iR ¼ IRm sinðot þ cÞ ● Ohm’s law: VRm ¼ IRm R ðpeak valueÞ VR ¼ IR R ðRMS valueÞ The sinusoidal expressions of resistor voltage (vR) and current (iR) are in the time domain. etc.20(a). instantaneous values.21(a).21 Inductor’s AC response We have learned from chapter 6 that the relationship between the voltage across the inductor and the current that flows through it is vL ¼ L di dt ð8:5Þ . iR ¼ 6 2sinðot À 30 ÞA in Figure 8. 8. Resistor’s AC response in phasor domain ● Ohm’s law: _ _ or VRm ¼ IRmR Peak value: V Rm ¼ I Rm R _R ¼ I_ R RMS value: V or VR ¼ I RR R Using conductance: I_ ¼ GV_R ðG ¼ 1=RÞ R Phasor diagram: À À ! ! (AC resistor voltage and current are in phase) _ IR _ VR ● Note that we can use Ohm’s law in AC circuits as long as the circuit quantities are consistently expressed. i. RMS values.e.4.2 Inductor’s AC response If an AC voltage source is applied to an inductor as shown in Figure 8.250 Understandable electric circuits pffiffiffi Example 8. pffiffiffi pffiffiffi Solution: vR ¼ Ripffiffiffi 10  6 2sinðot À 30 Þ ¼ 60 2ðsinot À 30 Þ R ¼ _ So VRm ¼ 60 2ff À 30 V. both the voltage and current are peak values.10: If R ¼ 10 O. determine the voltage across resistor in phasor domain. the current flowing through the inductor will be iL ¼ ILm sinðot þ cÞ iL vL e + L vL – 90° (a) (b) iL 0 wt ð8:4Þ Figure 8. 4) into (8. the reciprocal of reactance is called inductive susceptance and is denoted by BL. O Relationship of voltage and current of inductor in an AC circuit ● Instantaneous values (time domain) iL ¼ ILm sinðot þ cÞ vL ¼ XL ILm sinðot þ c þ 90 Þ . and skip the following mathematic derivation process. i. XL ¼ oL ¼ 2pfL ðo ¼ 2pf Þ So VLm ¼ XL ILm ðpeak valueÞ or XL ¼ VLm =ILm . and in an inductive circuit. XL ¼ VL =IL and VL ¼ XL IL ðRMS valueÞ where XL is measured in ohms (O) and is the same as resistance R. Recall that conductance G is the reciprocal of resistance R. i. the voltage and current have the same angular frequency (o) and a phase difference.21(b) if we assume that initial phase angle c ¼ 08. then just keep in mind that vL ¼ oLILm sinðot þ c þ 90 Þ is the sinusoidal expression of the inductor voltage.e.e. The inductor voltage vL leads the current iL by 908 as illustrated in Figure 8. where oL is called inductive reactance and is denoted by XL. The relationship between the voltage and current in an inductive sinusoidal AC circuit can be obtained from (8.6).Fundamentals of AC circuits 251 Note: If you haven’t learned calculus. BL ¼ 1=XL . which is given by VLm ¼ oLILm ðpeak valueÞ or VL ¼ oLIL ðRMS valueÞ This is also known as Ohm’s law for an inductive circuit.5) and applying differentiation gives vL ¼ L diL d½ILm sinðot þ cÞ ¼ oLILm cosðot þ cÞ ¼L dt dt ¼ oLILm sinðot þ c þ 90 Þ Therefore vL ¼ oLILm sinðot þ c þ 90 Þ ð8:6Þ Note: cos ¼ sinðot þ 90 Þ The sinusoidal expressions of the inductor voltage vL and current iL indicate that in an AC inductive circuit. Substituting (8. and is measured in siemens (S) or mho ( ). 21(a). An inductor can block DC (short-circuit equivalent). and the higher the angular frequency o. VL ! 1. In this case.252 Understandable electric circuits ● Ohm’s law: VLm ¼ XLILm (peak value) VL ¼ XLIL (RMS value) Inductive reactance: XL ¼ oL ¼ 2pfL Inductive susceptance: BL ¼ 1=XL ● ● In an AC inductive circuit. when the angular frequency approaches to infinite.e. i. The lower the angular frequency o.e. the relationship between the voltage and current is not only determined by the value of inductance L in the circuit. Recall that the inductor is equivalent to a short circuit at DC. VL ¼ 0. inductance L in the circuit is a constant. This indicates that an inductor can pass the high-frequency signals (pass AC) and block the low-frequency signals (block DC). Characteristics of an inductor ● ● An inductor can pass AC (open-circuit equivalent). If an inductor has a fixed value in the circuit of Figure 8. the lower the voltage across the inductor VL #¼ XL IL ¼ ðo # LÞIL When o ¼ 0. the AC voltage across the inductor now is equivalent to a DC voltage since the frequency (o ¼ 2pf Þ does not change any more. The peak and RMS values of the inductor voltage and the current in phasor domain also obey Ohm’s law as follows: _ _ Peak value: V Lm ¼ jXL I Lm _ _ RMS value: VL ¼ jXL IL or or V Lm ¼ jXL I Lm V L ¼ jXL I L . the greater the voltage across the inductor VL "¼ XL IL ¼ ðo " LÞIL When o ! 1. the inductor behaves as an open circuit in which the current is reduced to zero. the inductor is shortened because of zero voltage across the inductor. The sinusoidal expressions of the inductor voltage vL and current iL are in the time domain. but also related to the angular frequency o. i. pffiffiffi Example 8. iL +j e + L vL – (a) vL 90° 0 (b) IL + 0 (c) • • +j vL 90° IL y + • • Figure 8. and Figure 8. i.11: In an AC inductive circuit. determine the current through the inductor in time domain. given vL ¼ 6 2sinð60t þ 35 ÞV and L is 0.e. vL ¼ L 253 The relationship of the inductor voltage and current in an AC circuit can be presented by a phasor diagram illustrated in Figure 8.2 H. LjoIL (differentiating: multiply by jo) dt _ _ _ _ So VL ¼ ð joLÞIL or VL ¼ jXL IL ðXL ¼ oLÞ.Fundamentals of AC circuits This is because diL .22(b) is when the initial phase angle is zero.22 The phasor diagram of the AC inductive circuit Inductor’s AC response in phasor domain ● Ohm’s law: _ _ Peak value: VLm ¼ jXL ILm _ _ L ¼ jXL IL RMS value: V VL or or V Lm ¼ jXL I Lm V L ¼ jXL I L ● Phasor diagram: 0 90º IL ● Inductor voltage leads the current by 908. c ¼ 08.22(c) is when c 6¼ 08 (the inductor current lags voltage by 908).22(b and c). Solution: Inductive reactance XL ¼ oL ¼ ð60 rad=sÞð0:2 HÞ ¼ 12 O pffiffiffi pffiffiffi _ pffiffiffi _Lm ¼ V Lm ¼ 6 2ff35 V ¼ 6 2ff35 V ¼ 0:5 2ff À 55 A I jXL j12 O 12ff90 O Convert the inductor current from the phasor domain to the time domain pffiffiffi iL ¼ 0:5 2sinð60t À 55 Þ A . Figure 8. The capacitor current leads the voltage by 908 as illustrated in Figure 8. i.23(a).23 Capacitor’s AC response As we have learned from chapter 6.3 Capacitor’s AC response If an AC voltage source is applied to a capacitor as shown in Figure 8.e.23(b).7). The relationship between voltage and current in an inductive sinusoidal AC circuit can be obtained from (8. .4. if we assume that the initial phase angle c ¼ 08.254 Understandable electric circuits 8. the relationship between the voltage across the capacitor and the current through it is iC ¼ C dvC dt Substituting vC into the above expression and applying differentiation gives iC ¼ C That is iC ¼ oCVCm sinðot þ c þ 90 Þ ð8:7Þ d½VCm sinðot þ cÞ ¼ oCVCm sinðot þ c þ 90 Þ dt The sinusoidal expressions of the capacitor voltage vC and current iC indicated that in an AC capacitive circuit. which is given by ICm ¼ ðoCÞVCm or IC ¼ ðoCÞVC ðRMS valueÞ ðpeak valueÞ This is also known as Ohm’s law for a capacitive circuit. the voltage and current have the same angular frequency (o) and a phase difference. where oC is called capacitive reactance that is denoted by the reciprocal of XC. the voltage across the capacitor will be vC ¼ VCm sinðot þ cÞV Ic + e C vc – 90° (a) (b) Ic 0 • • vc wt Figure 8. 23(a). . when the angular frequency approaches infinite.Fundamentals of AC circuits XC ¼ So XC ¼ or XC ¼ VC IC ðRMS valueÞ VCm ICm ðpeak valueÞ 1 1 ¼ oC 2pfC ðo ¼ 2pf Þ 255 XC is measured in ohms (O) and that is the same as resistance R and inductive reactance XL. VC # ¼ XC IC ¼ IC o"C When o ! 1. the capacitor behaves as a short circuit in which the voltage across it will be reduced to zero. and it is also measured in siemens or mho (same as BL). the lower the voltage across the capacitor. the conductance C in the circuit is a constant. and the higher the angular frequency o. i. i. in an AC capacitive circuit not only is the relationship between voltage and current determined by the value of capacitive C in the circuit but it is also related to angular frequency o. The relationship of voltage and current of capacitor in an AC circuit ● ● ● ● Instantaneous values (time domain): vC ¼ VCm sinðot þ cÞ iC ¼ ocVCm sinðot þ c þ 90 Þ Ohm’s law: VCm¼ XCICm (peak value) VC ¼ XCIC (RMS value) Capacitive reactance: XC ¼ 1=oC ¼ 1=2pfC Capacitive susceptance: BC ¼ 1=XC Similar to an inductor. BC ¼ 1=XC . The reciprocal of capacitive reactance is called capacitive susceptance and is denoted by BC. VC ! 0.e. If there is a fixed capacitor in Figure 8.e. Recall that the inductive susceptance BL is the reciprocal of the inductive reactance XL. dt _ _ _ So I C ¼ joC V C ¼ jð1=XC ÞV C ðXC ¼ 1=oC Þ _ _ or V C ¼ ÀjXC I C ð1=j ¼ À jÞ: The relationship of the capacitor voltage and current in an AC circuit can be presented by a phasor diagram and is illustrated in Figure 8. The peak and RMS values of the capacitor voltage and the current in phasor domain also obey Ohm’s law as follows: _ _ Peak value: V Cm ¼ ÀjXC I Cm _ _ RMS value: V C ¼ ÀjXC I C This is because iC ¼ C or or VCm ¼ ÀjXC I Cm VC ¼ ÀjXC I C dvC . the AC voltage across the capacitor now is equivalent to a DC voltage since the frequency ðo ¼ 2pf Þ does not change any more. The characteristics of a capacitor are opposite to those of an inductor. Recall that a capacitor is equivalent to an open circuit at DC. the higher the voltage across the capacitor.24(b) is when the initial phase angle is zero.e. The sinusoidal expressions of the capacitor voltage vC and current iC are in the time domain. Characteristics of a capacitor ● ● A capacitor can pass DC (short-circuit equivalent). the capacitor is open because there will be no current flowing through the capacitor. i. In this case.256 Understandable electric circuits The lower the angular frequency o.24 The phasor diagram of an AC capacitive circuit . CjoVC (differentiating: multiply by jo). VC ! 1. A capacitor can block AC (open-circuit equivalent).24(b and c). and Figure 8.24(c) is when c 6¼ 08 Ic +j e C + vc – Ic 90° 0 (a) (b) Vc • • • Ic 90° + 0 (c) • +j vc + • Figure 8. c ¼ 08 (capacitor voltage lags current by 908). This indicates that a capacitor can block the high-frequency signal (block AC) and pass the low-frequency signal (pass DC). Figure 8. i. VC "¼ IC o#C When o ! 0.e. but does not change the polarity.Fundamentals of AC circuits 257 Capacitor’s AC response in phasor domain ● Ohm’s law: _ _ or V Cm ¼ ÀjXC I Cm Peak value: V Cm ¼ ÀjXC I Cm _ C ¼ ÀjXC IC _ RMS value: V or V C ¼ ÀjXC I C Phasor diagram: 0 IC 90º VC • • ● Capacitor current leads voltage by 908. saw-tooth wave. ● Sine AC varies over time according to the sine function.). pffiffiffi Example 8. ● Frequency f is the number of cycles of waveforms within 1 s: f ¼ 1=T . etc. Alternating current (AC) ● The voltage and current periodically change polarity with time (such as sine wave. capacitance is 5 mF and frequency is 500 Hz. ● The pulsing DC changes the amplitude of the pulse. ICm pffiffiffi pffiffiffi iC ¼ 786 2sinðot À 20 þ 90 Þ ¼ 786 2sinðot þ 70 ÞmA Summary ● ● ● ● Direct current (DC) ● The polarity of DC voltage and direction of DC current do not change. determine the capacitor current in the time domain. Period and frequency ● Period T is the time to complete one full cycle of the waveform. Solution: o ¼ 2pf ¼ 2pð500 HzÞ % 3142 rad=s XC ¼ 1 1 % 63:65 O ¼ oC ð3 142 rad=sÞð5  10À6 FÞ pffiffiffi pffiffiffi VCm 50 2 V ¼ ¼ % 786 2 mA 63:65 O XC . and is the most widely used AC. Three important components of the sinusoidal function f ðtÞ ¼ Fm sinðot þ cÞ ● Fm: Peak value (amplitude) ● o: Angular velocity (or angular frequency) o ¼ 2p=T ¼ 2pf ● c: Phase or phase shift .12: Given a capacitive circuit in which vC ¼ 50 2sinðot À 20 ÞV. square wave. instantaneous value and average value of sine waveform. ● ● ● iðtÞ ¼ Im sinðot þ ci Þ ● ● Vm V ¼ pffiffiffi ¼ 0:707 Vm . ● Peak value Fpk ¼ Fm: the amplitude ● Peak–peak value Fp–p: Fp–p ¼ 2Fpk ● Instantaneous value f(t): value at any time at any particular point of the waveform ● Average value: average value of a half-cycle of the sine waveform Favg ¼ 0:637Fm RMS value (or effective value) of AC sinusoidal function ● If an AC source delivers the equivalent amount of power to a resistor as a DC source. peak–peak value. which is the effective or RMS value of AC. ¼ c v À ci If ¼ 0: v and i in phase If 4 0: v leads i ● If 5 0: v lags i ● If ¼ Æp=2: v and i are orthogonal ● If ¼ +p: v and i are out of phase Peak value. 2 ● Im I ¼ pffiffiffi ¼ 0:707 Im 2 The general formula to calculate RMS value is sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Z 1 T 2 F¼ f ðtÞdt T 0 ● Complex numbers ● Rectangular form: A ¼ x þ jy ● Polar form: A ¼ affc ● Conversion between rectangular and polar forms: pffiffiffiffiffiffiffiffiffiffiffiffiffiffi y x2 þ y2 tanÀ1 ¼ affc x A ¼ affc ¼ x þ jy ¼ aðcosc þ jsincÞ A ¼ x þ jy ¼ ● ● Addition and subtraction: A1 Æ A2 ¼ ðx1 Æ x2 Þ þ jðy1 Æ y2 Þ Multiplication: A1 Á A2 ¼ a1 Á a2 ffðc1 þ c2 Þ ¼ ðx1 þ jy1 Þðx2 þ jy2 Þ .258 ● ● Understandable electric circuits c 4 0: Waveform shifted to the left side of 08 c 5 0: Waveform shifted to the right side of 08 Phase difference : For two waves with the same frequency such as vðtÞ ¼ Vm sin ðot þ cv Þ. ● The phasor notation is a method that uses complex numbers to represent the sinusoidal quantities for analysing AC circuits when all quantities have the same frequency.Fundamentals of AC circuits ● 259 Division: A1 a 1 x1 þ jy1 ¼ ffðc1 À c2 Þ ¼ A2 a 2 x2 þ jy2 ● Phasor ● A phasor is a vector that contains both amplitude and angle information. I ¼ Iffci ● ● ● Rotation factor: e jot or Æj ¼ Æ90 Differentiation and integration of the sinusoidal function in phasor notation: _ ● Differentiation: df ðtÞ=dt ¼ joF or joF (þj ¼ þ90 ) R _ ● Integration: f ðtÞdt ¼ F=jo or ð1=joÞF ð1=j ¼ Àj ¼ À90 Þ Characteristics of the inductor and capacitor: DC (v ¼ 0) Short circuit Open circuit AC (v ! `) Open circuit Short circuit Characteristics Pass DC and block AC Pass AC and block DC Element Inductor Capacitor ● Three basic elements in an AC circuit Time domain Phasor domain _ _ V R ¼ I RR _ V_L ¼ jXL I L Resistance and reactance R XL ¼ oL Conductance and susceptance G ¼ 1=R BL ¼ 1=XL vL 0 Element Phasor diagram IR • Resistor Inductor Capacitor vR ¼ RiR vL ¼ Lðdi=dtÞ VR 90º IL _ iC ¼ CðdvC =dtÞ V_C ¼ ÀjXC I C XC ¼ 1=oC BC ¼ 1=XC 0 IC 90º vC • • . and can be represented as a complex number. V ¼ V ffcv _ _ I m ¼ Im ffci . Time domain f ðtÞ ¼ Fm sinðot þ cÞ vðtÞ ¼ Vm sin ðot þ cÞ iðtÞ ¼ Im sin ðot þ cÞ Phasor domain F m ¼ Fm ffc or F_m ¼ Fm ffc ðpeak valueÞ _ ¼ Fffc ðRMS valueÞ F ¼ Fffc or F _ V_m ¼ Vm ffcv . 260 Understandable electric circuits Experiment 8: Measuring DC and AC voltages using the oscilloscope Objectives ● ● ● ● Become familiar with the operations of a function generator. 2. square waves. Its main function is to display waveforms to observe and analyse voltage. triangle waves. but most of their controls (knobs and buttons) in Table L8. Become familiar with the method to measure DC and AC voltages with an oscilloscope. period and phase difference of DC or AC signals. Function generator: The function generator is an electronic equipment that can generate various types of waveforms that can have different frequencies and amplitudes. Become familiar with the operations of an oscilloscope. The oscilloscope is a complex testing equipment and it is important to be familiar with its operations. A function generator can be used as an AC voltage source to provide time-varying signals such as sine waves. Background information 1. . frequency. Figure L8. Oscilloscope: The oscilloscope is one of the most important experimental and measurement instruments available for testing electric and electronic circuits. Become familiar with the settings and correction of an oscilloscope.1 shows the front panel of an oscilloscope.1 The main controls of an oscilloscope Display Horizontal control Vertical control Selecting switch Probe 61 610 INTENSITY Time base setting (TIME/DIV) FOCUS Volts per division Channel coupling (VOLTS/DIV) (CH I–DUAL– CH II) Vertical position Input coupling Horizontal position control control (DC–GND–AC) (X-POS $) (Y-POS) ● ● Intensity control (INTENSITY): It can adjust the brightness of the display.1 have similar functions. Focus control (FOCUS): It can adjust the sharpness and clarity of the display. Table L8. There are various types of oscilloscopes that may look different. We will use this oscilloscope as an example for a brief description of the operations of the oscilloscope. etc. Measured amplitude ¼ ðNumber of vertical divisionsÞ Â ðVOLTS=DIVÞ Note: There is a small calibration (CAL) knob in the centre of both the VOLTS/DIV and TIME/DIV knobs. ● ● ● ● Horizontal position control (X-POS$): It can adjust the horizontal position of the waveform. Channel coupling (CH I–DUAL–CH II): CH I: Displays the input signal from channel I. Volts per division selector (VOLTS/DIV – volts per division): It can set up the waveform amplitude value per vertical square (division) on the screen.1 An oscilloscope ● ● Time base control (TIME/DIV – seconds per division): It can set up the length of time displayed per horizontal square (division) on the screen. DUAL: Displays the input signals from both channels I and II. Vertical position control (Y-POSl): It can adjust the vertical position of the waveform. CH II: Displays the input signal from channel II. Input coupling (DC–GND–AC): The connection from the test circuit to the oscilloscope. It should be in the fully clockwise position for the accuracy of the measurement.Fundamentals of AC circuits 261 Figure L8. . 262 Understandable electric circuits DC: The DC position can display both DC and AC waveforms (the AC signal is superimposed on the DC waveform). The voltmeter reading should be 3 V now. and adjust DC power supply to 3 V. ● ● Equipment and components ● ● ● ● ● Digital multimeter Breadboard DC power supply Oscilloscope Function generator Procedure Part I: Measure DC voltage using an oscilloscope 1. then switch to DC ● TIME/DIV: 1 ms/DIV ● Trigger: Auto (The trigger can stabilize repeating waveforms and capture single-shot waveforms. and negative terminal of the multimeter (voltmeter function) should be connected together.2. 610 Probe: Needs to multiply by 10 for each measured reading (more accurate). The DC wave on the oscilloscope screen . ground of the oscilloscope probe. + E V – Oscilloscope Figure L8. The negative terminal of DC power. Connect a circuit as shown in Figure L8. GND: The GND position has a horizontal line on the screen that represents zero reference. Set up the oscilloscope probe to 61. AC: The AC position blocks the DC waveform and only displays AC waveform. 61 Probe: Can measure and read the signal directly but may load the circuit under test and distort the waveform.2 Measuring DC voltage using an oscilloscope 3. adjust VOLTS/DIV of the oscilloscope to 1 V/DIV.) 2. Set up the oscilloscope controls to the following positions: ● Channel coupling: CH I or CH II ● Input coupling: Set up to GND and adjust the trace to the central reference line (0 V) first. and record them in Table L8. ð3 vertical divisionsÞ Â ð1 V=DIVÞ ¼ 3 V 4.Fundamentals of AC circuits 263 occupies three vertical grids (squares) at this time.2. so the voltage measured by the oscilloscope is also 3 V. The ground of the function generator.2. respectively. ð8 vertical divisionsÞ Â ð0:5 V=DIVÞ ¼ 4 V Read the value on the voltmeter. and adjust DC power supply to 4 V. Keep the oscilloscope probe at 61. and record them in Table L8. adjust VOLTS/DIV of the oscilloscope to 0. Read the voltage values on the voltmeter and oscilloscope. Keep the oscilloscope probe at 61.5 V/DIV. adjust VOLTS/ DIV of the oscilloscope to 2 V/DIV. Replace the DC power supply by a function generator in Figure L8.5 2 Voltmeter (V) 3 Oscilloscope (V) 3 4 5. 6.2. adjust DC power supply to 8. Table L8. The DC wave on the oscilloscope screen occupies eight vertical divisions at this time. Set up the oscilloscope probe to 610. 12 and 16 V. and record it in Table L8. ground of the oscilloscope probe and negative terminal of multimeter (voltmeter function) should be connected together. and adjust VOLTS/DIV to suitable positions. and adjust DC power supply to 5 V.5 kHz DC offset: 0 V Amplitude knob: minimum (Fully counter clockwise) ● Set up the oscilloscope: VOLTS/DIV: 0.2 Probe 61 610 DC power supply (V) Example: 3 4 5 8 12 16 Vertical division (DIV) 3 8 VOLTS/DIV (V/DIV) 1 0. ● Set up the function generator: Waveform: sine Frequency: 1.2.5 V/DIV Channel coupling: CH I TIME/DIV: 0.2 ms/DIV . Part II: AC measurements using an oscilloscope 1. Read the voltage value on the voltmeter and oscilloscope. Determine the period T and frequency f of the sine wave. and record the values in Table L8.5 ms/DIV.3. Adjust the horizontal position control of the oscilloscope (X-POS) until the sine wave on the oscilloscope screen occupies six horizontal divisions (adjust the frequency knob on the function generator if necessary). and record the values in Table L8. Conclusion Write your conclusions below: .264 Understandable electric circuits Input coupling: Set up to GND and adjust the trace to the central reference line (0 V) first. Table L8. then switch to AC 2.3 Period T Step 4 Step 5 Frequency f 5. 3. Determine the period T and frequency f of the sine wave. Adjust the amplitude knob of the function generator until that sine wave on the vertical division of the oscilloscope screen occupies six divisions (squares).2 ms/DIV). and it can be converted to the peak value comparing with the waveform obtained from the oscilloscope. ● Determine the period of the sine wave T: Period ðTÞ ¼ ðNumber of horizontal divisionsÞ Â ðTIME=DIVÞ T ¼ ð4 divisionsÞ Â ð0:2 ms=DIVÞ ¼ 0:8 ms ● Determine the frequency f: 1 1 f ¼ ¼ ¼ 1:25 kHz T 0:8 ms 4. and adjust the horizontal position control of the oscilloscope (X-POS) until the sine wave on the oscilloscope screen occupies five horizontal divisions. Adjust the horizontal position control of the oscilloscope (X-POS) until the sine wave on the oscilloscope screen occupies four horizontal divisions. The voltage amplitude at this time is VPÀP ¼ ð6 DIVÞ Â ð0:5 V=DIVÞ ¼ 3 V Note that the reading of the multimeter is RMS value.3 (keep TIME/DIV ¼ 0. Adjust TIME/DIV of the oscilloscope to 0. we had learned that the phasor forms of relationship between voltage and current for resistor. inductor L and capacitor C determine the impedance and admittance of series and parallel AC circuits apply the voltage divider and current divider rules to AC circuits apply KCL and KVL to AC circuits understand the concepts of instantaneous power. It is similar to the concept of resistance in DC circuits. and it can be _ _ generally expressed as Z ¼ V =I. node voltage analysis. etc. _ IL _ VC ¼ ÀjXC _ IC The ratio of voltage and current is the impedance of an AC circuit. superposition theorem and Thevinin’s and Norton’s theorems. This equation is also known as Ohm’s law of AC circuits.1 Impedance In the previous chapter.1.Chapter 9 Methods of AC circuit analysis Objectives After completing this chapter. _ _ VC ¼ ÀjIC XC The above equations can be changed to a ratio of voltage and current _ VR ¼ R. apparent power. _ _ VL ¼ jIL XL . active power. The physical meaning of the impedance is that it is a measure of the opposition to AC current in an AC circuit. so the impedance is also measured in ohms. power triangle and power factor apply the mesh analysis. The impedance can be . _ IR _ VL ¼ jXL . inductor and capacitor in an AC circuit are as follows: _ _ VR ¼ IR R. you will be able to: ● ● ● ● ● ● ● understand concepts and characteristics of the impedance and admittance of AC circuits define the impedance and admittance of resistor R.1 Impedance and admittance 9. reactive power. to analyse AC circuits 9. 2 Admittance Recall that the conductance G is the inverse of resistance R. Admittance Y ● ● ● Y is the measure of how easily current can flow in an AC circuit. jXL XL YC ¼ 1 1 ¼j ÀjXC XC j¼ 1 Àj . and it is a measure of how easily current flows in a DC circuit. The impedances of resistor. it is denoted by Y. the admittance is the inverse of impedance Z. _ _ Ohm’s law in AC circuits: I ¼ V Y : Quantity Symbol Y Unit Siemens Unit symbol S Quantity Admittance The admittance of resistor. inductor and capacitor are as follows: YR ¼ 1 . It is a complex number that describes both the amplitude and phase characteristics. _ IL ZC ¼ ÀjXC ¼ _ VC _ IC Impedance Z ● ● Z is a measure of the opposition to AC current in an AC circuit. Y is the inverse of impedance: Y ¼ 1/Z. It is more convenient to use the admittance in a parallel AC circuit. inductor and capacitor are as follows: ZR ¼ R ¼ _ VR . _ _ Ohm’s law in AC circuits: Z ¼ V =I: Quantity Symbol Z Unit Ohm Unit symbol O Quantity Impedance 9. i. Y ¼ I=V . R YL ¼ 1 1 ¼ Àj . _ IR ZL ¼ jXL ¼ _ VL . It is more convenient to use the conductance in a parallel DC circuit. The admittance is a measure of how easily a current can flow in an AC circuit. It can _ _ be expressed as the ratio of current and voltage of an AC circuit.266 Understandable electric circuits extended to the inductor and capacitor in an AC circuit.e.1. and is measured in siemens (S). Y = 1/Z. Similarly. where the real part of the complex is the resistance R. X _ VR _ IR ¼ R . X and Z in the expression of the impedance is a right triangle. it can be expressed in both polar form and rectangular form (complex number) as follows: Z ¼ zfff ¼ R þ j X ¼ zðcos f þ j sin fÞ ð9:1Þ The rectangular form is the sum of the real part and the imaginary part. This can be illustrated in Figure 9. VX ¼ IX X . and can be described using the Pythagoras’ theorem. If we divide each side of the value by voltage V in the impedance triangle. Z _ VX _ IX ¼ . If we multiply each side of the quantity _ in the impedance triangle by current I the following expressions will be obtained: _ _ _ _ _ _ Vz ¼ Z Iz . voltage and current triangles Figure 9.1. z f R (a) X f VR (b) Vz VX f IR (c) Iz IX Figure 9. and the imaginary part is the reactance X.1(a).Methods of AC circuit analysis 267 9. which is z¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi R2 þ X 2 The corresponding angle f between the resistance R and reactance X is called the impedance angle and can be expressed as follows: f ¼ tanÀ1 X R The relationship between R.e. X ¼ XL À XC The lower case letter z in (9.1(a) is an impedance triangle.3 Characteristics of the impedance Since the impedance is a vector quantity.1) is the magnitude of the impedance.1(b). VR ¼ IX R These can form another triangle that is called the voltage triangle. which is _ illustrated in Figure 9.1 Impedance. the following expressions will be obtained: _ Vz _ Iz ¼ . i. The reactance is the difference of inductive reactive and capacitive reactance. 1(a) can be summarized as follows: ● ● ● If X 4 0 or X ¼ XL À XC > 0. ZR ZL R = 1.1(c). XC ¼ XL : The impedance angle f ¼ 0.1: Determine the impedance Z in the circuit of Figure 9. and the impedance angle f 4 0.268 Understandable electric circuits The above expression can form another triangle that is called the current triangle. the circuit will look like a purely resistive circuit (z = R) as shown in Figure 9.3 and plot the phasor diagram of the impedance. .2 The phasor diagrams of the impedance Example 9.5 Ω Z XC = 3 Ω ZC Figure 9. The characteristics of the impedance triangle in Figure 9. XC > XL : The reactance X is below the horizontal axis. +j XL XL – XC 0 +j XL 0 +j XL + z (XC > XL) (b) 0 z f R f R=z R + XC f >0 R (XL > XC) (a) + XL – XC XC f <0 XC f =0 (XL = XC) (c) Figure 9.5 Ω XL = 2.2(a).2(c).3 Circuit for Example 9. and it is illustrated in Figure 9. If X 5 0 or X ¼ XL À XC < 0. and the impedance angle f < 0. If X = 0 or X ¼ XL À XC ¼ 0. XL > XC : The reactance X is above the horizontal axis.2(b). The circuit is more capacitive as shown in Figure 9.1 Solution: The impedances in series in an AC circuit behave like resistors in series. The circuit is more inductive as shown in Figure 9. +j XL = 2. B ¼ BC À BL : The plower case letter y in (9.e.4 Impedance angle for Example 9. y ¼ G2 þ B2 .1 9. Z ¼ ZR þ ZL þ ZC ¼ R þ jX ¼ R þ jðXL À XC Þ ¼ 1:5 O þ jð2:5 À 3ÞO ¼ 1:5 O À j0:5 O qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi À0:5 ¼ 1:52 þ ðÀ0:5Þ2 tanÀ1 % 1:58ff À 18:44 O 1:5 269 Note: Since the imaginary term is 70.5 on y-coordinate. siemens (S). and the real term is þ1. it can be expressed in both polar and rectangular forms as follows: Y ¼ yfffy ¼ G þ jB ¼ yðcos f þ j sin fÞ ð9:2Þ The real part of the complex is the conductance G.1 is more capacitive since XC > XL . the impedance angle for this circuit is located in the 4th quadrant.5 Ω z f + XC = 3 Ω Figure 9. and the imaginary part is called the susceptance B.5 on the x-coordinate. i.1.4 Characteristics of the admittance The admittance is also a complex number. The circuit for Example 9. The susceptance is the difference of the capacitive susceptance and inductive susceptance.4.2) is the magnitude of the admittance. ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi i.e. i.5 Ω 0 XL – XC = –0. The susceptance is measured in the same way as the admittance.5 Ω R = 1. and f < 0 as shown in Figure 9.e.Methods of AC circuit analysis So. 1 Impedance and admittance Impedance _ Z ¼ V =I_ ZR ¼ R ZL ¼ jXL ZC ¼ 7jXC Z ¼ zfff ¼ R þ jX pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi z ¼ R2 þ X 2 X f ¼ tanÀ1 R Admittance Y ¼ 1/Z YR ¼ G YL ¼ 7jBL YC ¼ jBC Y ¼ yfffy ¼ G þ jB pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi y ¼ G 2 þ B2 B fy ¼ tanÀ1 G Component Resistor (R) Inductor (L) Capacitor (C) Conductance and susceptance Conductance: G ¼ 1/R Inductive susceptance: BL ¼ 1/XL Capacitive susceptance: BC ¼ 1/XC Reactance: X ¼ XL À XC Susceptance: B ¼ BC À BL The characteristics of the admittance triangle in Figure 9. inductor and capacitor are as follows: YR ¼ 1 ¼ G. R YL ¼ 1 1 ¼ Àj .1. XC ¼ .5 The phasor diagrams of the admittance .270 Understandable electric circuits The corresponding angle f between the conductance G and susceptance B is called the admittance angle and can be expressed as fy ¼ tanÀ1 B G The admittance of resistor.5 can be summarized as follows: 1 1 XL ¼ oL. Table 9. jXL XL YC ¼ 1 1 ¼j ÀjXC XC The impedance. admittance. j¼ oC Àj +j BC BC – BL 0 BL fy > 0 (BC > BL) (a) fy G Y + BC 0 BC – BL BL fy < 0 (BL > BC) (b) fy = 0 (BL = BC) (c) fy Y G +j BC + 0 BL +j G=Y G + Figure 9. susceptance and their relationship can be summarized as given in Table 9. 6 and plot the phasor diagrams of the admittance. If B ¼ 0 or B ¼ ðBC À BL Þ ¼ 0.075.1 and the real term is þ0. BL > BC : The susceptance B is below the horizontal axis. If B 5 0 or B ¼ ðBC À BL Þ < 0.2: Determine the admittance in the circuit of Figure 9.Methods of AC circuit analysis ● 271 ● ● If B 4 0 or B ¼ ðBC À BL Þ > 0. and the circuit is more capacitive as shown in Figure 9. B 4 0. If X ¼ 0. the circuit will look like a purely resistive circuit (Y ¼ G) as shown in Figure 9.5(c). Y R 13.3 Ω XL 5.5(a). f 4 0. f 5 0. Characteristics of impedance and admittance ● ● ● If X 4 0.6 Circuit for Example 9. So. If X 5 0. fy 5 0: The circuit is more inductive. and the circuit is more inductive as shown in Figure 9. the admittance angle fy < 0.72 Ω XC 13. the admittance angle fy 4 0. fy 4 0: The circuit is more capacitive. fy ¼ 0: The circuit is purely resistive. BC > BL : The susceptance B is above the horizontal axis. B 5 0. BL ¼ BC : The admittance angle fy ¼ 0. Y ¼ YR þ YL þ YC ¼ G þ jB ¼ G þ jðBC À BL Þ 1 1 1 ¼ þj À % 0:075 S þ jð0:075 À 0:175ÞS 13:3 O 13:3 O 5:72 O qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi À0:1 ¼ 0:125ff À 53:13 S ¼ 0:075 S À j0:1 S ¼ 0:0752 þ ðÀ0:1Þ2 tanÀ1 0:075 Note: The admittance angle for this circuit is located in the fourth quadrant since the imaginary term is 70.3 Ω Figure 9.2 Solution: The admittances in parallel in AC circuits behave like the conductances in parallel in DC circuits. . Example 9. f ¼ 0. B ¼ 0.5(b). 7 Admittance angle for Example 9. The equivalent impedance (or total impedance) for a series circuit in Figure 9.7.1° + Y = 0. the circuit is more inductive as shown in Figure 9.2 Impedance in series and parallel 9.8 Impedance of a series circuit The equivalent impedance (or total impedance) for a parallel circuit in Figure 9.2.9 is given as Zeq ¼ 1 ¼ Z1 ==Z2 == .9 Impedance of a parallel circuit . +j BC = 0.175.075) and fy 5 0.125 S Figure 9. BC ¼ 0. except the phasor form (complex number) is used.8 is given as Zeq ¼ Z1 þ Z2 þ Á Á Á þ Zn Z1 Zeq Z2 … Zn Figure 9.175 S fy = –53. ==Zn ð1=Z1 Þ þ ð1=Z2 Þ þ Á Á Á þ ð1=Zn Þ Yeq ¼ Y1 þ Y2 þ Á Á Á þ Yn Zeq Z1 Z2 … Zn Figure 9.272 Understandable electric circuits Since BL 4 BC (BL ¼ 0.075 S 0 BC – BL BL= 0.1 Impedance of series and parallel circuits The impedances in series and parallel AC circuits behave like resistors in series and parallel DC circuits. .075 S G = 0.2 9. . Z1 þ Z2 _ I2 ¼ Z1 _ IT Z1 þ Z2 Z1 V1 E • • E Z2 • IT Z1 • I1 Z2 • I2 • V2 • (a) (b) Figure 9.2. 9. Z1 þ Z2 _ V2 ¼ Z2 _ E Z1 þ Z2 _ I1 ¼ Z2 _ IT .2 Voltage divider and current divider rules The voltage divider and current divider rules in phasor form in AC circuits are very similar to the DC circuits as follows: _ V1 ¼ Z1 _ E.10 Voltage and current dividers . If only have two impedances in parallel. the equivalent impedance is given as Z1 Z2 ¼ Z1 ==Z2 Z1 þ Z2 Zeq ¼ Impedances in series and parallel ● ● ● Impedances in series: Zeq ¼ Z1 þ Z2 þ Á Á Á þ Zn Impedances in parallel: Zeq ¼ Z1 ==Z2 == Á Á Á ==Zn . use the same method that determines the equivalent resistance in series and parallel DC circuits. Yeq ¼ Y1 þ Y2 þ Á Á Á þ Yn Z1 Z2 Two impedances in parallel: Zeq ¼ Z1 þ Z2 To determine the equivalent impedance in series and parallel AC circuits. Zeq ¼ 1=Yeq .Methods of AC circuit analysis 273 The equivalent impedance is the reciprocal of equivalent admittance. Example 9.3 The phasor forms of KVL and KCL The phasor forms of Kirchhoff’s voltage law (KVL) and Kirchhoff’s current law (KCL) also hold true in AC circuits.11 Circuit for Example 9.11.3 Solution: (a) Zeq ¼ Z1 þ Z2 ==Z3 Z1 ¼ R1 ¼ 4 kO Z2 ¼ ÀjXC ¼ Àj8 kO Z3 ¼ RL þ jXL ¼ 4 kO þ j8 kO % 8:94ff63:44 kO Z2 ==Z3 ¼ Z2 Z3 ðÀj8Þð4 þ j8Þ 64 À j32 71:55ff À 26:57 ¼ kO kO ¼ kO % Àj8 þ 4 þ j8 4 4ff0 Z2 þ Z3 % ð16 À j8ÞkO ¼ 17:9ff À 26:57 kO ¼ 17:9½cosðÀ26:57 Þ þ j sinðÀ26:57 ÞkO Zeq ¼ Z1 þ Z2 ==Z3 ¼ ½4 þ ð16 À j8ÞkO ¼ ð20 À j8ÞkO % 21:54ff À 21:8 kO .2.274 Understandable electric circuits 9. Phasor forms of KVL and KCL ● ● KVL : KCL : _ _ _ _ _ SV ¼ 0 or V1 þ V2 þ Á Á Á þ Vn ¼ E _ _ _ SI ¼ 0 or Iin ¼ Iout The following examples show how to use the above equations in series-parallel AC circuits.3: Determine the following values for the circuit in Figure 9. I1 I2 R1 = 4 kΩ E = 100 V∠0° Z1 XC = 8 kΩ Z2 Z3 I3 RL = 4 kΩ XL = 8 kΩ Figure 9. ● ● the input equivalent impedance Zeq and _ the current I3 in the branch of RL and XL. Z1 ¼ R ¼ 120 O Z2 ¼ jXL ¼ jðoLÞ ¼ jð2  10 HÞ ¼ j20 O Z3 ¼ ÀjXC ¼ Àj 1 1 ¼ Àj ¼ Àj25 O oC 2  20 mF _ e ¼ 40 sinð2t À 30 ÞV ) E ¼ 40ff À 30 V _ _ VL ¼ E Z2 ==Z3 ðZ2 ==Z3 ¼ ?Þ Z1 þ Z2 ==Z3 Z2 Z3 j20ðÀj25Þ 500 Z2 ==Z3 ¼ ¼ O¼ O ¼ j100 O Z2 þ Z3 j20 À j25 Àj5 ● .12(b) first.Methods of AC circuit analysis _ (b) I3 ¼ Z2 _ I1 Z2 þ Z3 _ E 100ff0 V _ Here I1 ¼ ¼ % 4:64ff 21:8 mA Zeq 21:54ff À 21:8 O _ .12.12 Circuits for Example 9.4: Determine the voltage across the inductor L for the circuit in Figure 9. I3 ¼ Z2 _ I1 Z2 þ Z3 275 ¼ ð4:64ff21:8 ÞmA 8ff À 90 kO 37:12ff À 68:2 mA ¼ ðÀj8 þ 4 þ j8ÞkO 4ff0 ¼ 9:28ff À 68:2 mA Example 9.4 Solution: ● Convert the time domain to the phasor domain as shown in Figure 9. Z1 = 120 Ω R = 120 Ω E = 40 sin (2t-30°) V + + C = 20 mF E = 40 V ∠–30° L = 10 H VL – Z2 = j20 Ω VL – Z3 = –j25 Ω (a) (b) Figure 9. active power. 9. reactive power and apparent power. Im ¼ 2IÞ ¼ VI cos f À VIðcos 2ot cos f À sin 2ot sin fÞ ½. It is the product of instantaneous voltage v and current i at that particular moment (Figure 9. i. VL ¼ E ¼ Z2 ==Z3 j100 O ¼ ð40ffÀ 30 ÞV Z1 þ Z2 ==Z3 ð120 þ j100ÞO 4000ff 60 V % 156:2ff 39:8 % 25:61ff 20:2 V After converting the phasor form to the time form gives vL ¼ 25:61 sinð2t þ 20:2 ÞV 9.276 Understandable electric circuits _ _ .13 Instantaneous power If v ¼ Vm sinðot þ fÞ and i ¼ Im sin ot Then. such as instantaneous power.13).3.3. p ¼ À V m Im ½cosð2ot þ fÞ À cos f 2 pffiffiffi pffiffiffi ¼ VI cos f À VI cosð2ot þ fÞ ðVm ¼ 2V. cosðx þ yÞ ¼ cos x cos y À sin x sin y . instantaneous power can be expressed as p ¼ vi i + v – Load Figure 9. Power in AC circuits There are different types of power in AC circuits.1 Instantaneous power p The instantaneous power p is the power dissipated in a component of an AC circuit at any instant time. p ¼ vi ¼ Vm Im sin ot sinðot þ fÞ 1 .e. À sin x sin y ¼ ½cosðx þ yÞ À cosðx À yÞ 2 1 . instantaneous power is given as p ¼ VI cos fð1 À cos 2otÞ þ VI sin f sin 2ot 277 The waveform of the instantaneous power can be obtained from the product of instantaneous voltage and current at each point on their waveforms as shown in Figure 9. the component returns the stored energy to the source. Instantaneous power p p is the product of instantaneous voltage and current at any instant time: p ¼ vi ¼ VI cosf ð1 À cos 2otÞ þ VIðsinf sin 2otÞ p 4 0: The component absorbs (stores) energy. \ p ¼ vi 4 0 When instantaneous power p is 40 (p is positive). p v i 0 t2 t3 t1 t Figure 9. p 5 0: The component returns (releases) energy. \ p ¼ vi 5 0 between time t2 * t3 : v 5 0 and i 5 0. p ¼ vi ¼ 0 between time 0 * t1 : v 4 0 and i 4 0. ● Instantaneous power for a resistive component pR: Since voltage and current in a purely resistive circuit is in phase.e. the component stores energy provided by the source. p ¼ vi ¼ 0 at time t ¼ t1 : v ¼ 0. i. When instantaneous power p is 50 (p is negative).14 The waveform of instantaneous power Such as: ● ● ● ● ● at time t ¼ 0 : i ¼ 0. substituting this into the equation of the instantaneous power gives .Methods of AC circuit analysis Therefore. f ¼ 0. \ p ¼ vi 4 0 between time t1 * t2 : v 5 0 and i 4 0.14. When instantaneous power is positive.3) is average power dissipated in the resistive load (p 4 0. The second part in (9. both the instantaneous powers of inductive and capacitive loads are sinusoidal quantities with a double frequency 2o.3) and waveforms in Figure 9. this indicates that when voltage and current waveforms oscillate one full cycle in one period of time.3) is a sinusoidal quantity with a double frequency 2o.3) as follows: ● ● Instantaneous power for an inductive load: pL ¼ VI sin 2ot Instantaneous power for a capacitive load: pC ¼ ÀVI sin 2ot The diagrams of instantaneous power for inductive and capacitive loads are illustrated in Figure 9.16. power waveform will oscillate two cycles as illustrated in Figure 9.15 The waveform of instantaneous power for a R load ● Instantaneous power for capacitive and inductive components: In a purely inductive load circuit. or a resistor always dissipates power. pR v i 0 VI t Figure 9. Substituting f ¼ Æ90 into the equation of instantaneous power gives p ¼ VI cosðÆ90 Þð1 À cos 2otÞ þ VI sinðÆ90 Þ sin 2ot ¼ ÆVI sin 2ot ð9:3Þ The instantaneous power for inductive and capacitive loads can be obtained from (9. the component stores energy. when instantaneous power is . As seen from (9. The mathematical expression and the waveform all show that instantaneous power of a resistive load is always positive. voltage lags current by 908. voltage leads current by 908. the load absorbs power).15.16. indicating that the resistor is an energy consuming element. They have an average value of zero over a complete cycle since the positive and negative waveforms will cancel each other out. In a purely capacitive circuit.278 Understandable electric circuits pR ¼ vi ¼ VI cos 0 ð1 À cos 2otÞ þ VI sin 0 sin 2ot ¼ VI À VI cos 2ot ¼ VIð1 À cos 2otÞ ð9:3Þ The first part VI in (9. the inductor and capacitor do not absorb power. i. It is actually the average power dissipation on the resistive load.16 The waveforms of instantaneous power for L and C loads negative.2 Active power P (or average power) The active power is also known as average power.Methods of AC circuit analysis pL i 0 279 v t Store / Release / Store / Release (a) pC i 0 v t Release / Store / Release / Store (b) Figure 9.e. the component releases energy. The active power is also called true or real power since the power is really dissipated by the load resistor. Instantaneous power for R. which is the product of the RMS voltage and RMS current in an AC circuit.3. L and C components ● ● ● pR ¼ VI À VI cos 2ot pL ¼ VI sin 2ot pC ¼ ÀVI sin 2ot 9. and it can be converted to useful energy such as heat . the average power within one period of time (one full cycle) for a sinusoidal power waveform in an AC circuit. Therefore. they convert or transfer energy between the source and elements. This also indicates that the inductor and capacitor are energy storage elements. etc. Electric stoves and lamps are examples of this kind of resistive load. It can be obtained from integrating for instantaneous power in one period of time. Average power is easy to measure by an AC power metre (an instrument to measure AC power) in an AC circuit. Z Z 1 T 1 T P¼ pðtÞdt ¼ ½VI cos fð1 À cos 2otÞ þ VI sin f sin 2otdot T 0 T 0 Z . and skip the following mathematical derivation process. Since it is the actual power dissipated in the load. The instantaneous power always varies with time and is difficult to measure. average or active power P is used more often in AC sinusoidal circuits. then just keep in mind that P ¼ VI cos f is the equation for average power. Average power is the average value of instantaneous power in one period of time.280 Understandable electric circuits or light energy. so it is not very practical to use. Note: If you haven’t learned calculus. 1 1 T P ¼ VI cos fðotÞ. the voltage leads the current by 908. When active power P 4 0.e. the element absorbs power. when active power P 5 0. . cos 0 ¼ 1Þ Therefore. cos 90 ¼ 0Þ. PL ¼ 0. It consists of the product of RMS values of voltage and current VI and cos f where cos f is called power factor and it will be discussed at the end of this section. the element releases power. ● When f ¼ 0 . and PL ¼ VI cos 90 ¼ 0 ð. active or average power P is a constant. and PR ¼ VI cos 0 ¼ VI ð. so the first part of the integration is a constant VI cos f. since the average power value for a cosine function in one period of time is zero. and ot is a variable. PR ¼ VI ¼ I 2 R ¼ or Vm Im 1 PR ¼ VI ¼ pffiffiffi pffiffiffi ¼ Vm Im 2 2 2 V2 R ● When f ¼ 90 . the voltage and current are in phase. the circuit is a purely inductive circuit. The integration of the second part is zero (integrating for sine function). i. the circuit is a purely resistive circuit.T þ VI sin f sin 2 ot dot ¼ VI cos f ð9:4Þ 0 T T 0 where f is a constant. Therefore. sin 0 ¼ 0Þ .3 Reactive power Q Since the effect of charging/discharging in a capacitor C and storing/releasing energy from an inductor L is that energy is only exchanged or transferred back and forth between the source and the component and will not do any real work for the load. the circuit is a purely resistive circuit: QR ¼ VI sin 0 ¼ 0 ð. cosðÀ90 Þ ¼ 0. and average power dissipated on the load will be zero. Active power P (or average power. and measured in volt-amperes reactive (Var). real power and true power) The active power is the average value of the instantaneous power that is actually dissipated by the load. P ¼ VI cos f When f ¼ 0 When f ¼ 90 When f ¼ À90 1 V2 PR ¼ VI ¼ Vm Im ¼ I 2 R ¼ 2 R PL ¼ 0 PC ¼ 0 Quantity Active power Quantity symbol P Unit Watt Unit symbol W 9. PC ¼ 0. because the physical meaning of the reactive power is the maximum velocity of energy conversion between the energy storing element and the source.Methods of AC circuit analysis ● 281 When f ¼ À90 .e. Also. While energy is converting between the source and energy store elements. The integration of the second part of (9. the circuit is a purely capacitive circuit.3. So the average power dissipated on the load is zero.4) is active or average power. i. and PC ¼ VI cosðÀ90 Þ ¼ 0 ½. It can be expressed mathematically as Q ¼ VI sin f. The first part in (9. and that is the reactive power. ● When f ¼ 0 . denoted as Q. The reactive power Q can describe the maximum velocity of energy transferring between the source and the storage element L or C. the peak value of the second part is reactive power. the current leads the voltage by 908.4) is zero. the load will do not do any actual work. 17. the circuit is a purely capacitive circuit: QC ¼ VI sinðÀ90 Þ ¼ ÀVI Substituting V ¼ IXC or I ¼ ½sinðÀ90 Þ ¼ À1 V into QC gives XC V2 XC QC ¼ ÀVI ¼ ÀI 2 XC ¼ À Since QL is positive (QL 4 0) and QC is negative (QC 5 0). Since VI is the expression of the power equation. and the capacitor produces (releases) reactive power.282 ● Understandable electric circuits When f ¼ 90 . the circuit is a purely inductive circuit: QL ¼ VI sin 90 ¼ VI Substituting V ¼ IXL or I ¼ QL ¼ VI ¼ I 2 XL ¼ V2 XL ð. Apparent power is . If the load Z includes both the resistor and storage element inductor or capacitor. it is called apparent power. the power produced in the load is the product of voltage and current VI. Reactive power Q Q is the maximum velocity of energy conversion between the source and energy storing element.3. Q ¼ VI sin f When f ¼ 0 When f ¼ 90 When f ¼ À90 Quantity Reactive power QR ¼ 0 QL ¼ VI ¼ I 2 XL ¼ V2 XL QC ¼ ÀVI ¼ ÀI 2 XC ¼ À Unit V2 XC Unit symbol Var Quantity symbol Q Volt-amperes reactive 9. the inductor absorbs (consumes) reactive power.4 Apparent power S When the voltage V across a load produces a current I in the circuit of Figure 9. sin 90 ¼ 1Þ V into QL gives XL ● When f ¼ À90 . then VI will be neither a purely active power nor a purely reactive power. Methods of AC circuit analysis I Z + V 283 - Figure 9. Different types of power in AC circuits are summarized in Table 9. Table 9. i. The mathematical expression of apparent power is the product of the source current and voltage.e.2.17 Apparent power the maximum average power rating that a source can provide to the load or maximum capacity of an AC source and is denoted as S.2 Powers in AC circuits Power Instantaneous power Active power Reactive power Apparent power General expression p ¼ VI cos f (1 – cos 2ot) þ VI sin f sin 2ot P ¼ VI cos f R pR ¼ VI – VIcos 2ot PR ¼ VI ¼ 1/2(VmIm) ¼ I2R ¼ V2/R QR ¼ 0 L C pL ¼ pL ¼ VIsin 2ot –VIsin 2ot PL ¼ 0 PC ¼ 0 Q ¼ VI sin f QL ¼ VI ¼ I2XL ¼ V2/XL QC ¼ –VI ¼ –I2XC ¼ V2/XC S ¼ VI ¼ I2Z ¼ V2/Z . S ¼ IV and is measured in VA (volt-amperes). measured in VA. Substituting I ¼ V/Z or V ¼ IZ into apparent power S equation gives: S ¼ I 2Z ¼ V2 Z Usually the power listed on the nameplates of electrical equipment is the apparent power. S ¼ IV ¼ I 2 Z ¼ V 2 =Z where S represents apparent power. Apparent power S S is the maximum average power rating that a source can provide to an AC circuit. it will yield VI = S. If the circuit load is more capacitive (XC > XL . f < 0). It is called the impedance angle.3. and can be derived as follows. f is also in the power triangle. f > 0Þ.284 Understandable electric circuits 9. reactive power and apparent power. if the circuit is more inductive ðX ¼ XL À XC > 0.19 Circuit triangles for a more capacitive circuit The impedance triangle indicates that it has an angle f between resistance R and impedance Z of the circuit. .18(d). These three powers are actually related to one another in a right triangle. the circuit triangles will be opposite to the inductive circuit triangles as shown in Figure 9. Now the question is what are the relationships between these three powers. the active power. Later on. V f • • Z f X VX • I f IX • R (a) I • VR • IR (c) • (b) S f Q P (d) Figure 9. IVX = Q and VRI = P and this can be illustrated as a power triangle as shown in Figure 9.18(a–c) (refer to section 9. then the impedance triangle.18 Circuit triangles for a more inductive circuit If we multiply all quantities on each side of the voltage triangle by the current I. we’ll introduce the power factor cos f.1). and f is also called the power factor angle. voltage triangle and current triangle can be illustrated as shown in Figure 9. inductor and capacitor circuit.19. VR VX (b) • • f R X (a) f f P S (c) f IR • • Z Q V • I IX (d) • Figure 9. For a series resistor. is called the power triangle.5 Power triangle We have discussed three different powers in AC circuits. Impedance angle: f ¼ tanÀ1 Phasor power: Q ¼ S sin f. and represented by cos f. it will give: S ¼ P þ jQ This is known as the phasor power. It also can be obtained from the power triangle as .18 or 9. Q X _ _ _ _ ¼ tanÀ1 ¼ tanÀ1 ðVX =VR Þ ¼ tanÀ1 ðIX =IR Þ P R _ __ _ _ S ¼ V I ¼ I 2 Z ¼ V 2 =Z.3. S¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi P2 þ Q2 . The phasor apparent power can also be expressed as _ V2 _ __ _ S ¼ V I ¼ I 2Z ¼ Z The impedance angle f can be obtained from circuit triangles (either inductive or capacitive circuit) and can be expressed as f ¼ tanÀ1 _ _ Q X VX IX ¼ tanÀ1 ¼ tanÀ1 ¼ tanÀ1 _R _ P R V IR Active power P and reactive power Q can be expressed with the impedance angle f and obtained from the power triangle in Figures 9.e. i.19 as: P ¼ S cos f and Q ¼ S sin f Power triangle S f Q P P ¼ S cos f. _ S ¼ P þ jQ 9. S ¼ P2 þ Q2 _ If expressed by complex numbers.6 Power factor (PF) The ratio of active power P and apparent power S is called the power factor PF.Methods of AC circuit analysis 285 The relationship between different powerspffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi in the power triangle can be obtained from the Pythagoras’ theorem. so the power factor is also zero.20(b)) can increase the power factor of the load. i. the reactive power Q is zero. so the apparent power S is equal to the active power P. and the amount of the active power P can be determined by the power factor cos f. the active power P produced by the source will also decrease accordingly ðP # ¼ S cos f #Þ. For a purely reactive load ðf ¼ Æ90 Þ. i. So increasing the power factor can increase the real power in a circuit.e. the active power produced by the source is the maximum capacity of the source.20 Increasing the power factor . The circuit source will produce active power P to the load. cos f ¼ P=S ¼ 0=S ¼ 0. Therefore. But how to increase the power factor of a circuit? A method called power-factor correction can be used. the range of the power factor cos f is between 0 and 1. cos f ¼ 1 ). S¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi P2 þ Q2 ¼ P2 þ 02 ¼ P and the power factor is 1. This is indicated in the equation of P ¼ S cos f. I R L R L C fʹ (a) (b) S f P (c) QC Q Qʹ Figure 9. cos f ¼ P=S ¼ P=P ¼ 1 This is the maximum value for the power factor cos f.286 Understandable electric circuits PF ¼ P cos f S For a purely resistive circuit ðf ¼ 0 Þ. an inductive load in parallel with a capacitor (Figure 9. i. The power factor is an important factor in circuit analysis. active power P in the circuit is zero.e. and the impedance angle f is between 08 and +908. If the power factor cos f decreases. Since most of the loads of the electrical systems are inductive loads (such as the loads that are driven by a motor). This method can increase the power factor and does not affect the load voltage and current. . and all the energy supplied by the source will be consumed by the load (P ¼ S. If the power factor cos f of the load is the maximum value of 1.e. Methods of AC circuit analysis 287 The power triangle in Figure 9. PT PFT ¼ cos fT ¼ ST Total power ● ● Total active power: PT ¼ P1 þ P2 þ Á Á Á þ Pn Total reactive power: QT ¼ QLT À QCT ¼ ðQL1 þ QL2 þ Á Á ÁÞ À ðQC1 þ QC2 þ Á Á ÁÞ where QLT is the total reactive power for inductors and QCT the total reactive power for capacitors. Power factor (cos f) ● ● ● cos f ¼ P=S ð0 cos f 1. i.3. the total current I will also decrease.7 Total power When calculating the total power in a complicated series–parallel circuit. cos f À dimensionlessÞ: When cos f ¼ 1: All energy supplied by the source is consumed by the load. for instance cos 30 ¼ 0:866 is > cos 60 ¼ 0:5. Since f#! cos f". Power-factor correction: An inductive load in parallel with a capacitor can increase cos f. The total apparent power S can be determined by using QT and PT using the Pythagoras’ theorem.e. and the sum of all the active powers is the total active power PT. .20(c) indicates that when a capacitor C is in parallel with the inductive load. Therefore. and the power factor cos f will increase to cos f0 . 9. the impedance angle will reduce from f to f0 . The difference between QLT and QCT is the total reactive power QT. ST ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi P2 þ Q2 T T The total power factor can be determined by using the total active and reactive power. since I #¼ P=ðV cos f "Þ ðP ¼ S cos f ¼ VI cos fÞ. QLT is the sum of all reactive powers for the inductors and QCT is the sum of all reactive powers for the capacitors.e. This is why increasing the power factor has a significant meaning. determine the active power P and reactive power Q in each branch first. i. This can reduce the source current and line power loss (I2R). the reactive power Q in the circuit will be reduced to Q0 (Q0 ¼ Q 7 QC).. 6: Determine the following values in the circuit shown in Figure 9.21 and plot the power triangle for this circuit. ST ¼ PT ?2 þ QT ?2 (the symbol ST ? ‘?’ indicates an unknown).21 Circuit for Example 9.22 The power triangle for Example 9.39 VA f = 42. QT and ST for the circuit power factor cos f power triangle source current I the capacitance C needed to increase the power factor cos f to 0.87 the source current I0 after increasing the power factor .22. ST = 81.5 Solution: ● ● ● pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi PT ? Total power factor PFT ¼ cos fT ¼ .3° QT = 55 Var PT = 60 W Figure 9. Z1 P1 = 10 W E = 10 V ∠ 0° Z2 P2 = 30 W QC = 15 Var Z3 P3 = 20 W QL = 70 Var Figure 9.5 Example 9.288 Understandable electric circuits pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Total apparent power: ST ¼ P2 þ Q2 T T Total power factor: PFT ¼ cos fT ¼ PT =ST ● ● Example 9.5: Determine the total power factor cos f in the circuit of Figure 9. Total active power: PT ¼ P1 þ P2 þ P3 ¼ ð10 þ 30 þ 20ÞW ¼ 60 W Total reactive power: QT ¼ QLT À QCT ¼ ð70 À 15ÞVar ¼ 55 Var pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Total apparent power: ST ¼ P2 þ Q2 ¼ 602 þ 552 ¼ 81:39 VA T T PT 60 W Total power factor: PFT ¼ cos fT ¼ ¼ % 0:74 ST 81:39 VA Impedance angle: f ¼ cosÀ1 fT ¼ cosÀ1 0:74 % 42:3 The power triangle is shown in Figure 9.23: ● ● ● ● ● ● the total power PT. 24): f ¼ cosÀ1 0:77 % 39:7 ST = 9 084. ðf ¼ tanÀ1 ðQ=PÞÞ Total power: PT ¼ P1 þ P2 þ P3 ¼ ð500 þ 500 þ 6 000ÞW ¼ 7 000 W QT ¼ Q1 þ Q2 þ Q3 ¼ 0 þ 500 Var þ 5 290 Var ¼ 5 790 Var pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ST ¼ PT þ QT ¼ 7 0002 þ 5 7902 % 9 084:3 VA PT 7 000 (b) Power factor PFT ¼ cos fT ¼ ¼ % 0:77 ST 9 084:3 (c) Power triangle (as shown in Figure 9.ST ¼ EI . I ¼ 82:6ff À 39:7 A (Voltage leads current or current lags voltage in the inductive load.6 (d) Source current I: .25) .75 PF ••• E =110 V∠ 0° f = 60 Hz • 5 lamps 100 W/each C Lamps Heating Electric stove Figure 9.I¼ ST 9 084:3 VA ¼ % 82:6 A E 110 V _ Therefore.23 Circuit for Example 9.Methods of AC circuit analysis ••• 289 I XL = 5Ω R = 5Ω 10 A 60 kW 0.6 Solution: (a) Lamp: P1 ¼ 5  100 W ¼ 500 W Heating: P2 ¼ I 2 R ¼ ð10 AÞ2 ð5 OÞ ¼ 500 W Q2 ¼ I 2 XL ¼ ð10 AÞ2 ð5 OÞ ¼ 500 Var Electric stove: P3 ¼ 6 kW ¼ 6 000 W.3 VA f = 39.24 Power triangle for Example 9.7° PT = 7 000 W QT = 5 790 Var Figure 9.) (e) The capacitance C that needs to increase the power factor to 0. so f ¼ À39:7 . f ¼ cosÀ1 0:75 % 41:4 Q3 ¼ P3 tan f ¼ ð6 000 WÞðtan 41:4 Þ % 5 290 Var.87 can be determined by the following way: 1 V2 V2 1 C¼ ) QC ¼ À ) XC ¼ ) XC ¼ À XC QC ? 2pfXC ? 2pfC QT ¼ QC þ Q0T ? ) Q0T ¼ PT tan f0 ? ) f0 ¼ cosÀ1 0:87 (as shown in Figure 9. 9. (f) The source current I 0 after increasing the power factor can be determined by the following expression: P ¼ S cos f ¼ IE cos f So.25. So the source current can decrease 9. after a capacitor is in parallel and the power factor is increased.6 A from step (d).5 A). the source current is I 0 = 73. I0 ¼ PT 7000 W ¼ % 73:1 A E cos f0 110 V cos 29:5 Comparing with the original source current I = 82.5° Qc Q′T = 3 960 Var QT P Figure 9.1 A = 9.1 A.3 VA f = 29. such as the branch .87.4 Methods of analysing AC circuits All analysis methods that we have learned for analysing DC circuits with one or two more sources can also be used for analysing AC circuits.87 should be 400 µF. QC can be obtained from Figure 9.25 New power factor angle Therefore. This can reduce the line power loss (I2R) and utilize the capacity of the source more efficiently. The reactive power can be determined from the above expression as Q0T ¼ PT tan f0 ¼ 7 000 tan 29:5 % 3 960 Var The new power factor angle f0 is shown in Figure 9.) XC ¼ Therefore C¼ 1 1 ¼ % 0:0004 F ¼ 400 mF 2pfXC 2pð60 HzÞð6:61 OÞ That is the capacitance C needed to increase the power factor to 0. to increase the power factor to 0. the power factor angle should be reduced to f0 ¼ cosÀ1 0:87 % 29:5 .6 A 7 73.290 Understandable electric circuits ST = 9 084.5 A (I 7 I 0 = 82.25: QC ¼ QT À Q0T ¼ 5 790 À 3 960 ¼ 1 830 Var ÀV 2 E2 À1102 V ¼À ¼ % 6:61 O QC QC 1 830 Var (The voltage across XL and R is equal to E. 9. But the phasor form will be used to represent the circuit quantities. so the number of KVL equations can be reduced. Thevenin’s and Norton’s theorems. and the numbers of KVL equations should be equal to the numbers of mesh (windowpanes).26. superposition theorem. Since these analysis methods have been discussed in detail for DC circuits (chapters 4 and 5). Example 9. etc. as shown in Figure 9.Methods of AC circuit analysis 291 current analysis. E2 ¼ IR3 ¼ ð2:5 Aff0 Þð4 OÞ ¼ 10 Vff 0 : _ _ 1. mesh analysis. and determine each mesh current. Identify each mesh and label the reference directions for each mesh current clockwise. _ _ _ Mesh 1: ðZ1 þ Z2 ÞI1 À Z2 I2 ¼ ÀE1 _ _ _ Mesh 2: ÀZ2 I1 þ ðZ2 þ Z3 ÞI2 ¼ ÀE2 .1 Mesh current analysis The procedure for applying the mesh current analysis method in an AC circuit: 1. The procedure for applying the mesh current analysis method in an AC circuit is demonstrated in the following example. Label all the reference directions for each mesh current I1 and I2 (clockwise). and the number of KVL is equal to the number of meshes (there are two meshes in Figure 9.26(b). Apply KVL around each mesh of the circuit. and each mutual-impedance _ _ voltage as negative in KVL ðSV ¼ SEÞ. There.26(b)). Sign each self-impedance voltage as positive.7: Use the mesh current analysis method to determine the mesh current I1 in the circuit of Figure 9. 2. Solve the simultaneous equations resulting from step 2. Solution: Convert the current source to the voltage source (connect R and X to Z) as _ _ shown in Figure 9. Write KVL around each mesh (windowpane). ● If the circuit has a current source. some examples will be presented to use these methods in AC circuits or networks.26(b). 3. ● Mutual-impedance: An impedance that is located on the boundary of two meshes and has two mesh currents flowing through it. the source current will be the same with the mesh current. Reviewing chapters 4 and 5 before reading the following contents is highly recommended. ● Self-impedance: An impedance that only has one mesh current flowing through it. Sign each selfimpedance voltage as positive and each mutual-impedance voltage as negative in KVL equations. node voltage analysis. if there is any.4. 2. Note: ● Convert the current source to the voltage source first in the circuit. Z2 = 2 O and Z3 = (4 þ j8)O _ _ so.7 Substitute the following values of Z1. Solve the simultaneous equations resulting from step 2 using the determinant _ method. Z2 and Z3 into the above equations: Z1 = (6 7 j6)O.292 Understandable electric circuits • E1 = 10 V∠ 0° R1 = 6 Ω XC = 6 Ω – + R2 = 2 Ω XL = 8 Ω R3 = 4 Ω • I = 2.5 A ∠ 0° (a) Z3 • E1 = 10 V∠ 0° R1 = 6 Ω XC = 6 Ω – + I1 Z1 • XL = 8 Ω R3 = 4 Ω Z2 R2 = 2 Ω + I2 • – E2 = 10V∠ 0° • (b) Figure 9.26 Circuits for Example 9. and determine the mesh current I1 : . ð8 À j6ÞI1 À 2I2 ¼ À10 V _ _ À2I1 þ ð6 þ j8ÞI2 ¼ À10 V 3. . . À10ff 0 À2 . . . . À10ff 0 6 þ j8 . _ . % 1:18ff À 151:9 A I1 ¼ . . 8 À j6 À2 . . . . À2 6 þ j8 . . ● Assign an arbitrary reference direction for each branch current (this step can be skipped if using the inspection method). 9.2 Node voltage analysis The following is the procedure for applying the node analysis method in an AC circuit. 1.4. Label the circuit: ● Label all the nodes and choose one of them to be the reference node. 2. Label nodes a.5 Ω + º –j1 Ω –2 V ∠0 c b 1Ω a i2 = 2 ∠0 º = –2j A 1∠90 º 5Ω d 5Ω 2. Solve the simultaneous equations and determine each nodal voltage.27(b). Example 9. ● Method 1: Write KCL equations and apply Ohm’s law to the equations (assign a positive sign (þ) to the self-impedance voltage and entering node current. Determine the total contribution by calculating the algebraic sum of all contributions due to single sources.3. ● Method 2: Convert voltage sources to current sources and write KCL equations using the inspection method. c and d.27(a) to Figure 9. .3 Superposition theorem The following is the procedure for applying the superposition theorem in an AC circuit: 1. replace the voltage source with the short circuit (placing a jump wire). 9. KCL equations are shown in Table 9.8: Write node equations for the circuit in Figure 9. Convert two voltage sources to current sources from Figure 9. 3. Analyse and calculate this circuit by using the single source method. Apply KCL to all n71 nodes except for the reference node (n is the number of nodes). Turn off all power sources except one.27(a). and choose ground d to be the reference node as shown in Figure 9.8 1.27 Circuits for Example 9.Methods of AC circuit analysis 293 2. Redraw the original circuit with a single source.5 Ω d –j1 Ω j1 Ω b 5 V ∠0º –j1 Ω º i1 = 5 ∠0 º = 5j A 1∠–90 (b) (a) Figure 9. 3. b. 3. and negative sign (7) for the mutual-impedance voltage and exiting node current).4. and repeat steps 1 and 2 for the other power sources in the circuit. Three equations can solve three unknowns that are node voltages.27(a). i. 2. and replace the current source with an open circuit. The procedure for applying the node voltage analysis method in an AC circuit is demonstrated in the following example. a –j1 Ω c 1Ω – + j1 Ω 2. and write KCL equations to n 7 1 ¼ 4 7 1 ¼ 3 nodes by inspection (method 2).e. 3 _ Vb 7 KVL Equations for example 9.8 _ Vc 7 7 _ Va _ ð1=j1ÞVb þ _ I ¼ ¼ ¼ Node a 1 1 1 _ þ þ Vb 2:5 j1 Àj1 _ ð1= À j1ÞVc þ 1 1 1 _ þ þ Va 5 Àj1 j1 _ ð1= À j1ÞVc À2j 5j À ðÀ2jÞ À5j Node b _ ð1= À j1ÞVa 7 Node c _ ðÀ1= À j1ÞVa _ ð1= À j1ÞVb 1 1 1 _ þ þ Vc 1 Àj1 Àj1 After simplifying þ þ 7 _ 0:2 Va 7 7 þ _ jVb _ 0:4Vb _ jVb _ jVc _ jVc _ ð1 þ 2jÞVc ¼ ¼ ¼ Àj2 j7 Àj5 _ jV a _ ÀjVa .Table 9. 9 Solution: _ 1.28(c).Methods of AC circuit analysis 295 (The result should be positive when the reference polarity of the unknown in the single source circuit is the same with the original circuit.28(b).5 Ω I = 2A ∠0° • (a) R = 10 Ω R = 10 Ω = - Z1 • + E = 10 V ∠0° (b) ' VC • + XC = 7.5 Ω Z2 + Z1 ' VC' • + XC = 7.28(c): _ _0 VC ¼ E _ 00 _ VC ¼ IðZ1 ==Z2 Þ Z1 ==Z2 ¼ Z1 Z2 10ðÀj7:5Þ ¼ O 10 À j7:5 Z1 þ Z2 75ff À 90 O ¼ 6ff À 53:13 O % 12:5ff À 36:87 _ 00 _ VC ¼ IðZ1 ==Z2 Þ ¼ ð2ff0 AÞð6ff À 53:13 OÞ ¼ 12ff À 53:13 V . and calculate Vc0 : Z2 7:5 Off À 90 ¼ À10Vff0 Z1 þ Z2 10 O À j7:5 O À75ff À 90 ¼ V ¼ À6ff À 53:13 V 12:5ff À 36:87 _ 2. When the current source I is applied to the circuit only and the voltage source _ E is replaced by a jump wire.9: Determine Vc in circuit as shown in Figure 9. the circuit is as shown in Figure 9. _ Example 9. Cal_ culate Vc00 in Figure 9. Choose E to apply to the circuit first. otherwise it should be negative. and use an open circuit to replace the _ _ current source I as shown in Figure 9.28 Circuits for Example 9.28(a) by using the superposition theorem.5 Ω Z2 I = 2A ∠0° • - (c) Figure 9. R = 10 Ω E = 10 V ∠0 • Z1 ° + VC • + - Z2 XC = 7.) The procedure for applying the superposition theorem in an AC circuit is demonstrated in the following example. and mark the letter a and b on the two terminals. i. Calculate the sum of voltages VC and VC : _ _0 _ 00 VC ¼ VC þ VC ¼ À6ff À 53:13 V þ 12ff À 53:13 V ¼ ½À6 cosðÀ53:13 Þ À 6j sinðÀ53:13 Þ þ 12 cosðÀ53:13 Þ þ 12j sinðÀ53:13 ÞV % ½À3:6 þ j4:8 þ 7:2 À j9:6V ¼ ð3:6 À j4:8ÞV ¼ 6ff À 53:13 V 9.4 Thevenin’s and Norton’s theorems The following is the procedure for applying Thevenin’s and Norton’s theorems in an AC circuit.4. Determine the equivalent impedance ZTH or ZN: It should be equal to the equivalent impedance when you look at it from the a and b terminals when all sources are turned off or equal to zero.29(a) by using Thevenin’s theorem. and connect the load (or unknown current or voltage branch) to a and b terminals of the equivalent circuit. and a current source should be replaced by an open circuit. Then the load voltage or current can be calculated. Solution: 1. Determine Norton’s equivalent current IN: It equals the short circuit current for the original linear two-terminal network of a and b. 4. Open the load branch and remove ZL. and use Norton’s theorem to check the answer. IN ¼ Isc. _ 2.e.) i. VTH ¼ Vab.29(b): ZTH ¼ Zab ¼ Z3 þ Z4 þ Z1 ==Z2 Àj2:5ð2:5 þ j2:5Þ 6:25 À j6:25 ZTH ¼ 1 À j1 þ O ¼ 1 À j1 þ O Àj2:5 þ ð2:5 þ j2:5Þ 2:5 ¼ ð1 À j1 þ 2:5 À j2:5ÞO ¼ ð3:5 À j3:5ÞO % 4:95ff À 45 O .e. Determine Thevenin’s equivalent impedance ZTH (the voltage source E is replaced by a short circuit) in Figure 9. i.29(b). Open and remove the load branch (or any unknown current or voltage branch) in the network.296 Understandable electric circuits _0 _ 00 3. The procedure for applying Thevenin’s and Norton’s theorems method in an AC circuit is demonstrated in the following example. Plot Thevenin’s or Norton’s equivalent circuits. 2. 1. ZTH ¼ ZN ¼ Zab 3.10: Determine the current IL in the load branch of Figure 9. (A voltage source should be replaced by a short circuit. ● ● Determine Thevenin’s equivalent voltage VTH: It equals the open circuit voltage from the original linear two-terminal network of a and b. and label a and b on the terminals of the load branch as shown in Figure 9. _ Example 9.e. 5 Ω b Z4 VTH • b E = 5 V∠0° d Z3 1 Ω (c) Figure 9. .5 Ω + - 2. Connect the load ZL to a and b terminals of the equivalent circuit and calculate the load _ current IL .5 Ω c -j2. VTH ¼ Vcd ¼ E Z2 2:5 þ j2:5 ¼ 5ff0 V O Àj2:5 þ ð2:5 þ j2:5Þ Z1 þ Z2 % 5ff0 ð1:414ff45 Þ % 7:07ff45 V 4.29(c).5 Ω Z4 Z2 Z3 1Ω (b) RTH = Zab • • 297 -j2.5 Ω Z1 Z2 -j Ω a j2.5 Ω E = 5V∠0° • j2.5) Ω 1Ω (a) -j Ω a -j2. Determine Thevenin’s equivalent voltage V TH by using Figure 9.29(d).10(a)–(c) _ 3.5 Ω Z1 j2.Methods of AC circuit analysis -j Ω IL ZL = (1.29(c) to calculate the open circuit voltage across terminals a and b: _ _ _ VTH ¼ Vab ¼ Vcd _ Since I ¼ 0 for Z3 and Z4 in Figure 9.5 Ω 2.29 Circuits for Example 9. voltages across Z3 and Z4 are also zero _ _ _ .5 + j3. Plot Thevenin’s equivalent circuit as shown in Figure 9.5 Ω 2. 29(e. Z2 _ _ _ (the current-divider rule). IN ¼ ISC ¼ I Z2 þ ðZ3 þ Z4 Þ There.5 + j3. Use Norton’s theorem to check the load current IL : Determine the load current _ IL on the terminals of a and b in Figure 9. ● Norton’s equivalent impedance ZN: ZN ¼ ZTH ¼ 3.5) Ω b • + - Figure 9. .95 ff7458 ● Norton’s equivalent current IN: It is equal to the short circuit current for the original two-terminal circuit of a and b (as shown in Figure 9.07 ∠45 V ° • IL ZL = (1.29(a) as seen by ZL.298 Understandable electric circuits a ZTH = 4.29(e. Z2 _ _ IN ¼ I Z2 þ ðZ3 þ Z4 Þ ¼ 1:54ff68:2 A ð2:5 þ j2:5ÞO % 1:43ff90 A ½2:5 þ j2:5 þ ð1 À j1ÞO _ 6.29(d) Thevenin’s equivalent circuit for Example 9.5 7 j3. IL is the same by Norton’s theorem as the method by using Thevenin’s theorem (checked).5 = 4. II) by using Norton’s equivalent circuit. Determine Norton’s equivalent circuit in Figure 9.10 _ IL ¼ _ VTH 7:07ff45 V 7:07ff45 V ¼ ¼ ZTH þ ZL ð3:5 À j3:5ÞO þ ð1:5 þ j3:5ÞO 5O % 1:4ff45 A 5. I). _ E 5ff0 V _ I¼ ¼ ð2:5 þ j2:5Þð1 À j1Þ Z1 þ Z2 ==ðZ3 þ Z4 Þ O Àj2:5 þ ð2:5 þ j2:5Þ þ ð1 À j1Þ 5ff0 V % 1:54ff68:2 A ¼ ðÀ j2:5 þ ðj5Þ=ð3:5 þ j1:5ÞÞO Therefore. ZN 4:95ff À 45 O _ _ IL ¼ IN ¼ 1:43ff90 A Z N þ ZL ½ð3:5 À j3:5Þ þ ð1:5 þ j3:5ÞO 4:95ff À 45 % 1:4ff45 A ¼ 1:43ff90 A 5 _ Therefore.95 Ω∠-45° VTH = 7. 5 Ω E = 5 ∠0° V • ZN = 4.95 ∠-450 Ω IN = 1.5 Ω • 299 -j Ω a Z4 Z2 IN Z3 b 1Ω (I) a • Z1 j 2.Methods of AC circuit analysis I -j 2.5 Ω 2.10 Summary Impedance and admittance Component R L Impedance _ _ Z ¼ V =I ZR ¼ R ZL ¼ j XL ZC ¼ 7j XC Z ¼ z fff ¼ R þ jX pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi z ¼ R2 þ X 2 f ¼ tanÀ1 X R Admittance Y ¼ 1/Z YR ¼ G YL ¼ 7jBL YC ¼ jBC Y ¼ y ff fy ¼ G þ jB pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi y ¼ G 2 þ B2 fy ¼ tanÀ1 B G Conductance and susceptance Conductance: G ¼ 1/R Inductive susceptance: BL ¼ 1/XL Capacitive susceptance: BC ¼ 1/XC Reactance: X ¼ XL À XC and susceptance: B ¼ BC À BL C XL ¼ oL.5) Ω • b (II) Figure 9. oC j¼ 1 Àj .29(e) Norton’s equivalent circuit for Example 9.43 ∠90. XC ¼ 1 .10 A • IL ZL = (1.5 + j3. B ¼ 0 (BL ¼ BC). Impedances in series and parallel ● Impedances in series: Zeq ¼ Z1 þ Z2 þ Á Á Á þ Zn Impedances in parallel: Zeq ¼ Zeq ¼ 1 Yeq 1 ¼Z1 ==Z2 ==ÁÁÁ==Zn ð1=Z1 Þþð1=Z2 ÞþÁÁÁþð1=Zn Þ ● Yeq ¼Y1 þY2 þÁÁÁþYn ● ● Z1 Z2 ¼ Z1 ==Z2 Z1 þ Z2 Z1 _ _ Z2 _ _ E V2 ¼ E Voltage-divider rule for impedance: V1 ¼ Z1 þ Z2 Z1 þ Z2 Z2 _ Z1 _ _ _ Current-divider rule for impedance: I1 ¼ IT I2 ¼ IT Z1 þ Z2 Z1 þ Z2 _ _ _ The phasor forms of KVL and KCL: SI ¼ 0 Iin ¼ Iout Two impedances in parallel: Zeq ¼ _ SV ¼ 0 _ _ _ _ V1 þ V2 þ Á Á Á þ Vn ¼ E . current and power triangles ● For a more inductive circuit: • Z f R ● X V f I • • VX • I • f IR • IX • • S f Q VR P For a more capacitive circuit: R f f VR f • • P f IR Q IX • • Z X V VX • S I • Impedance angle: f ¼ tanÀ1 ● _ _ VX IX X Q ¼ tanÀ1 ¼ tanÀ1 ¼ tanÀ1 _ _ R P VR IR Characteristics of impedance and admittance: ● The inductive load: X 4 0 (XL 4 XC). f 40. fy 4 0 ● The resistive load: X ¼ 0 (XC ¼ XL). f ¼ 0. fy ¼ 0.300 ● Understandable electric circuits Impedance. f 50. B 4 0 (BC 4BL). fy 5 0 ● The capacitive load: X 5 0 (XC 4 XL). B 5 0 (BL 4BC). voltage. Total power ● Total active power: PT ¼ P1 þ P2 þ Á Á Á þ Pn ● Total reactive power: QT ¼ QLT À QCT ¼ ðQL1 þ QL2 þ Á Á Á Þ À ðQC1 þ QC2 þ Á Á Á Þ (QLT is the total reactive power of inductors. Q ¼ S sin f.) pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ● Total apparent power: ST ¼ P2 þ Q2 ST ¼ S1 þ S2 þ Á Á Á þ Sn T T PT ● Total power factor: PFT ¼ cos fT ¼ ST P . S ¼ S f pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi P2 þ Q2 Q ● ● _ __ _ _ Phasor power: S ¼ R þ jQ ¼ V I ¼ I 2 Z ¼ V 2 =Z Power factor P ● Power factor: PF ¼ cos f ¼ ð0 cos f 1Þ S ● Power-factor correction: A capacitor in parallel with the inductive load can increase the power factor (the power factor angle f# ! cos f") P _ _ # I #¼ _ ! line power loss ðI 2 RÞ # ! cos f" ! line current I V cos f " utilize capacity of the source more efficiently. and QCT is the total reactive power of capacitors.Methods of AC circuit analysis 301 Power of AC circuits Power Instantaneous power Active power Reactive power Apparent power General form p ¼ ui ¼VI cos f (1 – cos 2ot) þ VI sin f sin 2ot P ¼ VI cos f Q ¼ VI sin f R pR ¼ VI – VI cos 2ot PR ¼ VI ¼ 1/2VmIm ¼ I2R ¼ V 2/R QR ¼ 0 S ¼ VI ¼ I2Z ¼ V2/Z L C pL ¼ pC ¼ VI sin 2ot –VIsin 2ot PL ¼ 0 PC ¼ 0 QL ¼ VI QC ¼ –VI ¼ I 2 XL ¼ –I2XC 2 ¼ V /XL ¼ V2/XC Quantity Instantaneous Power Active Power Reactive power Apparent power Quantity Symbol p P Q S Unit Watt Watt Volt-amperes reactive Volt-Amperes Unit symbol W W Var VA ● Power: P ¼ S cos f. i. B A 2π 0 π Figure L9.5 grids.1. Experiment 9: Sinusoidal AC circuits Objectives ● ● ● To become familiar with the operation of an oscilloscope for measuring the sinusoidal AC voltage. and then calculate the branch current using Ohm’s law. Since there is one grid difference between waveform A and B as shown in Figure L9.1. To verify theoretical calculations of the AC series–parallel circuits through experiment.5 6 608 ¼ 308. 0. the phase difference will be 308. its impact on the circuit measurement and calculation may be negligible.) Use the oscilloscope to measure the current in the inductive or capacitive branch (indirect measurement): Connect a small resistor.302 ● Understandable electric circuits Analysis methods for AC sinusoidal circuits All analysis methods that are used to analyse DC circuits with one or two more sources can also be used to analyse AC circuits. determine the phase difference of waveforms A and B in Figure L9. in series with the inductor or the capacitor. ZL ¼ jXL. Since the sensing resistance is very small. 1 1 ¼ oC 2pfC ZR ¼ R.e. called a sensing resistor. (If the distance between A and B is 0. . ZC ¼ 7jXC XL ¼ oL ¼ 2pfL.1 Phase difference Solution: Each grid is 608 ð360 =6 grids ¼ 60 =gridsÞ. To become familiar with the operation of an oscilloscope for measuring the phase difference of two waveforms. XC ¼ Use the oscilloscope to measure the phase difference f (with dual-channel CH I and CH II): Background information ● ● ● Example L9. the phase difference of waveforms A and B is 608. Measure the voltage across the sensing resistor.1. If these two waveforms appear on the screen of the oscilloscope and occupy six horizontal grids.1: There are two sinusoidal waveforms A and B with complete cycles of 2p(3608) as shown in Figure L9. inductor and capacitor.2 XC Formula for calculation Calculated value * XL Zeq* Refer to chapter 9. Use the measured R.1 mH C 3 600 pF 2.Methods of AC circuit analysis 303 Equipment and components ● ● ● ● ● ● ● ● ● Multimeter Breadboard Function generator Oscilloscope Z meter or LCZ meter Switch Resistors:15 O (two) and 510 O Inductor: 1.1.2 Experiment circuit Table L9.2 .2 (at frequency of 80 kHz) and record the results in Table L9.1 mH Capacitor: 3 600 pF Procedure 1.2 on the breadboard. XL and Zeq of the circuit shown in Figure L9.2.1 R Nominal value Measured value 510 O RL 15 O RC 15 O L 1. and record in Table L9. Table L9. Connect a circuit as shown in Figure L9. L and C values to calculate XC. section 9. Use the multimeter (ohmmeter function) and Z meter or LCZ meter to measure the values of the resistor.1 mH R = 510 Ω B C Figure L9. A C = 3 600 pF E = 3 sin ωt V f = 80 kHz L = 1. Connect the oscilloscope probe CH II to point E of the circuit in Figure L9. Note: The oscilloscope probe CH I is still connected to point A of the circuit in Figure L9.4. . Connect the oscilloscope probe CH I to the point A in the circuit of Figure L9.304 Understandable electric circuits 3. Adjust the output voltage of the function generator to 6 V peak–peak value (Ep7p ¼ 6 V). Then use the oscilloscope to observe and determine the phase difference fR of resistor voltage VR relative to source voltage E. Connect a 15 O resistor RL (sensing resistor) to the inductive branch as shown in Figure L9.3. and record in Table L9.3. 8.3. Use the measured VR value to calculate the current IR in the resistive branch (Ohm’s law) and record it in Table L9.3 Resistive branch Parameter VR Measured value IR fR Inductive branch VRL IL fL Capacitive branch VRC IC fC 5. Adjust the frequency of the function generator to 80 kHz. Record the result in Table L9. Use the measured VRL to calculate the current IL in the inductor branch (Ohm’s law) and record it in Table L9.3.3. and then measure the output sinusoidal voltage of the function generator. Inductor branch 6.2. connect the probe ground of the oscilloscope to the ground of the function generator.2 (choose DUAL channel coupling for the oscilloscope). Connect the oscilloscope probe CH II to the point D in the circuit of Figure L9. Table L9.4. Connect the oscilloscope probe CH II to point B in the circuit of Figure L9.3. Capacitor branch 9. measure the peak voltage on resistor RC and record the measurement as VRC in Table L9. Resistor branch 4. and measure the peak voltage on resistor RL and record the result as VRL in Table L9.3. Connect a 15 o resistor RC to the circuit as shown in Figure L9. 7.3. 10. Then use the oscilloscope to observe and determine the phase difference oL of VRL relative to the source voltage E.3.3. then measure the voltage across the resistor VR (peak value) and record in Table L9. 3. _ _ _ 12.Methods of AC circuit analysis 305 11. Are there any significant differences? If so. Calculate the branch currents IR . Then use the oscilloscope to observe and determine the phase difference cC of capacitor voltage VRC relative to source voltage E and record in Table L9.2 in phasor form (use peak values) and record in Table L9.3. Table L9. IL and IC in Table L9. Compare the measured values and calculated values.4. Use the measured VRC to calculate the capacitor current IC (Ohm’s law) and record it in Table L9.3 to the phasor form and record in Table L9. IL and IC in the circuit of Figure L9.4 (as the measured value). Convert the measured values IR. Conclusion Write your conclusions below: .4 I_R Formula for calculations Calculated value Measured value I_L I_C Phasor form 13. explain the reasons. . Resonant circuits are simple combinations of inductors.Chapter 10 RLC circuits and resonance Objectives After completing this chapter. since the capacitor or inductor voltage/current in a resonant circuit could be much higher than the source voltage or current. capacitors. This is why the resonant circuit is one of the most important circuits in electronic communication systems. as shown in Figure 10. So it is very important to analyse and study resonant phenomena and to know its pros and cons. resonant voltage. Resonance may also damage the circuit elements if it is not used properly. you will be able to: ● ● ● ● ● ● understand concepts and characteristics of series and parallel resonance determine the following quantities of series and parallel resonant circuits: resonant frequency. when the capacitor reactance XC is equal to the inductor reactance XL. voltage and impedance for series and parallel resonant circuits understand characteristics of the selectivity in series and parallel of resonant circuits understand the actual parallel resonant circuits understand the applications of the resonant circuits The resonant phenomena that will be introduced in this chapter have a wide range of applications in electrical and electronic circuits. 10. bandwidth and quality factor plot the frequency response curves of current. resonant current.1. However.1 Series resonance 10.1. resonant impedance. inductor and capacitor (RLC) circuit. a small input signal can produce a large output signal when resonance appears in a circuit. . particularly in communication systems.1 Introduction Resonance may occur in a series resistor. resistors and a power source. Series resonance XL ¼ XC. X ¼ 0. This is the characteristic of the series resonant circuit. when resonance occurs in a series RLC circuit. i. or when reactance X is zero (X ¼ XL 7 XC ¼ 0). and the equivalent impedance of the series RLC circuit will be the lowest (Z ¼ R).e. the energy of the reactive components in the circuit will compensate each other (XL ¼ XC). That is. resonance will occur in the RLC series circuit. Resonant frequency: fr ¼ 1=2p LC pffiffiffiffiffiffiffi Resonant angular frequency: or ¼ 1= LC Frequency of series resonance ffiffiffiffiffiffiffi p . the equivalent or total circuit impedance Z is equal to the resistance R. (The subnotation ‘r’ stands for resonance.1 An RLC series circuit When the magnitudes of inductive reactance XL and capacitive reactance XC are equal (XL ¼ XC).2 Frequency of series resonance The angular frequency of the series resonant circuit can be obtained from 1 XL ¼ XC or oL ¼ oC pffiffiffiffiffiffiffi Solving for o gives or ¼ 1= LC. Z ¼ R 10. _ Z ¼ R þ jðXL À XC Þ ¼ R Under the above condition. solving for f gives the series resonant frequency as 1 fr ¼ pffiffiffiffiffiffiffi 2p LC This is a very important equation for the series resonance. The resonant frequency fr is dependent on the circuit elements inductor (L) and capacitor (C).1.308 Understandable electric circuits R L • I • VR • VL C • VS VC • Figure 10.) Since o ¼ 2pf . meaning that it may produce or remove resonance by adjusting the inductance L or capacitance C in the RLC series circuit. and the resonant current will be _ _ _ V V I¼ ¼ Z R Therefore.3 The response curve of I vs. the only opposition to the flow of the current is resistance R. f for series resonance 10. XL ¼ XC.1. i. Z 0 fr f Figure 10.e. the impedance is minimum and current is maximum in a series resonant circuit. This is illustrated in Figure 10. the circuit’s equivalent impedance is at the minimum (Z ¼ R). When f ¼ fr. when f ¼ fr.3 Impedance of series resonance As previously mentioned.1.2 The response curve of Z vs. the impedance Z is at the lowest point on the curve. Figure 10. when series resonance occurs.3 illustrates the response curve of current I versus frequency f in the series resonant circuit. and the current is at the highest point on the curve when f ¼ fr. which is the response curve of the impedance Z versus frequency f in the series resonant circuit.2.RLC circuits and resonance 309 10. I • 0 fr f Figure 10.4 Current of series resonance When resonance occurs in a series RLC circuit. f for series resonance I and Z of series resonance ● ● Impedance is minimum at series resonance: Z ¼ R _ _ _ _ Current is maximum at series resonance: I ¼ V =Z ¼ V =R . the impedance of the circuit is equal to the resistance (Z ¼ R). XL < XC XL > XC XL Z ( Z = R) 0 XC fr f Figure 10. A phasor diagram of the series resonant circuit is illustrated in Figure 10.1. and the phase difference between E and I is in phase (since V zero (f ¼ 0). _ I 10. V _ _ The resistor voltage is equal to the source voltage (VR ¼ E) since X ¼ 0 _ and source voltage E are also _ when series resonance occurs.5 Phasor diagram of series resonance An RLC series resonant circuit is equivalent to a purely resistive circuit since Z ¼ R. +j VL 0 VC • • I • VR = E • • + Figure 10.310 Understandable electric circuits 10. The capacitor and inductor voltages in the series resonant circuit are _ _ equal in magnitude but are opposite in phase since XL ¼ XC .5.1. Thus. f .6 Response curves of XL. _ and E are in phase.5 Response curves of XL. and f ¼ 0.4. VL ¼ jXL IL and _ _ C ¼ ÀjXC IC . XC and Z vs. capacitive reactance XC and impedance Z versus frequency f are illustrated in Figure 10.4 Phasor diagram of the series resonant circuit Phasor relationship of series resonance ● ● _ _ VL and VC are equal in magnitude but opposite in phase. the current I _ _ _ _ R and I in phase). XC and Z versus f The response curves of the inductive reactance XL. The phase response of the series resonant circuit can be illustrated in Figure 10.6. When f 4 fr. XC and Z versus f show that when the circuit frequency is below the resonant frequency fr. the inductive reactance XL is lower than the capacitive reactance XC and the circuit appears capacitive. as frequency increases. XC and f are inversely proportional ðXC ¼ 1=2pfCÞ. When f ¼ fr. the circuit is more capacitive XL 5 XC. The following characteristics of the series resonant circuit can also be obtained from Figure 10. XL ¼ 0. 10. Z ¼ Zmin ¼ R: the circuit is purely resistive and resonance occurs. Only when the circuit frequency is equal to the resonant frequency fr. XL 4 XC: the circuit is more inductive. the voltage lags current. When frequency f is zero in the circuit. i. qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðZ ¼ R2 þ ð2pfL À ð1=2pfCÞÞ2 ¼ R2 þ ð0 À 1Þ2 ¼ R þ 12 ) 1Þ. When the circuit frequency is above the resonant frequency fr. I ¼ Imax. The response curves of XL. voltage and current are in phase. voltage leads current. Impedance Z is equal to the circuit resistance R and has a minimum value.6. and the phase difference is between zero and negative 908 (7908 f 0). as frequency increases XC decreases. Z ¼ R. and the circuit appears purely resistive. and the circuit appears more inductive. . ● ● ● When the frequency of the circuit is above the resonant frequency fr.1. the resonance occurs in the circuit.5.e. and XL increases. XL ¼ XC. XC and Z approach infinite. These can be summarized as follows: Characteristics of series resonance ● ● ● When f 5 fr.7 Phase response of series resonance The phase response of the series resonant circuit can also be obtained from Figure 10. and the phase difference is zero (f ¼ 0).RLC circuits and resonance ● 311 ● ● XL and f are directly proportional ðXL ¼ 2pfLÞ. XL ¼ XC. ● When the frequency increases from the resonant frequency fr to infinite. the inductive reactance XL is higher than the capacitive reactance XC. XL 5 XC: the circuit is more capacitive. When the frequency of the circuit is below the resonant frequency fr. and the phase difference is between zero and positive 908 (0 f 908). When the frequency of the circuit is equal to the resonant frequency fr. the phase angle approaches positive 908. the circuit is more inductive XL 4 XC. i.e. e. The lower the energy consumption of a resistor (power loss) in a circuit. i.8 Quality factor There is an important parameter known as quality factor in the resonant circuit. and the better the quality of the resonant circuit. . the phase angle approaches negative 908. If substituting the equations of the reactive power and average power into the quality factor equation (10.6 Phase response of the series resonant circuit ● When the frequency decreases from the resonant frequency fr to zero. which is denoted as Q.1). The expression of the phase angle is f ¼ tanÀ1 X X L À XC 2pfL À ð1=2pfCÞ ¼ tanÀ1 ¼ tanÀ1 R R R Phase response of series resonance ● ● When f ! 1. f ! þ90 When f ! 0. Quality factor Q ¼ Reactive power=average power ð10:1Þ The quality factor can be used to measure the energy that a circuit stores and consumes. The quality factor is defined as the ratio of stored energy and consumed energy in physics and engineering. the quality factor of the series resonance will be obtained as follows: Q¼ I 2 XL XL oL ¼ ¼ R I 2R R where R is the total or equivalent resistance in the series circuit. the higher the quality factor.1. so it is the ratio of the reactive power stored by an inductor or a capacitor and average power consumed by a resistor in a resonant circuit.312 Understandable electric circuits f +90° 0 –90° fr f Figure 10. f ! À90 10. the quality factor Q can also be expressed by the capacitive reactance and the resistance as Q¼ XC 1 ¼ oCR R The quality factor can be used to judge the quality of an inductor (or coil). Quality factor Q ● ● ● Quality factor: is the ratio of the reactive power and average power. which can be used to distinguish between these two quantities. The quality factor Q for a coil is defined as the ratio of the inductive reactance and the winding resistance. Note: Both the quality factor and reactive power are denoted by the letter Q. gives Q¼ _ _ XL IXL VL ¼ ¼ _ _ R IR E Similarly.1. Quality factor of the coil: Q ¼ XL =Rw . for Q ¼ XC =R Q¼ _ _ IXC V C ¼ _ _ IR E Therefore. The quality factor is a dimensionless parameter. Quality factor of the series resonance: Q ¼ XL =R ¼ XC =R. when the resonance occurs in an RLC series circuit: _ _ _ VL ¼ VC ¼ EQ ð10:2Þ The quality factor Q is always greater than 1.9 Voltage of series resonant _ Multiplying current I for both the denominator and nominator of the quality factor equation Q ¼ XL =R. Q¼ XL Rw The lower the winding resistance Rw of a coil. the higher the quality of the coil.) 10. (The lower the Rw.e. so be careful not to confuse them. which is the resistance of the wire in the winding. A coil always contains a certain amount of winding resistance Rw. the higher the quality of the coil. and the unit of reactive power is Var. i. as can . so the inductor or capacitor voltage may greatly exceed the source voltage in a series resonant circuit.RLC circuits and resonance 313 Similarly. For example. such as pushing a child in a playground swing to the resonant frequency. Voltage of series resonance ● ● A lower input voltage may produce a higher output voltage. quality factor. Inductor or capacitor voltage may greatly exceed the supply voltage _ _ _ VL ¼ VC ¼ EQ ðQ > 1Þ Example 10. therefore. the bridge collapsed since the uniform pace reached resonant frequency that resulted in a small force producing a large vibration. which makes the swing go higher and higher to the maximum amplitude with very little effort. Once the ball is bounced to the resonant frequency.7. When choosing the storage elements L and C for a series resonant circuit. Determine the total equivalent impedance.2). Resonance may also cause damage. and inductor voltage of this circuit.1: A series resonant circuit is shown in Figure 10. There are many examples of resonance in daily life. it will yield a smooth response. which is defined as a system oscillating at maximum amplitude at resonant frequency. The concept of circuit resonance is similar to resonance in physics. Solution: Z ¼ RT ¼ R þ Rw ¼ ð2 þ 0:5ÞO ¼ 2:5ff0 O 1 1 ffi fr ¼ pffiffiffiffiffiffi ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi % 6 366 Hz 2p LC 2p ð2:5 mHÞð0:25 mFÞ XL ¼ 2pfL ¼ 2pð6 366Þð2:5 mHÞ % 100 O XL 100 O ¼ ¼ 40 Q¼ RT 2:5 O _ _ _ E VL ¼ jXL I ¼ jXL Z 2:5ff0 V  100ff90 O ¼ 100ff90 V ¼ 2:5ff0 O . so a small input force can produce a large output vibration. This means that a lower input voltage may produce a higher output voltage. and the ball will reach maximum height since a small force produces a large vibration. a legend says that when a team of soldiers walking a uniform pace passed through a bridge. or else the high resonant voltage may damage circuit components. the affordability of their maximum voltage should be taken into account. the series resonance is also known as the voltage resonance. That is one of the reasons that series resonant circuits have a wide range of applications. Another example is bouncing a basketball.314 Understandable electric circuits be seen from the equation (10. . cutoff or half-power frequencies.8 Bandwidth of a series resonant circuit The characteristic of the resonant circuit can be expressed in terms of its bandwidth (BW) or pass-band. As displayed in the diagram. The curve of the current versus frequency of the series resonant circuit is illustrated in Figure 10. I Imax 0.1 The bandwidth of series resonance When an RLC series circuit is in resonance. _ _ ðVL ¼ 100ff90 VÞ > ðE ¼ 2:5ff0 VÞ 10.RLC circuits and resonance R = 2 Ω C = 0. the current reaches the maximum value Imax as the frequency closes in on the resonant frequency fr.707 Imax BW 0 1 P 2 max f1 fr f2 f Figure 10.1 This example shows that the inductor voltage of the series resonant circuit is indeed greater than the supply voltage.2 Bandwidth and selectivity 10.8.5 Ω L = 2. The bandwidth of the resonant circuit is the difference between two frequency points f2 and f1.7 Circuit for Example 10. its impedance will reach the minimum value and the current will reach the maximum value. which is located at the centrer of the curve.5 V ∠0° • + – Rw = 0.2.5 mH Figure 10.25 μF 315 E = 2. BW ¼ f2 À f1 where f2 and f1 are called critical. 9. 10. or 70. the circuit power is only one-half of the maximum power that it is produced by the source at resonance frequency fr.2 The selectivity of series resonance Figure 10. A series resonant circuit with a wider bandwidth is good for passing the signals.2. in Figure 10. the bandwidth of the resonant circuit is a frequency range between f2 and f1 when current I is equivalent to 0. such as the selectivity curve 2 (BW2) in Figure 10.2 ¼ 0:5Pmax . The selectivity is the capability of a series resonant circuit to choose the maximum current that is closer to the resonant frequency fr. The curve in the Figure 10. Therefore. The steeper the selectivity curve. at both points f2 and f1. which has a better current selectivity than selectivity curve 2 or 3. where f2 is the upper critical frequency. and the better the circuit selectivity.707Imax.8. This means that the series resonant circuit of curve 1 has a higher quality and can be expressed as Q ¼ fr =BW. Sometimes in order to take into account both aspects. the higher the maximum current value. and f1 is the lower critical frequency. where Q is the quality factor of the series resonant circuit.8 is called the selectivity curve of the series resonant circuit. or cutoff or half-power frequencies: Pf 1. The reason to define the term ‘half-power’ frequency can be derived from the following mathematical process. The power delivered by the source at the points f1 and f2 can be determined from the power formula P ¼ I2R 2 Pf 1 ¼ If21 R ¼ ð0:707Imax Þ2 R % 0:5Imax R ¼ 0:5Pmax and 2 Pf 2 ¼ If22 R ¼ ð0:707Imax Þ2 R % 0:5Imax R ¼ 0:5Pmax Therefore.707 of its maximum value Imax.316 Understandable electric circuits As shown in Figure 10. Bandwidth (pass-band) ● ● Bandwidth (BW ¼ f2 7 f1) is the range of frequencies at I ¼ 0. the concepts of bandwidth and selectivity may apply to different circuits with different design choices.7 per cent of the maximum value of the curve. the selectivity curve between narrow and wide curves may be chosen. f2 and f1 are critical. A series resonant circuit with a narrower bandwidth has a better current selectivity. and the series resonant circuits can select frequencies in this range.9. . The bandwidth BW is an important characteristic for the resonant circuit. the faster the signal attenuation (reducing). For example. the selectivity curve 1 has a bandwidth of BW1 and a maximum current I1max.8 shows the frequency range between f2 and f1 at which the current is near its maximum value. 9 Selectivity of a series resonant circuit Selectivity of the series resonance The capability of the circuit to choose the maximum current Imax closer to the resonant frequency fr. and 200 O. the disadvantage of the narrower BW or higher Q is that the ability for passing signals in the circuit will be reduced. the narrower the bandwidth ðBW #¼ fr =Q "Þ. which is desirable in many applications. the wider the bandwidth ðBW "¼ fr =Q #Þ. 10. it will have a poor current selectivity.2: Given a series resonant circuit shown in Figure 10. The lower the Q value. and the better the current selectivity. Solution: When R ¼ 50 O: Q¼ XL 2 kO fr 50 Hz ¼ 40. Example 10.2. deter_ mine the bandwidth BW and current I (phasor-domain) of this circuit with three resistors that are 50. This is the reason that Q is denoted as the ‘quality’ factor since it represents the quality of a resonant circuit. As mentioned earlier. however. 100. the higher the maximum current.10(a). and plot their selectivity curves.3 The quality factor and selectivity The quality factor Q in the resonant circuit is a measure of the quality and selectivity of a resonant circuit. and the better the ability to pass signals. BW1 ¼ ¼ ¼ 1:25 Hz ¼ 50 O 40 R Q _ _ _ E E 10ff0 V ¼ 0:2ff0 A I¼ ¼ ¼ 50 O Z R .RLC circuits and resonance I I1max I2max I3max 1 2 3 317 0 BW1 BW2 BW3 f Figure 10. The higher the Q value. 1 A 0.05 A 0 BW 1 = 1.2 When R ¼ 100 O: XL 2 kO fr 50 Hz ¼ ¼ 20.2.5 Hz BW 3 = 5 Hz fr = 50 Hz f Figure 10. (b) Selectivity curve for Example 10. BW2 ¼ ¼ ¼ 2:5 Hz R Q 100 O 20 _ _ _ E E 10ff0 V ¼ 0:1ff0 A I¼ ¼ ¼ 100 O Z R Q¼ When R ¼ 200 O: Q¼ XL 2 kO ¼ ¼ 10. the selectivity curve of the circuit is .2 shows that the selectivity curve of a resonant circuit depends greatly upon the amount of resistance in the circuit.10 (a) Circuit for Example 10.2 A I • 0.318 Understandable electric circuits (a) R = 50 Ω E = 10 V∠0° fr = 50 Hz – • + XL = 2 kΩ XC = 2 kΩ (b) 0. When resistance R in a series resonant circuit has a smaller value.25 Hz BW 2 = 2. R 200 O BW3 ¼ fr 50 Hz ¼ ¼ 5 Hz Q 10 _ _ _ E E 10ff0 V ¼ 0:05ff0 A I¼ ¼ ¼ Z R 200 O Example 10. 1 Series resonance summary Series resonance XL ¼ XC. but the worse the ability to pass signals. the worse the current selectivity.3. The analysis method of the series resonant circuit can also be applied to the parallel resonant circuits. as shown in Figure 10.11 A parallel RLC circuit .1 Introduction Resonance may occur in a parallel resistor. However. fr ¼ 1= 2p LC Z ¼ R minimum (admittance Y maximum) _ _ IT ¼ V =R (maximum) BW ¼ f2 7 f1 ¼ fr/Q Q ¼ XL =R ¼ XC =R _ _ _ VL ¼ VC ¼ EQ Characteristics Condition of resonance Resonant frequency Impedance Current Bandwidth Quality factor Relationship of voltage and quality factor 10.2. the passband (BW) of the circuit with a smaller R value is narrower. the quality factor Q has a higher value.3. pffiffiffiffiffiffiffiÁ Z ¼ R À X ¼ 0.11. the current at the resonant frequency fr has a higher value. Q #¼ XL =R") BW"¼ fr =Q#: the flatter the selectivity curve. when the circuit inductive susceptance BL is equal to the capacitive susceptance BC. and the ability to pass signal will be poor. and the selectivity is better. the better the current selectivity.3 Parallel resonance 10. Quality factor and selectivity ● ● ● Quality factor: a measure of the quality and selectivity of a resonant circuit Q ¼ fr =BW: Q "¼ XL =R#) BW#¼ fr =Q": the steeper the selectivity curve.RLC circuits and resonance 319 steeper. 10. but the better the ability to pass signals. IT E • • • • • IR R IL jwL IC 1 jwC Figure 10. inductor and capacitor (RLC) circuit. Y¼ G 10.3. the energy of the reactive components in the circuit will compensate each other (BC ¼ BL).320 Understandable electric circuits The analysis method of the parallel resonance is similar to series resonance. which is the response curve of the impedance Z versus the frequency f in the parallel resonant circuit. resonance may be produced or removed. You may have noticed that the parallel resonant angular frequency or and resonant frequency fr are the same with those in the series resonant circuit.12. when parallel resonance occurs. When f ¼ fr. That is. the circuit input equivalent (total) admittance Y is equal to the circuit conductance G. The resonant frequency fr is dependent on the circuit elements L and C. B ¼ BC À BL ¼ 0 so the circuit equivalent impedance Z is at a maximum ðZ "¼ 1=Y #Þ. i.3. resonance will occur in the RLC parallel circuit. Frequency of parallel resonance pffiffiffiffiffiffiffiÁ À ● ● Resonant frequency: fr ¼ 1= 2p LC pffiffiffiffiffiffiffi Resonant angular frequency: or ¼ 1= LC 10. the impedance Z is at the highest point on the curve and this is opposite to the series resonance. . B ¼ 0. This is shown in Figure 10. when the resonance occurs in an RLC parallel circuit. This is the characteristic of the parallel resonant circuit.3 Admittance of parallel resonance As previously mentioned. meaning that if adjusting the inductance L or capacitance C in the RLC parallel circuit. or when the susceptance B is zero (B ¼ BC 7 BL ¼ 0). and the equivalent admittance of the parallel RLC circuit is at the lowest (Y ¼ G). the equivalent admittance Y of the circuit is at the minimum (Y ¼ G).pffiffiffiffiffiffiffi Since o ¼ 2pf . When the magnitudes of the capacitive susceptance BC and the inductive susceptance BL are equal (BC ¼ BL). Parallel resonance BC ¼ BL.2 Frequency of parallel resonance The angular frequency of the parallel resonant circuit can be obtained from 1 1 Y ¼ G þ jðBC À BL Þ ¼ þ j oC À R oL pffiffiffiffiffiffiffi From BC ¼ BL or oC ¼ 1=oL. the parallel resonant frequency is fr ¼ 1=ð2p LC Þ.e. Y ¼ G þjB ¼ G Under the above condition. solving for o gives or ¼ 1= LC . f for parallel resonance 10. the impedance Z is at the maximum. Y ¼ GÞ. and the total current in the circuit will be _ _ V V _ IT ¼ ¼ Z R Therefore. BC ¼ BL. This is also opposite of series resonance.Y ¼ G þ jðBC À BL Þ ● _ _ _ _ Current is minimum at parallel resonance: I ¼ V =Z ¼ V =R . when f ¼ fr. Y ¼ G. Figure 10.12 The response curve of Z vs.13 illustrates the response curve of current I versus frequency f in the parallel resonant circuit and current is at the lowest point on the curve when f ¼ fr.3. and the _ _ _ current is at the minimum in the parallel resonant circuit.4 Current of parallel resonance When resonance appears in a parallel RLC circuit. IT #¼ V =Z "¼ V =R. Y ¼ G þ j (BC 7 BL). the impedance of the circuit is equal to the resistance (Z ¼ R). .RLC circuits and resonance Z 321 0 fr f Figure 10. I • 0 fr f Figure 10.13 The response curve of I vs. f for parallel resonance I and Z of parallel resonance ● Impedance is maximum at parallel resonance: Z ¼ R ðB ¼ 0. the admittance Y is at the minimum . I 10. IL ¼ ÀIC . IL ¼ ÀIC .322 Understandable electric circuits 10. jXL XL VC VC _ IC ¼ ¼j ÀjXC XC À1 þj ¼ j _ _ i. and fy ¼ 0.VL ¼ VR ¼ EÞ . _ _ The resistor voltage is equal to the source voltage ðVR ¼ EÞ in the parallel _ _ resonant circuit of Figure 10. i. BC ¼ 1/XC) and VL VL _ IL ¼ ¼ Àj . i. quality factor Q ¼ reactive power/ average power. If we substitute the expressions of the reactive power and average power in Figure 10. the admittance angle fy ¼ 0.13.e. The capacitor and inductor branch currents in the parallel resonant circuit are equal in magnitude but opposite in phase.6 Quality factor From the previous description.14 Phasor diagram of the parallel resonant circuit Phasor relationship of parallel resonance ● ● _ _ _ _ IL and IC are equal in magnitude but opposite in phase.e. A phasor diagram of the parallel resonant circuit is illustrated in Figure 10.14. the quality factor of a parallel resonance will be obtained as follows: _ R E2 =XL ¼ _ 2 =R XL E Q¼ _ _ _ ð.11 into the quality factor equation.3. IC 0 IL • • IT E = VR • • • Figure 10.e.3. _ _T and E are in phase. The total current ðIT Þ and the source voltage E _ _ _ are in phase (since VR and E are in phase) and the phase difference between E _ and IT is zero. since BL ¼ BC (BL ¼ 1/XL. we know that the quality factor is the ratio of the reactive power stored by an inductor or a capacitor and the average power dissipated by a resistor in a circuit.5 Phasor diagram of parallel resonance An RLC parallel resonant circuit is equivalent to a purely resistive circuit since Y ¼ G and Z ¼ R. When choosing the storage elements L and C for a parallel resonant . Q¼ _ 2 =R _ _ E E=R IT _ _ E=X L IL ¼ _ _ E=R IT _ E 2 =XL _ E2 =R Therefore. the quality factor Q can be expressed by the capacitive reactance and the resistance as Q¼ R XC The quality factor of a parallel resonant circuit is inverted with the series resonant circuit. Recall the quality factor of a series resonant circuit: Q¼ XL XC ¼ R R Quality factor Q Quality factor of the parallel resonance: Q ¼ R=XL ¼ R=XC 10. It is similar to series resonance. and there are benefits and disadvantages to using parallel resonance. and therefore the parallel resonance is also known as current resonance.3. when resonance occurs in an RLC parallel circuit _ _ _ IL ¼ IC ¼ IT Q ð10:3Þ Usually the quality factor Q is always greater than 1. the inductor or capacitor branch current may greatly exceed the total supply current in a parallel resonant circuit.RLC circuits and resonance 323 Similarly. and this can be seen from the equation (10.3).7 Current of parallel resonance _ Dividing the voltage E for both the denominator and the numerator of the quality factor equation Q¼ gives Q¼ Similarly For Q ¼ _ _ _ E2 =XC E=XC IC ¼ . This means that a lower input current may produce a higher output current. 707Imax BW 0 f1 fr f2 f Figure 10.8 Bandwidth of parallel resonance The characteristic of the parallel resonant circuit can be expressed in terms of its bandwidth (BW) or pass-band.1 Parallel resonance summary Parallel resonance BL ¼ BC.707 of its maximum value Imax.15 The bandwidth of the parallel resonance 10. Current of parallel resonance ● ● A lower input current may produce a higher output current. its current reaches the minimum value. or 70.15.8. The bandwidth of the parallel resonant circuit is illustrated in Figure 10. I Imax 0. The BW of the parallel resonant circuit is a frequency range between the critical or cutoff frequencies f2 and f1. B ffiffiffiffiffiffiffi0. or else the higher resonant current may damage circuit components. The inductor or capacitor current may greatly exceed the supply current _ _ _ IL ¼ IC ¼ IT Q ðQ > 1Þ 10.324 Understandable electric circuits circuit.3.7 per cent of the maximum value of the curve.3. the affordability of their maximum current should be taken into account. when the current is equivalent to 0. When the RLC parallel circuit is in resonance. Y ¼ G À p¼ Á fr ¼ 1= 2p LC Z ¼ R maximum (admittance Y minimum) _ _ IT ¼ V =R (minimum) BW ¼ f2 À f1 ¼ fr =Q Q ¼ R=XL ¼ R=XC _ _ _ IL ¼ IC ¼ IT Q Characteristics Conditions of resonance Resonant frequency Impedance Current Bandwidth Quality factor Relationship of current and quality factor . Recall that BW ¼ f2 À f1 or BW ¼ fr =Q. the parallel resonant circuit is usually is formed by an inductor (coil) in parallel with a capacitor.16 A practical parallel circuit 10.Y ¼ G þ jBÞ .16.16 is Y¼ 1 1 þj R þ jXL XC Multiplying (R 7 jXL) to the numerator and denominator of the first term in the above expression gives Y¼ or Y¼ R 1 XL þj À 2 2 2 XC R þ XL R2 þ XL ð10:4Þ R XL 1 2 À j R2 þ X 2 þ j X R2 þ XL C L The parallel resonance occurs when the circuit admittance Y is equal to the circuit conductance G (Y ¼ G).1 Resonant admittance The input equivalent admittance of the practical parallel circuit shown in Figure 10.RLC circuits and resonance 325 10. I VS • • IL • IC • R C L Figure 10.16. so when the resonance occurs for the practical parallel circuit in Figure 10. an actual parallel resonant circuit will look like the one illustrated in Figure 10.4 The practical parallel resonant circuit In practical electrical or electronic system applications.4. Since a practical coil always has internal resistance (winding resistance). the resonant admittance should be Y ¼G¼ R 2 R2 þ X L ð. resonance will not occur. or R < ● ● pffiffiffiffiffiffiffiffiffi L=C resonance occurs.e. R < .326 Understandable electric circuits 10.2 Resonant frequency According to the parallel resonant conditions.5) as follows: Resonance angular frequency: rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi L À CR2 1 CR2 or ¼ ¼ pffiffiffiffiffiffiffi 1 À 2C L L LC Resonance frequency: rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 CR2 fr ¼ pffiffiffiffiffiffiffi 1 À L 2p LC ðo ¼ 2pf Þ This indicates that resonance will occur in the circuit of Figure 10. resonance occurs when the capacitive susceptance BC is equal to the inductive susceptance BL. or R < 1À C C L L If 1 À ðCR2 =LÞ < 0. i.4.4) gives R2 or oL R2 þ ðoLÞ2 ¼ oC ð10:5Þ XL 1 2 ¼X þ XL C The resonance frequency and angular frequency for the circuit in Figure 10. BC ¼ BL Thus. 1 > . .16 can be obtained from (10. (10.16 only when rffiffiffiffi CR2 CR2 L L 2 > 0. Practical parallel resonance ● Resonant admittance: Y¼ R2 R 2 þ XL ● Resonant angular frequency: rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 CR2 or ¼ pffiffiffiffiffiffiffi 1 À L LC Resonant frequency: rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 CR2 fr ¼ pffiffiffiffiffiffiffi 1 À L 2p LC When 1 À ðCR2 =LÞ > 0. the desired input signal will be passed and amplified. The purpose of resonant circuits are the same 7 to select a specific frequency (resonant frequency fr ) and reject all others. such as filters. this signal of the corresponding station can be clearly heard. R I • VS • L C 0 fr f Figure 10.18 is a simplified series radio tuning circuit. Once fr matches the desired input signal frequency with the highest current. The combination of a practical parallel resonant circuit and an amplifier can select the appropriate signal to be amplified. etc. adjusting the switch of the radio channel). the circuit resonant frequency fr will consequently change. When adjusting the capacitance of the variable capacitor in the series resonant circuit. After it is amplified by the amplifier in the circuit. The key circuit of a communication system is a tuned amplifier (tuning circuit).RLC circuits and resonance 327 10. resonant circuits are used in a wide range of applications in communication systems.17 A simplified parallel radio tuner The input signals in the radio tuner circuit have a wide frequency range. Figure 10. Once fr matches the desired input signal frequency with the highest input impedance. the desired input signal will be passed.e.4. Figure 10. When adjusting the capacitance of the variable capacitor in the practical parallel resonant circuit (i.3 Applications of the resonance As previously mentioned. because there are many different radio signals from different radio stations. and this is the only signal that will be amplified. It is similar to the parallel tuning circuit. Z • RW L C 0 fr f Figure 10.17 is a simplified radio tuning circuit for a radio circuit. or select signals over a specific frequency range that is between the cutoff frequencies f1 and f2. tuners.18 A simplified series radio tuner . the circuit resonant frequency fr will change. _ _ IT and E in phase jy ¼ 0. +j ● _ _ IL and IC are equal in magnitude but opposite in phase. X ¼ 0. Y ¼ G ● ● _ _ VL and VC are equal in magnitude but opposite in phase. _ _ I and E in phase f ¼ 0.328 Understandable electric circuits Summary Series/parallel resonance Characteristics Conditions of resonance Phasor relationship Series resonance XL ¼ XC. IC • Phasor diagram VL 0 • I • • VC • VR = E • + 0 IL • IT • E = VR • • Resonant frequency Impedance pffiffiffiffiffiffiffi fr ¼ 1=ð2p LC Þ Z ¼ R minimum (admittance Y maximum) Z pffiffiffiffiffiffiffi fr ¼ 1=2p LC Z ¼ R maximum (admittance Y minimum) Z 0 fr f 0 fr f Current _ _ I ¼ V =R (maximum) I • _ _ IT ¼ V =R (minimum) I • 0 fr f 0 Bandwidth Quality factor Relationship of voltage/current and Q BW ¼ f2 À f1 ¼ fr =Q Q ¼ XL =R ¼ XC =R or Q ¼ fr =BW _ _ _ VL ¼ V C ¼ EQ BW ¼ f2 À f1 ¼ fr =Q Q ¼ R=XL ¼ R=XC or Q ¼ fr =BW _ _ _ IL ¼ IC ¼ IT Q fr f . Z ¼ R ● Parallel resonance BL ¼ BC. B ¼ 0. a lower input current may produce a higher output current.RLC circuits and resonance ● 329 ● ● ● ● ● _ _ Bandwidth (pass-band): the frequency range corresponding to I ¼ 0:707Imax . f2 and f1: critical frequencies or cutoff frequencies or half-power point frequencies. IL or IC may greatly exceed the total current IT. Determine the resonant frequency fr of the series resonant circuit by experiment.2 ¼ 0:5Pmax Quality factor: a measure of the quality and selectivity of a resonant circuit. Pf 1. i. Selectivity: the capability of the circuit to choose the maximum current closer to the resonant frequency fr. VR ¼ E _ _ Current reaches maximum at the series resonance: I ¼ VR =R I • ● ● pffiffiffiffiffiffiffi Resonant frequency: fr ¼ 1=2p LC Quality factor: Q ¼ XL =R ¼ XC =R 0 fr f Equipment and components ● ● Multimeter Breadboard . i. VL or VC may greatly exceed the supply voltage E. and BW ¼ f2 7 f1. a lower input voltage may produce a higher output voltage. Experiment 10: Series resonant circuit Objectives ● ● ● Observe and analyse the characteristics of a resonant circuit by using the oscilloscope and function generator.e. In a parallel resonant circuit. Observe and analyse the frequency versus current curve and verify that the circuit current reaches the maximum in a RLC series resonant circuit by experiment. In a series resonant circuit.e. Background information ● ● ● Impedance of RLC series circuit: Z ¼ R þ jðXL À XC Þ _ _ Resonance: XL ¼ XC . VL ¼ VC . 1 on the breadboard. Is the value of VR and supply voltage E approximately equal? . 7.1 using Q ¼ XL =ðR þ Rw Þ (inductor has a winding resistance RW).1.1. Record the value in Table L10.1 Rw Value Q fr VR VC VL 2. Use multimeter (ohmmeter function) to measure the winding resistance Rw of the 12-mH inductor.1. 5. Use the oscilloscope CH II to measure the voltage across resistor VR (peak value). L = 12 mH e = 4 sin ωt V C = 0. Calculate the quality factor Q of the series resonant circuit in Figure L10.0013 μF R = 300 Ω Figure L10. Table L10. Note that the ground of the oscilloscope. Record the value in Table L10.1 Series RLC resonant circuit 3. function generator and circuit should be connected together. Record the value in Table L10.1.1. Construct a circuit as shown in Figure L10. Compute the resonant frequency fr and record the value in Table L10. 4.330 ● ● ● ● ● ● Understandable electric circuits Function generator Oscilloscope LCZ meter or Z meter Resistor: 300 O Inductor: 12 mH Capacitor: 0. 6. Set the sinusoidal output voltage of the function generator to 4 V (peak value). and then adjust the frequency of the function generator to the circuit resonant frequency fr. Use the oscilloscope CH I to measure the peak source voltage E from the function generator in Figure L10.0013 µF Procedure 1. 3.2.1. Then measure the voltage across the inductor (VL).7 fr ¼ ¼ ¼ ¼ I VR VC VL . Exchange the position of the capacitor and inductor in the circuit as shown in Figure L10.2.RLC circuits and resonance 331 8.0013 μF Figure L10.3 The circuit to measure VL 10. L = 12 mH e = 4 sin ωt V R = 300 Ω C = 0.0013 μF R = 300 Ω e = 4 sin ωt V L = 12 mH Figure L10. Exchange the position of the capacitor and resistor in the circuit as shown in Figure L10.6 fr fr ¼ 1. Then measure the voltage across the capacitor (VC).3 fr 0. Record the values in Table L10. Calculate the currents and voltages of the RLC series resonant circuit at different frequencies given in Table L10. function generator and circuit (inductor) are connected together.2.2 F 0. C = 0.1.4 fr 1. and so that ground of the oscilloscope. and record the value in Table L10. so that the ground of the oscilloscope. function generator and circuit (capacitor) are connected together. and record the value in Table L10.2 The circuit to measure VC 9. Table L10. plot the current versus frequency curve (with current I in the vertical axis and frequency f in the horizontal axis). explain the reason.7 fr ¼ ¼ ¼ ¼ VR VC VL 12.7 fr.3 fr 0.332 Understandable electric circuits 11. fr. Conclusion Write your conclusions below: .4 fr 1. Table L10. 13. Does the circuit current reach the maximum at resonant frequency fr? If not.3 f 0. Repeat steps 7–9 and record the results in Table L10. 0.6 fr. Based on the data in Table L10. Adjust the frequency of function generator to 0.3. 1.3 fr.6 fr fr ¼ 1.2. respectively.4 fr and 1. instrumentation and many other electrical and electronics fields.1. when a changing current flows through a coil (inductor). Transformers have a very wide range of applications in power systems. impedance and power of the primary and secondary of a transformer understand the concept of impedance matching of a transformer know applications of transformers The concept of self-inductance has been introduced in chapter 6. 11. . A transformer is a device that is built based on the principle of mutual inductance and can be used to increase or decrease the voltage or current. It also can be used for impedance matching.1 Mutual inductance and coefficient of coupling As discussed in chapter 6 (section 6.1 Mutual inductance 11. and transfer electric energy from one circuit to another. it will produce an electromagnetic field around the coil. radio.3). Mutual inductance is the ability of a coil to produce an induced voltage due to the changing of the current in another coil nearby. The changing current in a coil that produces the ability to generate an induced voltage is called self-inductance. and as a result an induced voltage vL will flow across it. telecommunications.Chapter 11 Mutual inductance and transformers Objectives After completing this chapter. voltage. This chapter will introduce the mutual inductance and transformer. you should be able to: ● ● ● ● ● ● ● ● ● understand the concept of mutual inductance understand the dot convention concept know the basic construction of a transformer know different types of transformers understand the characteristics of transformers determine the turns ratio of an ideal transformer calculate the current. that is produced by changing the electromagnetic field linked to the coil L2.334 Understandable electric circuits In Figure 11. a coil L1 is placed close to another coil L2. and consequently produces the induced voltage v2 across the second coil L2.2(a). the changing of alternating current will produce a changing electromagnetic field and flux f1. The coefficient of coupling k determines the degree of the coupling between the two coils. pffiffiffiffiffiffiffiffiffiffi LM ¼ k L1 L2 There are three factors that affect mutual inductance: inductances of the two coils L1. f1–2 is called the crossing link flux. resulting in a selfinduced voltage v1 across the first coil L1. Since the two coils are very close. and this is the principle of mutual inductance. L2 and the coupling coefficient k.1 Magnetic coupling Mutual inductance is denoted bypffiffiffiffiffiffiffiffiffiffi can be expressed mathematically LM and using the following formula: LM ¼ k L1 L2 : Mutual inductance An inducted voltage in one coil due to a current change in a nearby coil. .1. i1 e v1 L 1 L2 v2 Figure 11. f1–2. When AC current i1 flows through the first coil L1. and f1–2 is the portion of the magnetic flux that is generated by the current i1 in the first coil L1 and linked to the second coil as shown in Figure 11. The phenomenon of a portion of the flux of a coil linking to another coil is called inductive coupling. there is also a portion of magnetic flux. and it is the ratio of f1–2 and f1: k¼ f1À2 f1 f1 is the magnetic flux generated by the current i1 in the first coil L1. This method can be used to indicate whether the induced voltage in the second coil is in phase or out of phase with the voltage in the first coil.1. k is in the range between 0 and 1 (0 k 1).e. and the coupling coefficient k to decrease.3. the induced voltage in _ _ the coil L2 is given by v2 ¼ LM(di1)/dt. not all of the magnetic flux generated by current i1 will pass through L1 and L2. and there will be no leakage flux.3(b)). In practice. one on the coil L1 and the other on the coil L2. So when the AC current i1 flows through the second coil L2. The closer the two coils are placed (or if the two coils have a common core as shown in Figure 11. f1 : The flux generated by the current i1 in the first coil L1.) or asterisks (*). as shown in Figure 11. the higher the cross-linking flux f1–2 and the lower the leakage flux. i. The full-coupling occurs when k ¼ (f1–2)/(f1) ¼ 1. when all of the flux link coils 1 and 2. If the gap between the two coils is large. or V2 ¼ jLM I1 in the phasor form. the leakage flux to increase. it will cause the cross-linking flux to decrease.Mutual inductance and transformers i1 LM LM 335 e f1 L1 f12 L2 L1 L2 k (a) k (b) Figure 11. The dot convention method places two small phase dots (. Coefficient of coupling ● ● ● The coefficient of the coupling: k ¼ ðf1À2 Þ=f1 (0 k 1). f1–2: The flux generated by the current i1 in the coil L1 cross-linking to coil L2. This means that the dotted terminals of coils . 11. to indicate that polarities of the induced voltage v1 in the coil L1 and v2 in the adjacent coil L2 are same at these points. and the portion of the magnetic flux that does not link with L1 and L2 is known as a leakage flux. f1–2 ¼ f1.2 Mutual inductance The induced voltage generated by a changing current (AC) that flows through the self-inductance coil L1 is given by v1 ¼ L(di1)/dt (chapter 6).2 Dot convention The polarity of the induced voltage across the mutually coupled coils can be determined by the dot convention method. 2 Basic transformer 11. The first coil is called primary winding.3 Dot convention Dot convention Dotted terminals of coils have the same voltage polarity. Figure 11. (a) (b) * Figure 11.1 Transformer A transformer is an electrical device formed by two coils that are wound on a common core. 11. A changing current from the AC voltage source in the first coil produces a changing magnetic field. e L1 L2 ZL e ZL (a) (b) Figure 11. the transformers are categorized as two main types: the aircore and iron-core transformers. Recall that mutual inductance is the ability of a coil to produce induced voltage due to the changing of current in another coil nearby.4 shows two simplified transformer circuits. . v1 v2 v1 . inducing a voltage in the second coil. respectively (inside the dashed lines). and dotted terminals are known as corresponding terminals. A transformer uses the principle of mutual inductance to convert AC electrical energy from input to output. and the second coil connected to the load ZL is called secondary winding. (b) iron-core .4 Implified transformer circuits (a) air-core. Structurally. .2.4 (a and b). v2 v1 * * v2 v1 * . The symbols for them are shown in Figure 11. You may have seen transformers on top of utility poles.336 Understandable electric circuits should have the same voltage polarity at all time. 5 Air-core transformer The air-core transformer is also known as a linear transformer.2 Air-core transformer The air-core transformers are usually used in high-frequency circuits. such as air. Air-core transformer A transformer uses the principle of mutual inductance to convert AC electrical energy from input to output. such as in instrumentation.5(a). radio and TV circuits. R1 R2 L1 L1 L2 vs ZL L2 (a) (b) Figure 11. it is a liner transformer. so it can be obtained by placing the two coils L1 and L2 close to each other. wood. The circuit of an air-core transformer is shown in Figure 11.Mutual inductance and transformers 337 Transformer A transformer uses the principle of mutual inductance to convert AC electrical energy from input to output. . An air-core transformer does not have a physical core.2.2.3 Iron-core transformer Iron-core transformers are usually used in audio circuits and power systems. 11.. there R1 and R2 represent the primary and secondary winding resistors of the transformer. When the core of the transformer is made by the insulating material with constant permeability. 11.6. plastic. The coils of the iron-core transformer are wound on the ferromagnetic material that are laminated sheets insulated to each other.5(b). etc. as illustrated in Figure 11. or by winding both the coils L1 and L2 on a hollow cylindrical-shaped core with isolating material as illustrated in Figure 11. e. Also the primary and secondary windings are wound on a common core. and this is the reason that iron-core transformer is usually considered as the ideal transformer (k ¼ 1). ● Transformer parameters: The parameters of an ideal transformer in Figure 11.6 Iron-core transformer When two coils are wound on a common core. 11.338 Understandable electric circuits iP ZP Z NP NS iS f vS vP e (a) ZS ZL (b) Figure 11. and the load. Furthermore. i.2.6(a) is a circuit of an ideal transformer with the voltage source. The ferromagnetic materials can provide an easy path for the magnetic flux. An iron-core transformer is considered the ideal transformer because it uses ferromagnetic materials with high permeability as its core. Table 11. ideal full-coupling. it will have higher crosslinking flux and lower leakage flux. if two coils are wound on a common core. This means that the coupling coefficient k is close to 1.4 Ideal transformer The coupling coefficient k of an ideal transformer is 1.1 Parameters vP vS Np NS ip iS ZP ZS ¼ ZL Name Primary voltage Secondary voltage Number of turns on the primary coil Number of turns on the secondary coil Primary current Secondary current Primary impedance Secondary or load impedance . the flux generated in the coil L1 will almost all link with the coil L2.6(a) are listed in Table 11. which have near zero leakage flux and can achieve a full-coupling (k ¼ 1). Figure 11.1. and the portion within the dashed line is the symbol of the ideal transformer. neglecting winding resistance and magnetic losses in the coils of the transformer. current and impedance.e. i. pP ¼ pS or vP iP ¼ vS iS . current and impedance: The expressions of the transformer’s turns ratio indicate that a transformer can be used to convert voltage. or ZL ¼ n2 ZP pffiffiffiffiffiffiffiffiffiffiffiffiffi _ _ _ _ Phasor form: n ¼ NS =NP ¼ VS =VP ¼ IP =IS ¼ ZL =ZP Power ● ● Instantaneous form: pS ¼ iS vS . the transformer has no power loss itself. vL ¼ N df we can get: dt df ● the primary voltage vP ¼ NP dt df ● the secondary voltage vS ¼ NS dt Dividing vS by vP gives the transformer’s turns ratio n: vS NS ¼ ¼n vP NP If the transformer is an ideal transformer.e. so iP =iS ¼ vS =vP ¼ n ð11:1Þ vP ¼ vS/n can be obtained from (11. the number of turns on the secondary coil NS to the number of turns on the primary coil NP.1). The primary impedance can be obtained by substituting vP and iP into ZP as follows: ZP ¼ or n2 ¼ ZL =ZP . PP ¼ IP VP ● Conversion of the voltage.Mutual inductance and transformers ● 339 Turns ratio n: The turns ratio of a transformer is the ratio of the number of turns. . the input power is equal to the output power.e. From Faraday’s law described in chapter 6. pP ¼ iP vP _ _ _ _ _ _ Phasor form: PS ¼ IS VS . and also iP ¼ n iS. i. which can be derived from the voltage ratio of the secondary and primary voltages. i. vS iS n¼ vP vS =n 1 ¼ ¼ 2 ZL n iS n iP pffiffiffiffiffiffiffiffiffiffiffiffiffi ZL =ZP where the secondary impedance is the load ZL ZL ¼ Turns ratio ● ● pffiffiffiffiffiffiffiffiffiffiffiffiffi Instantaneous form: n ¼ NS =NP ¼ vS =vP ¼ iP =iS ¼ ZL =ZP . vP ¼ ð1=nÞðvS Þ Current conversion: iS ¼ ð1=nÞðiP Þ. IP ¼ 5 A and ZL ¼ 2 O.) Transformer parameters conversion ● ● ● Voltage conversion: vS ¼ nvP . Solution: NS 100 ● n¼ ¼ ¼ 2:5 40 NP _ _ ● VS ¼ nVP ¼ ð2:5Þð50 VÞ ¼ 125 V _ IP 5 A _ ● ¼ 2A IS ¼ ¼ n 2:5 ZL 2O ● ZP ¼ 2 ¼ ¼ 0:32 O 2:52 n _ _ _ ● PS ¼ IS VS ¼ð2 AÞð125 VÞ ¼ 250 W ¼ 0:25 kW 11. Determine the transformer’s turns ratio. and the number of turns on the secondary is 100. vS 4 vP. secondary voltage. iP ¼ n iS Impedance conversion: ZL ¼ n2 ZP . secondary current.3.340 ● Understandable electric circuits Voltage conversion: ● From the primary to the secondary. multiplying by 1/n: vP = (1/n)vS ● Current conversion: ● From the primary to the secondary. VP ¼ 50 V.1: The number of turns on the primary is 40 for an ideal trans_ _ former. ZP ¼ ð1=n2 ÞðZL Þ Example 11. Since a step-up transformer always has more secondary winding turns than the primary.3 Step-up and step-down transformers 11. multiplying by n: vS = nvP ● From the secondary to the primary.e. i. multiplying by 1/n: iS ¼ (1/n)(iP) ● From the secondary to the primary. multiplying by 1/n2: ZP ¼ (1/n2)(ZL) ● From the secondary to the primary.1 Step-up transformer A step-up transformer is a transformer that can increase its secondary voltage. multiplying by n2: ZL ¼ n2ZP (The converted impedance is also called the reflected impedance. multiplying by n: iP ¼ n iS ● Impedance conversion: ● From the primary to the secondary. the secondary voltage of a step-up transformer (vS) is always higher than the primary voltage (vP ). primary impedance (reflected from the secondary) and the primary power (the amplitude only). The value of the secondary voltage . meaning the reflection of the primary impedance results in the secondary impedance. . the number of turns on the secondary coil must be lesser than primary. calculate its turns ratio and determine if it is a stepup or a step-down transformer. meaning the turns ratio n ¼ (NS/NP) 5 1. the number of turns on the secondary winding must be greater than that of the primary.e. NS 5 NP as illustrated in Figure 11. NS NP vP vS vP NP NS vS (a) (b) Figure 11. The equation n ¼ NS/NP ¼ vS/vP indicates that to have a voltage that is lower in secondary than primary. Since a step-down transformer always has less turns on the secondary winding than the primary.e. i. which is opposite of a step-up transformer. i. meaning the turns ratio n ¼ (NS/NP) 4 1.7 (a) Step-up and (b) step-down transformers Step-up transformer ● ● ● vS 4 vP NS 4 N P n41 11.7(a).7(b). NS 4 NP as illustrated in Figure 11.2: If a transformer has 125 turns of secondary windings and 250 turns of primary windings. Step-down transformer ● ● ● vS 5 vP NS 5 N P n51 Example 11.Mutual inductance and transformers 341 depends on the turns ratio (n). the secondary voltage of a step-down transformer (vS) is always lower than the primary voltage (vP). This is an important characteristic of a step-up transformer. i. The equation of n ¼ NS/NP ¼ vS/vP indicates that to have a higher secondary voltage.e. The value of the secondary voltage depends on the turns ratio (n).3. This is an important characteristic of a step-down transformer.2 Step-down transformer A step-down transformer is a transformer that can decrease its secondary voltage. vS 5 vP. current and impedance. and the loss of the power due to the winding resistance in the line will reduce to 10 000 W.e.3 Applications of step-up and step-down transformers As mentioned in the previous section. If the voltage is increased through the step-up transformer before the transmission. This can reduce the loss of energy or power created due to the winding resistance in the line (I2Rw ¼ PLoss) for a long-distance-line transmission. but reduce the power loss on the line. Power plant 22 kV 500 kV Step-up transformer Step-down transformer 66 k V Industries Residential areas V /110 k 220 kV Step-down transformer Commercial areas 4800 kV Step-down transformer Figure 11. transformers can be used to convert voltage. since I 2 Rw ¼ PLoss and I 2 # Rw ) PLoss #. n 5 1 it is a step-down transformer. NP ¼ 250 and NS 5 NP or n ¼ NS =NP ¼ 125=250 ¼ 1=2 ¼ 0:551. iS #¼ ð1=n "ÞðiP Þ.8 Power transmission system The voltage generated from the generator of a power plant needs to rise to a very high value through the step-up transformer so that it can be delivered through long-distance transmission lines. which will require converting voltage or current from primary to secondary winding. The functions of step-up and step-down transformers are to increase or decrease the voltage of their secondary windings. In the power system. A simplified power transmission system is illustrated in Figure 11. 11.: ~ ¼ ðI # ÞðV " Þ ) ðI 2 # ÞðRÞ ¼ PLoss # P If a step-up transformer is used to increase the voltage by 100 V. the basic usage of transformers is stepping up or stepping down the voltage or current. and improve the efficiency of the electricity transmission. The local distribution stations require step-down transformers to reduce the very high voltage by the long distance transmission and can send it to commercial or residential areas.e. i.8. . Decreasing the current to reduce the power loss on the transmission line may reduce the output power (P ¼ IV) of the transmission system. and have important applications in the power transmission system.3.342 Understandable electric circuits Solution: Since NS ¼ 125. i. then the current will reduce by 100 A ½vS "¼ ðn " ÞðvP Þ. it can maintain the same output power. 9(b).3. and it can provide two balanced output voltages with the same value. which is a common coil for both the primary and the secondary coils.9 (b) Multiple-tapped transformer ● Adjustable (or variable) transformer: The output voltage of adjustable (or variable) transformer across the secondary winding is adjustable.9(c).9(a). An adjustable transformer is shown in Figure 11. Figure 11. Figure 11.9 (a) Center-tapped transformer ● Multiple-tapped transformer: It has multiple taps in the secondary winding. and it can provide several output voltages with different values. as shown in Figure 11.Mutual inductance and transformers 343 11. as shown in Figure 11.9(d). The secondary winding of the adjustable transformer can provide an output voltage that may be variable in a range of zero to the maximum values. and a portion of the common coil acts as part of both the primary and secondary coils. An auto-transformer can be made smaller and lighter. as shown in Figure 11.9 (c) Adjustable transformer ● Auto-transformer: It is a transformer with only a single winding. .4 Other types of transformers There are other types of commonly used transformers listed as follows: ● Center-tapped transformer: It has a tap (connecting point) in the middle of the secondary winding. Figure 11. this is shown in Figure 11. When the load impedance ZL is equal to the source internal impedance ZS. the power delivered from the source to the load reaches the maximum. ZS e ZL Figure 11.1 Maximum power transfer The theory of maximum power transfer in the DC circuits was introduced in chapter 5. or when the load resistance is equal to the Thevenin/ Norton equivalent resistance of the network (RL ¼ RTH ¼ RN). .4 Impedance matching In addition to stepping-up and stepping-down voltages.9 (d) Auto-transformer 11. the maximum power delivered from a source to a load in a circuit can be achieved when the load resistance is equal to the internal resistance of the source (RL ¼ RS). matching the load and source impedance in a circuit to achieve the maximum power transfer from the source to the load.344 Understandable electric circuits Figure 11.e. the power received by the load from the source reaches the maximum. 11. i. This theory can also be applied to an AC circuit by replacing the resistance with impedance. a transformer has another important application. It is known as impedance matching.10.4.10 Impedance matching Maximum power transfer When ZL = ZS. i. a transformer with an appropriate turns ratio n can be placed between the load and source to make the load impedance and the source internal resistance equal. i. The circuit within dashed lines is Thevenin’s equivalent circuit for the amplifier circuit. usually is not matching with the load impedance.Mutual inductance and transformers 345 11. and also not adjustable. In this case.3 Solution: ● Since the load impedance (ZL ¼ RL ¼ 4 O) does not match with the source internal impedance (ZS ¼ RS ¼ 100 O) currently. Example 11. and to achieve the maximum power pffiffiffiffiffiffiffiffiffiffiffiffiffi transfer.3: A simplified amplifier circuit is illustrated in Figure 11. n¼ rffiffiffiffiffi rffiffiffiffiffiffiffiffi ZL 4 1 ¼ 0:2 ¼ ¼ 100 5 ZP . ● Choose an audio transformer with the appropriate turns ratio n.11 Circuits for Example 11.11(a). How do we deliver the maximum power to the speaker if the resistance of the speaker is 4 O (so the speaker can have the maximum volume)? n = 1/5 100 Ω 4Ω 100 Ω e 4Ω e (a) (b) Figure 11. and its internal resistance is 100 O. n ¼ ZL =ZP .4.e. Impedance matching pffiffiffiffiffiffiffiffiffiffiffiffiffi Place a transformer with n ¼ ZL =ZP between the source and the load to achieve maximum power transfer. the maximum power cannot be delivered to the speaker if the source and load are connected directly.2 Impedance matching In the practical circuits (or Thevenin’s equivalent circuits).e. the internal resistance of the source is fixed. if placing an impedance matching transformer with the turns ratio of 1/5 between the amplifier and speaker as illustrated in Figure 11. The parameters of an ideal transformer (k ¼ 1): Parameters vP vS NP NS iP iS ZP ZS ¼ ZL Name Primary voltage Secondary voltage Number of turns on the primary coil Number of turns on the secondary coil Primary current Secondary current Primary impedance Secondary or load impedance ● ● ● ● ● pffiffiffiffiffiffiffiffiffiffiffiffiffi Turns ratio: n ¼ NS =NP ¼ vS =vP ¼ iP =iS ¼ ZL =ZP or ZffiL ¼ n2 ZP pffiffiffiffiffiffiffiffiffiffiffiffi _ _ _ _ In phasor form: n ¼ NS =NP ¼ VS =VP ¼ IP =IS ¼ ZL =ZP Power: pS ¼ iS vS . iP ¼ n iS ● Impedance conversion: ZL ¼ n2 ZP . vP ¼ ð1=nÞðvS Þ ● Current conversion: iS ¼ ð1=nÞðiP Þ.346 ● Understandable electric circuits Therefore. Summary Mutual inductance ● ● ● ● Mutual inductance: An inducted voltage in one coil due to a change current in pffiffiffiffiffiffiffiffiffiffi nearby coil. f Coefficient of coupling: k ¼ 1À2 ð0 k 1Þ.11(b). f1 Dot conversion: Dotted terminals of coils have the same voltage polarity. the speaker will have the maximum volume. Basic transformer ● ● Transformer: It uses the principle of mutual inductance to convert AC electrical energy from input to output. I and Z): ● Voltage conversion: vS ¼ n vP . the other LM ¼ k L1 L2 . pP ¼ iP vP _ _ _ _ _ _ PS ¼ IS VS PP ¼ IP VP Transformer conversion (V. ZP ¼ ð1=n2 ÞðZL Þ Step-up transformer: ● vS 4 vP ● NS 4 N P ● n41 Step-down transformer: ● vS 5 vP . circuit behaviour and performance. and compare them to the theoretical equivalents. Experiment 11: Transformer Objectives ● ● ● Experimentally verify the equation for calculating the turns ratio of the transformer. Experimentally verify the theory of impedance matching transformer. Analyse experimental data.Mutual inductance and transformers ● ● 347 ● NS 5 NP n51 pffiffiffiffiffiffiffiffiffiffiffiffiffi Impedance matching: Place a transformer with the turns ratio n ¼ ZL =ZP between the source and the load to achieve maximum power transfer from the source to the load. Choose a small transformer with lower secondary voltage (12–25 V) in the laboratory. Table L11. Formula for the transformer’s impedance matching: ZL ¼ n2 ZP Multimeter Breadboard Function generator A small transformer with lower secondary voltage (12–25 V) A small audio transformer Continuously adjustable auto-transformer Several resistors A small audio speaker Background information ● ● ● Equipment and components ● ● ● ● ● ● ● ● Procedure Part I: Turns ratio of a transformer 1. pffiffiffiffiffiffiffiffiffiffiffiffiffi The turns ratio of the transformer: n ¼ NS =NP ¼ vS =vP ¼ iP =iS ¼ pZL =ZPffi ffiffiffiffiffiffiffiffiffiffiffiffi Impedance matching: Place a transformer with the turns ratio n ¼ ZL =ZP between the source and the load (if Zin 6¼ ZL ) to achieve maximum power transfer from the source to the load. Record the values in Table L11.1.1 Primary resistance RP Secondary resistance RS Calculated turns ratio n ¼ VS /VP Measured turns ratio n . and measure the primary (RP) and secondary (RS) resistances of the transformer using a multimeter (ohmmeter function). Note: The continuously adjustable auto-transformer is used as a variable AC voltage source.1(b). If the primary voltage of the small transformer is not too high. 3. and record the value in the column ‘Measured turns ratio’ in Table L11.1 (a) Auto-transformer and small transformer 4.1. Calculate the primary current IP (IP = PP/VP) and record the value in Table L11. IP IS VP VS RL Figure L11.348 Understandable electric circuits 2.1.2. Record the values in Table L11. and connect the secondary adjustable auto-transformer to the primary of the small transformer.2 Primary current IP Secondary current IS Turns ratio: n ¼ IP /IS .1(a). Calculate the turns ratio n of the small transformer using measured primary and secondary voltages. Adjust the output voltage of the auto-transformer until it is slightly lower than the rated primary voltage of the small transformer. and make sure that IS and PS will not exceed the rated current and power of the small transformer after connecting RL to the secondary. a function generator can be used to replace the adjustable auto-transformer. Determine the value of resistor RL by calculating the secondary current IS and power PS.1 (b) The circuit for step 5 Table L11. then measure the primary and secondary voltages (RMS values) of the small transformer using a multimeter (voltmeter function). as shown in Figure L11. 5. Calculate the turns ratio n of the transformer using the primary and secondary voltages of the transformer (from the nameplate). vP vS Figure L11. Connect a suitable load resistor RL to the secondary of the small transformer as shown in Figure L11. Plug the primary of the adjustable auto-transformer into the AC power outlet. Table L11.2(a) e e (a) (b) Figure L11. and record the value. Record the value in Table L11. Record the value in Table L11.2(a). Measure the secondary current IS using either the direct or indirect method.2. Measure the voltage across the two terminals of the speaker in Figure L11. Part II: Impedance matching 1.2(b). 5. Adjust the output voltage of the function generator to a suitable value so that the speaker reaches a comfortable listening volume.1 and L11.2. Find a small audio transformer that has a turns ratio n closer to the calculated n in step 4.3. 7. Compare the turns ratio n of the transformer in Tables L11.Mutual inductance and transformers 349 6.2. . Record the value in Table L11. ● The output impedance of the function generator ZP usually is about 600 O. Are there any significant differences? If so.2 Circuit for Part II 2. Then connect a small audio speaker to the two terminals of the function generator as shown in Figure L11. ● The impedance of the small speaker ZL usually is about 8 O.3 in ‘Without transformer’ row. explain the reasons. Set the frequency of the function generator to 2 kHz. 3.3 Voltage across the speaker Without transformer: With transformer: pffiffiffiffiffiffiffiffiffiffiffiffiffi Transformer turns ratio: n ¼ ZL =ZP pffiffiffiffiffiffiffiffiffiffiffiffiffi 4. Calculate the turns ratio ðn ¼ ZL =ZP Þ of a transformer that can be used as the impedance matching. Calculate turns ratio n of the transformer using primary and secondary currents in Table L11. and connect it between the function generator and speaker as shown in Figure L11. Measure the voltage across the speaker in Figure.350 Understandable electric circuits 6. and explain the reason in the conclusion. Compare the volume of the speaker with and without the transformer. 7. L11. Conclusion Write your conclusions below: .2(b).3 in the row ‘With transformer’. Record the value in Table L11. V v I i Figure 12. The circuits we will analyze in this chapter have dependent (or controlled) sources. amplifiers. in which the source voltage or current is a function of other voltage or current in the circuit. filters.1) and will not be affected by other voltages and currents in the circuit. etc. such as transistors. you should be able to: ● ● ● ● understand the concept of the dependent source define four types of dependent sources know how to convert dependent sources equivalently understand the methods for analysing circuits with dependent sources The DC and AC circuits we have discussed in the previous chapters have independent voltage/current sources (Figure 12. Dependent sources are a useful concept in modelling and analysing electronic components.Chapter 12 Circuits with dependent sources Objectives After completing this chapter.1 Independent sources . k1.1 Dependent (or controlled) sources As the name suggests. and the current I is a controlling current through the 5 O resistor branch in the _ same circuit. k1V + – V + – k2I (VCVS) – k3V (VCCS) + – I k4I (CCVS) (CCCS) (a) (b) Figure 12. .2. Its control coefficient _ is 8. _ 5I in the circuit of Figure 12. as well as whether the dependent source itself is a voltage source or current source: ● ● ● ● voltage-controlled voltage-controlled current-controlled current-controlled voltage source (VCVS) current source (VCCS) voltage source (CCVS) current source (CCCS) The above dependent sources can be represented by the symbols in Figure 12. k3 and k4 are called control coefficients or gain parameters.3(b) is a VCVS.1 Dependent sources 12.352 Understandable electric circuits 12. Its control coefficient k _ is 5. 8 V in the circuit of Figure 12. and the voltage V is a controlling voltage across resistor R2 in the same circuit.3(a) represents a CCCS.1. The dependent sources can be categorized into the following four types according to whether it is controlled by a circuit voltage or current. A current-controlled source has a current in its branch that equals to a control coefficient k multiplied by a voltage or current elsewhere in the same circuit. it is called a dependent (or controlled) source. A voltage-controlled source has a voltage across its two terminals that equals to a control coefficient k multiplied by a controlling voltage or current elsewhere in the same circuit. k2.2 Dependent sources: (a) controlling sources and (b) controlled sources In this figure. when a source voltage or current is controlled by or dependent on other voltage or current in the circuit. 4(a) can be converted equivalently to a current-controlled source as shown in Figure 12. the voltage-controlled source in Figure 12. I_S ¼ VS =RS _ VS ¼ I_S RS and . just apply Ohm’s law to convert the source. you will understand that a good example for modelling a CCCS is a transistor circuit.4(b). Internal resistance RS of the source does not change before and after the conversion. its large collect current ic is proportional to the small base current ib according to the relationship ic ¼ bib. the current gain b is the same as the control coefficient k in the dependent source. 12.3 Circuits with dependent sources After you take an analogue electronics course. Rs a a VS • + - ⇔ b (a) IS • RS b (b) Figure 12.4 Equivalent conversion: (a) RS ¼ RS .Circuits with dependent sources 2Ω I 5V • 353 3Ω • R1 8V • + E2 • 5Ω 5I • E1 R2 + V • - Figure 12. In this equation.2 Equivalent conversion of dependent sources Equivalent conversion of dependent sources is the same as the equivalent conversion of independent sources. For instance. Dependent (controlled) sources The source voltage or current is a function of the other voltage or current in the same circuit.1. and vice versa. _ (b) RS ¼ RS . a current amplifier. Based on the property of a bipolar transistor. 2: The CCCS in Figure 12. (a) Voltage-controlled voltage source (VCVS) and (b) voltage-controlled current source (VCCS). ● VS ¼ IS RS controlled voltage source!controlled current source: RS ¼ RS .1.2.6 Circuit for Example 12.5 Circuit for Example 12.1: The VCVS in Figure 12. VS IS ¼ RS . _ _ V ¼ IRS ¼ ð10ff0 mAÞð5 kOÞ ¼ 50ff0 V Equivalent conversion of dependent sources The same as the equivalent conversion of independent sources: ● controlled current source!controlled voltage source: RS ¼ RS . (a) Current-controlled current source (CCCS) and (b) current-controlled voltage source (CCVS).6(a) can be converted equivalently to a CCVS as shown in Figure 12.5(a) can be converted equivalently to a VCCS as shown in Figure 12.5(b). _ _ I ¼ V =R ¼ 10 Vff0 =2 kO ¼ 5ff0 mA Example 12.354 Understandable electric circuits Example 12.6(b). R = 2 kΩ + V = V∠0° • a a + - + 10∠0° V V b • V∠0° I = 5∠0° mA • R = 2 kΩ (a) (b) b Figure 12. 5 kΩ a • a I = I∠0° A 10∠0° mA 5 kΩ ⇒ I • I∠0° A + V = 50∠0° V • - b (a) (b) b Figure 12. 7(b). Write the KCL equation for the node a in Figure 12.7(a) to that in Figure 12.7(b): ÆI ¼ 0 : 1 A À 2I À I þ 0:5I ¼ 0 . Example 12. There. 2I 2A 4Ω 2Ω I 2Ω 4Ω 0.Circuits with dependent sources 355 12.7(a). 3A À 2A ¼ 1A 2 þ 4 ==4 ¼ 4 Note: This circuit has a CCCS. simplify the circuit without changing the CCCS (both controlling branches and controlled source). The following examples will describe these methods.7 Circuits for Example 12.5I 3A (a) a 2I 1A 4Ω I 2Ω 0.3 Solution: Simplify and convert the circuit of Figure 12.2 Analysing circuits with dependent sources The analysing methods for circuits with dependent sources are similar to that for circuits with independent sources.3: Determine the current I in the circuit of Figure 12.5I (b) Figure 12. 4 Solution: Applying KVL.8 Circuit for Example 12.5: Write node voltage equations for the circuit in Figure 12. Solving for V: V ¼ À3I ¼ ðÀ3 Þð0:29 AÞ ¼ À0:87 V Example 12. 4V + 8V 3Ω + V - 2Ω I 6V Figure 12. I ¼ 0:4 A Example 12. P V = 0: À6 þ 2I þ 4V þ 8 þ 3I ¼ 0 That is 2 þ 5I þ 4V ¼ 0 Substituting V¼ 73I into the above equation gives: 2 þ 5I þ 4ðÀ3IÞ ¼ 0 2 þ 5I À 12I ¼ 0 Solving for I: 2 À 7I ¼ 0.9 using the node voltage analysis method. and then represent the control quantity as node voltages.4: Determine the voltage V in the circuit of Figure 12.) I % 0:29 A . (Write KVL by treating the dependent source as an independent source first.356 Understandable electric circuits Current I can be solved from the above equation: À2:5I ¼ À1 A .8. Circuits with dependent sources a R1 E1 c R2 R3 + V - 357 b R4 3V (VCCS) Figure 12.) Solution: The procedure for applying the mesh current analysis method (chapter 4. and then represent the controlling quantity as mesh current. b and c. Write KCL equations to n71 = 371 = 2 nodes (nodes a and b) by inspection.6: Use the mesh current analysis method to write mesh equations for the circuit in Figure 12. R2. R3 and E1 are given).5 Solution: The procedure for applying the node voltage analysis method (chapter 4. 2. I2 and I2 (clockwise) as shown in Figure 12. Node a: 1 1 1 1 1 þ þ Va À Vb ¼ E1 R1 R2 R3 R3 R1 ð12:1Þ Node b: À 1 1 1 Vb ¼ 3V Va þ þ R3 R3 R 4 Substituting the control voltage V ¼ Va À Vb to (12.9. Label all the reference directions for each mesh current I1. Example 12. (Write KVL by treating the dependent source as an independent source first.10. Solving (12.10. Label nodes a.2) gives: À 1 1 1 Vb ¼ 3ðVa À Vb Þ Va þ þ R3 R3 R4 ð12:3Þ ð12:2Þ 3. .9 Circuit for Example 12.3) can determine the node voltage Va and Vb (if R1. section 3) to the above circuit is as follows: 1.1) and (12. and choose ground c as the reference node as shown in Figure 12. section 4) to the above circuit is as follows: 1. 7 .10).7: Determine the branch current I for the circuit in Figure 12. resulting from step 2 can determine the three mesh currents I1.5) yields ÀR2 I1 þ ðR1 þ R3 ÞI2 À R3 I3 ¼ À7ðI1 À I2 Þ ð12:6Þ ð12:5Þ ð12:4Þ 3.4). I2 and I3. Solving all the three simultaneous equations.11 Circuits for Example 12.358 Understandable electric circuits I I1 E R2 I0 R1 I2 + - 7I0 (CCVS) + - R3 I3 R4 Figure 12. (12.5) and (12. Apply KVL around each mesh (windowpane). Mesh 1: ðR1 þ R2 ÞI1 À R1 I2 À R2 I3 ¼ E Mesh 2: ÀR1 I1 þ ðR1 þ R3 ÞI2 À R3 I3 ¼ À7I0 Mesh 3: ÀR2 I1 À R3 I2 þ ðR2 þ R3 þ R4 ÞI3 ¼ 0 Substituting the controlling current I0 = I17I2 to (12.11 by using the superposition theorem. Example 12. (12. and ensure the number of KVL equations is equal to the number of meshes (there are three meshes in Figure 12. (The dependent source will not act separately I=? 2Ω 20 V + − (a) 6A 4I (CCVS) I' 2Ω 20 V I" 2Ω + − (c) 4I 4Ω 4Ω = + − + 4I 4Ω 6A (b) Figure 12.6).10 Circuit for Example 12.6 2. replace the 20 V voltage source with a short circuit as shown in Figurec 12.1) to the above circuit is as follows: 1. and calculate I 00 : I 00 ¼ À 6 A 2O ¼ À2 A ðthe current À divider ruleÞ 2 Oþ4 O (The 6 A current is negative due to its assumed direction to be opposite to I 00 . When a 6 A current source is applied to the circuit. Do not change the dependent source in the circuit when another independent source is acting in the circuit. replace the 6 A current source with an open circuit as shown in Figure 12.11(c).8 .Circuits with dependent sources 359 in the superposition theorem.12(a) by using Thevenin’s theorem.) Solution: The procedure for using the superposition theorem (chapter 5. Choose 20 V voltage source applied to the circuit first. Calculate the sum of currents I 0 and I 00 : I ¼ I 0 þ I 00 ¼ 3:33 A þ ðÀ2 AÞ ¼ 1:33 A Example 12.12 (a–c) Circuits for Example 12. and calculate I 0 : I0 ¼ 20 V % 3:33 A 4O þ 2O 2.11(b).) 3.8: Determine the voltage across the two terminals a and b in Figure 12. section 5. 2Ω I 5V 3Ω a 2Ω I =0 3Ω a CCCS 2I (CCCS) RL + – 5V 2I = 0 b (b) b (a) 2Ω I 3Ω (2I × 3) = 6I + – CCVS a b (c) Figure 12. 12(b). Summary ● Dependent (controlled) sources: The source voltage or current is a function of other voltage or current in the circuit. and mark a and b on the terminals of the load branch as shown in Figure 12. ● VCVS ● VCCS ● CCVS ● CCCS .360 Understandable electric circuits Solution: The procedure for using Thevenin’s theorem (chapter 5.2) to the above circuit is as follows: 1. 2. RTH ¼ Rab ¼ a RTH = 11 Ω RL VTH = 5 V b (d) Figure 12. Determine Thevenin’s equivalent voltage VTH: Since the branches a and b are open.12(d).12 (d) Thevenin’s equivalent circuit Analysing circuits with dependent sources Analysing circuits with dependent sources is similar to the methods of analysis for circuits with independent sources. Open and remove the load branch resistor RL. Plot Thevenin’s equivalent circuit as shown in Figure 12. I = 0. and the CCCS is also 0 (2I = 0) in the circuit of Figure 12.12(c). Determine Thevenin’s equivalent resistance RTH: Replace the 5 V voltage source with a short circuit and convert CCCS to CCVS as shown in Figure 12. section 5. Vab 6I þ ð2 þ 3 ÞI ¼ 11 ¼ I I 4.12(b). so: VTH ¼ Vab ¼ 5 V 3. .Circuits with dependent sources ● 361 Equivalent conversion of dependent sources is the same as the equivalent conversion of independent sources: ● controlled current source!controlled voltage source: RS ¼ R S . ● _ _ VS ¼ IS RS controlled voltage source!controlled current source: RS ¼ R S . _ V _ IS ¼ RS ● The method of analysis for circuits with dependent sources is similar to the methods of analysis for circuits with independent sources (ideal sources). . Appendix A Greek alphabet Uppercase/lowercase A B À D E Z H Y I K à M a b g d " z Z y i k l m Letter Alpha Beta Gamma Delta Epsilon Zeta Eta Theta Iota Kappa Lambda Mu Uppercase/lowercase N Ä O Å P S T Y È X É n x o p r s or B t u f w c o Letter Nu Xi Omicron Pi Rho Sigma Tau Upsilon Phi Chi Psi Omega . i. taking the derivative of f(t) with respect to t gives df ðtÞ ¼ Fm o cosðot þ cÞ dt and df ðtÞ ¼ Fm o sinðot þ c þ 90 Þ dt ¼ Jm ½oFm ejðotþcþ90 Þ ¼ Jm ðoFm ejot ejc ej90 Þ ¼ Jm ðjoFejot Þ There F¼Fm ejc and ej90 ¼ j (From Euler’s formula. ej90 ¼ cos 90 þ jsin 90 ¼ j) Therefore. the derivative of the sinusoidal function with respect to time can be obtained by its phasor F multiplying with jo.e. the phasor of df ðtÞ=dt is joF (there ejot is the rotating factor). df ðtÞ .Appendix B Differentiation of the phasor For a sinusoidal function f ðtÞ ¼ Fm sinðot þ cÞ. joF dt Therefore. . this is equivalent to a phasor that rotates counterclockwise by 908on the complex plane (since þj ¼ þ90 ). FL: Saunders College Publishing. Basic Circuit Analysis for Electronics: Through Experimentation.A. Qian. Mark Edward Hazen. 2003. Jianping. David Buchla. Floyd. Cook. NJ: Prentice Hall. Electronics Technician Common Core Learning Guider. 7th edition. Boylestad. NJ: Prentice Hall. Electric Circuit Analysis. 3rd edition. Nanjing: Nanjing University of Science and Technology. 2003. 2003. Miller. CA: Technical Education Press.Bibliography Robert L. Orlando. 1995. NY: Merrill (an imprint of Macmillan Publishing Company). 1994. Transfer and Technology. Charles K. Bell. . Beijing: People’s Education Press. Laboratory Manual for Fundamentals of DC and AC Circuits. Circuit Analysis: Theory and Practice (Laboratory Manual). Allan H. Upper Saddle River. Burnaby. Laboratory Manual for Electric Circuits. Englewood Cliffs. David A. Sadiku. Introductory Circuit Analysis. NJ: Prentice Hall. 2000. Englewood Cliffs. 1983. 6th edition. Electric Circuits: Principles. 1999. Alexander and Mathew N. Allan H. Upper Saddle River. 1990. 4th edition. Lorne MacDonald. NJ: Prentice Hall. Robbins and Wilhelm C. 1997. 1998. Circuit Analysis: Theory and Practice. Chico. New York. 1987. Upper Saddle River. Boctor. Englewood Cliffs. MA and London: McGraw Hill. Boston. Electric Circuits (Revised edition). NJ: Prentice Hall. NJ: Prentice Hall. 2004. Applications. Experiments in Electronics Fundamentals and Electric Circuits Fundamentals. 6th edition. Albany. David A. Robert T. 1995. Guangyuan. Fowler. 1999. and Computer Analysis.O. Robbins and Wilhelm C. 1995. Fundamentals of Electric Circuits. Qiu. Paynter. Richard J. Electric Circuits. NY: Delmar Publishers. New York. Thomas L. S. Principles of Electric Circuits. BC: Ministry of Education. Miller. 7th edition. Introductory Electric Circuits. Introductory DC/AC Circuits. 1995. Electricity: Principle and Application. Nigel P. Upper Saddle River. Skills and Training and the Centre for Curriculum. NY: Delmar Publishers. NY: McGraw-Hill. Bell. NJ: Prentice Hall. Albany. ucsd.asp http://books.ca/books?id=avEjv8zAhQkC&pg=PT615&lpg=PT615 &dq=control+coefficients+voltage+controlled+sources&source=web& ots=QjyOKEW-Dv&sig=H_gLC4hWUTo192z9n9B5djI24w8&hl=en&sa =X&oi=book_result&resnum=9&ct=result#PPT615. and does not guarantee that any content on such websites is.kpsec.com/content/neets/14193/css/14193_138.html http://www.com/Oscilloscope/DisplayControls.nanotech-now.com/metric-prefix-table.freeuk.il.ee.M1 http://books.google.us/gbssci/phys/Class/circuits/u9l3b.html http://en.pdf http://www.electronics-radio.wikipedia.org/wiki/Siemens_(unit) http://www.ieeeusa.M1 The publisher has no responsibility for the persistence or accuracy of URLs for external or third-party Internet websites referred to in this book.tpub.366 Understandable electric circuits Web resources http://www.com/kits/instructions/breadboard.com/keyword/electric-electronic-engineering? page=11 http://www.org/wiki http://www.k12.htm http://www.com/breadb.org/careers/yourcareer. or will remain.wikipedia.org/wiki/Electronic_color_code http://www.job-search-engine.html http://en.htm http://www.google.com/vol_1/chpt_13/4.com/neets/book2/3b.php http://www.allaboutcircuits.wisc-online.org/wiki/Electrical_resistivity#Table_of_resistivities http://en.wikipedia.htm http://www.htm http://en.asp?objID=DCE1802 http://www-ferp.htm http://ourworld.htm http://www.nmt.wikipedia.com/objects/index_tj.com/articles/test-methods/meters/multimeterresistance-measurement. .edu/*rhb/ee101/labs/2lab/2lab.compuserve.ca/books?id=ZzHudUMb7WAC&pg=PA26&lpg=PA26 &dq=Dependent+voltage+Source&source=web&ots=78V2Df3xV5&sig= ZKXLpVNMsqZyxiRlYdI858fd84s&hl=en&sa=X&oi=book_result&resnum =2&ct=result#PPA26.hiviz.glenbrook.tpub. accurate or appropriate.com/homepages/Bill_Bowden/resistor.edu/najmabadi/CLASS/MAE140/NOTES/analysis-2.html http://oscilloscope-tutorials. 257 Apparent power 282–3. 190 ac response 254–7 charging 165–6. 24. 56 Branch current analysis 108–9. 258 Balanced bridge 90–1. 285–6 Adjustable transformer 343 Admittance 266. 47. 209–10 in parallel 176–7 in series 174–5 Capacitors in series–parallel 178 Center-tapped transformer 343 Charging equations 209 Charging process of an RC circuit 199–201 Characteristics of a capacitor 191. 122 triangle 267. 190 Capacitance 167–8. 263 Ampere 3. 92 Conductance 17. 259 a resistor 191 Chassis ground 70–1. 25. 24 source 50–1 sources in series 107–8. 229. 24 Autotransformer 343 Average power 279–81 Average value 236–7. 299–300 Air-core transformer 337 Alternating current (ac) 227. 269–72. 24 Critical frequencies 315–16 Current 8 direction 9–10. 259 admittance 269–71. 300 Current-controlled current source (CCCS) 352. 270. 329 Blocking ac 256 Branch 41.Index Active power 279–81. 346 Complex number 240–2. 259. 299 Capacitors 164–7. 92 Circuit ground 70 Circuits quantities and their SI units 54 Circuit symbol 5–7 Closed-loop circuit 36–7 Coefficient of the coupling 335. 301 Applied voltage 13. 25 Controlled source 352–3. 258 Conversion of dependent sources 353–4 Conventional current flow version 10. 300 impedance 271. 232 Ammeter 9. 267 Capacitive susceptance 255. 190. 300 an inductor 191. 260. 8 Amplitude 230. 91 Angular frequency 231–2. 258–9 Common ground 70–1. 190 Capacitive reactance 254–5. 17. 252. 122 Breadboard 26–7 Breakdown voltage 170–1. 94 Bandwidth 315–16. 190. 257 Angular velocity 231–2. 209–10 discharging 166–7. 360 . 360 Conversion between rectangular and polar forms 241–2. 256. 299 Conductance form of Ohm’s law 20. 284. 239. 24 Faraday’s law 180. 360 Current divider rule (CDR) 76–7. 190 Inductive reactance 251–2 Inductive susceptance 251–2. 24 Electric current 8. 244. 252 D and p configuration 83–4. 300 angle 284–5. 190 First-order circuit 195–6. 300 Inductance 182–3. 300 in parallel 272–3. 24 Energy 32. 257 Discharging equations 207. 56 Ideal transformer 338–9. 259 Direct current (dc) 228. 300 matching 344–5. 191 Factors affecting resistance 15–16. 220. 190 ac response 250–1. 271. 94 Delta to wye conversion (D ! Y) 84–6. 92 Equivalent series inductance 188 Equivalent parallel resistance 74–5. 93–4. 24 Cutoff frequency 315–16 DC Blocking 172. 300 triangle 284. 93 Euler’s formula 241–2 Excitation 197 Factors affecting capacitance 169. 221 Initial state 197 Input 197 Iron-core transformer 337–8 Instantaneous power 276–9. 270. 222 Discharging process of the RC circuit 204–5 Dot convention 335–6. 190 Factors affecting inductance 184–5. 222 Frequency 229–30. 258 Electric circuit 4. 347 in series 272–3. 122. 24 Electric power 33 Electrolytic capacitor 192 Electron flow version 10. 346 Ideal voltage source 48. 13. 253 in series 188 in series–parallel 189 in parallel 188–9 Initial conditions 198–9. 361 Circuit symbol 6. 300 Current source ! Voltage source 102. 56 Impedance 265–6. 190 Electromotive force (EMF) 11. 92 Effective value 237. 94 Dependent sources 352–4.368 Understandable electric circuits Equivalent (total) series resistance 65. 301 Instantaneous value 236–7. 257 Frequency of series resonance 308 Function generator 260 Half-power frequency 315–16 I–V characteristic 19–20 Ideal current source 50–1. 56 Energy storage element 166 Energy releasing equations for RL circuit 218 Energy storing equations for RL circuit 214 Energy stored by a capacitor 173 Energy stored by an inductor 185–6 Equivalent parallel capacitance 177 Equivalent parallel inductance 189 Equivalent resistance 65 Equivalent series capacitance 175 . 346 Double-subscript notation 70–1. 24 Electromagnetic field 179–80. 258 International system of units (SI) 53–4 Current-controlled voltage source (CCVS) 352. 299 Inductors 182. 93 Earth ground 70–1. 360 Dielectric constant 169–70 Differentiation of the sine function in phasor notation 246. 23 Multimeter 28–9. 24 Power 32–3. 322 notation 239–40. 267. 56–7 of ac circuits 276–7. 93 Parallel resonance 319. 257 Phasor 239–40.Index Integration of the sine function. 58–60 Multiple-tapped transformer 343 Mutual inductance 333–4. 93 Pass-band 316 Passing dc 184 Peak value 235–7. 329 Resonant circuit 307 RC circuit 199 RC time constant 208–9 Reactance 259. 322–3. 274. 56. 24 triangle 284–5. 326. 259 Kirchhoff’s current law (KCL) 41–3. 156. 56 Mesh current analysis 113. 257 Polar form 240–2. 344 Mesh 47. 322. 57 Milestones of the electric circuits 3–4. 301 Real current source 52–3. 258 Phase shift 230–2. 317. 296 369 Ohmmeter 16–17. 250. 301 source 5. 56 Node voltage analysis 116–17. 24 Load voltage 13. 301 Phase difference 232–4. 190 Linear circuits 128. 258 Potential difference 11–2. 24 Ohm’s law 19–20. 56 Real power 279–81 Real voltage source 48–9. 155 Linear network 134 Linear two-terminal network with the sources 135. 123. 300 KVL extension 40 LCZ meter 192 Leakage current 170. polarity of V and I 22–3. 25 Natural response 197–8 Network 134 Network with the power supplies 134 Node 47. 346 Millman’s theorem 151–2. 155 Linearity property 128 Load 5. 299 Reactive power 281–2. 122–3. 257–8 Peak-peak value 235–7. 300 Kirchhoff’s voltage law (KVL) 36–8. 242–4. 56. 156 Mutually related ref. 328 Parallel voltage 73. 274. 327 Quality factor 312–13. 93 Parallel current 73–4. 259 domain 244. 93 Parallel power 75. 258 Period 229–30. 190 Lenz’s law 181–2. 259 diagram 243. 292–3 Norton’s theorem 135–6. 301 Power-factor correction 286–7. 252 for a capacitor 172 for an inductor 183 Operations on complex numbers 241 Oscilloscope 260–4 Output 197 Parallel circuit 71–3. 56 Maximum power 148–9 Maximum power transfer 147–8. 300 Power factor 285–7. 270. 301 Practical parallel circuit 325. 155–6. 259 power 285. 56 . in phasor notation 246. 291 Metric prefix 54–5. 25. 24 Loop 47. 92 Series–parallel circuit 79. 221 t = 0+ 198. 299 Switching circuit 198 Symbols and units of electrical quantities 25 . 25 Reference direction of power 34–5. 300 Voltage-controlled current source (VCCS) 352. 56 Superposition theorem 128–9. 346–7 Step response 196–7. 93 Source equivalent conversion 102. 24 Resistors 14 ac response 248–50 colour code 26–7 Requirements of a basic circuit 5. 258 Reference direction of current 20–1. 220 Source voltage 13. 57 SI prefixes 54–5. 122 sources in series 104–5. 155. 122 Voltmeter 13. 221 Thevenin’s theorem 135–6. 301 Total series resistance 65. 296 Time constant 208–9. 346 Two-terminal network 134–5 Viewpoints 139 Voltage 11–13. 220 True power 279. 156 Supernode 46. 300 Voltage source ! Current source 102. 281 Turns ratio 339. 24 Selectivity 316–17. 56 Wye and delta configurations 83–4 Wye to delta conversion (Y ! D) 86–7 Y ! D 86–7. 191 Wires 5. 58 Wheatstone bridge 89 Winding resistance 186–7. 216 RL time constant 218–19 RMS value 237–9. 329 Self-inductance (inductance) 182–3. 92 Transformer 336–7 Transformer parameters conversion 340. 24 rise 13. 94 Y or T configuration 83–4. 92–3 Series current 66. 24 Steady-state 196. 269–70. 24 divider 273 drop 13. 94 Z meter 192 Rectangular form 240–2. 259 Total power 287–8. 155–6. 317. 94 Series power 66 Series resonance 307–8. 24. 122 triangle 284. 218. 24 Resistivity 15–16. 24 Work 31. 122 Source-free response 197.370 Understandable electric circuits t = 0– 198. 259 Schematic 5–6. 221–2 Time domain 244. 328 Short circuit 50 SI units 53–4. 56 Resistance 14. 57 Single-subscript notation 70–1. 25 Reference polarity of voltage 21–2. 92 Total series voltage 65. 360 Voltage-controlled voltage source (VCVS) 352. 258 Rotating factor 244–5. 24 source 48–50 sources in parallel 105–6. 360 Voltage divider rule (VDR) 67–9. 190 Series circuit 63–4. 346 Transient state 196. 220 Step-up transformer 340–1. 92. 220 Step-down transformer 341–2. 24 Response 196 Right-hand spiral rule 179 RL circuit 211–2. 293–5 Susceptance 259. 346 Substitution theorem 152–3. 17.