Fundamentals of Electrical Engineering / ElectronicsAlternating Current Technology 4th Edition – V 0102 Fundamentals of Electrical Engineering / Electronics © hps SystemTechnik Lehr- + Lernmittel GmbH Altdorfer Strasse 16 88276 Berg / Germany Phone: + 49 751 / 5 60 75 80 Telefax: + 49 751 / 5 60 75 17 Internet: www.hps-systemtechnik.com E-mail:
[email protected] Order no.: V 0102 All rights including those of translation reserved. No part of this document may be reproduced in any way (printed, copied or by any other method) or edited, reproduced or distributed by means of electronic systems without the written consent of hps SystemTechnik. Code: 0.3.6 Alternating Current Technology Preface Preface This manual Alternating Current Technology is designed for use in conjunction with hps training equipment, to provide information through experimentation on the proper- ties of the most important standard circuits. The individual topics are divided into the following sections: • General • Experiments, consisting of the set tasks (experiments) and the procedures to be used. The section General contains a short introduction to the re- spective experiments. Because of the intensive subject matter involved, a detailed theoretical description has been purposely dispensed with. We recommend you to read the commercially available text- books for more profound theoretical knowledge and as back- ground reading to the experiments. All tables and diagrams required in the experiments are pro- vided, in order to simplify procedures. The Appendix to this manual contains a detailed section of so- lutions to the tasks and questions set in the experiments and enables you to check your own results. hps SystemTechnik offers several training systems for con- ducting the experiments. These systems are geared to individ- ual requirements. www.hps-systemtechnik.com Alternating Current Technology I Contents Contents 1 Establishing and Displaying Characteristics in AC Technology ....................................................5 1.1 General...................................................................................................................................................5 1.2 Characteristics of the Sine-Wave Voltage .............................................................................................5 1.3 Active Power with Sine-Wave Voltage ...................................................................................................9 1.4 Characteristics of Square-Wave AC Voltage .......................................................................................13 2 Three-Phase Alternating Current..................................................................................................... 13 2.1 General................................................................................................................................................ 13 2.2 Potential Gradient in Three-Phase Current Systems.......................................................................... 15 2.3 Loads in Star Circuit ............................................................................................................................ 19 2.4 Loads in Delta Circuit .......................................................................................................................... 23 2.5 Measurements on a Defective Star Circuit ......................................................................................... 27 2.6 Measurements on a Defective Delta Circuit........................................................................................ 31 3 Capacitor in the AC Circuit .............................................................................................................. 35 3.1 General................................................................................................................................................ 35 3.2 Charging and Discharging Process of a Capacitor ............................................................................. 37 3.3 Phase Shift between Current and Voltage on the Capacitor .............................................................. 41 3.4 Capacitive Reactance of a Capacitor.................................................................................................. 43 3.5 Series Circuiting of Capacitors............................................................................................................ 45 3.6 Parallel Circuiting of Capacitors .......................................................................................................... 47 3.7 Reactive Power of a Capacitor............................................................................................................ 49 4 Coil in the AC Circuit ........................................................................................................................ 53 4.1 General................................................................................................................................................ 53 4.2 Switch-On and Switch-Off Process of a Coil....................................................................................... 55 4.3 Phase Shift between Current and Voltage on a Coil........................................................................... 59 4.4 Inductive Reactance of a Coil ............................................................................................................. 61 4.5 Series Circuiting of Coils ......................................................................................................................63 4.6 Parallel Circuiting of Coils ....................................................................................................................65 4.7 Reactive Power of a Coil......................................................................................................................67 V 0102 www.hps-systemtechnik.com II Alternating Current Technology Contents 5 Interconnecting Resistor, Capacitor and Coil .................................................................................71 5.1 General .................................................................................................................................................71 5.2 Series Circuiting of Resistor and Capacitor..........................................................................................71 5.3 Parallel Circuiting of Resistor and Capacitor........................................................................................75 5.4 Series Circuiting of Resistor and Coil...................................................................................................79 5.5 Parallel Circuiting of Resistor and Coil .................................................................................................83 5.6 Series Circuiting of Capacitor and Coil.................................................................................................87 5.7 Parallel Circuiting of Capacitor and Coil...............................................................................................91 5.8 Series Circuiting of Resistor, Capacitor and Coil .................................................................................95 5.9 Parallel Circuiting of Resistor, Capacitor and Coil ...............................................................................99 5.10 Active, Reactive and Apparent Power................................................................................................103 6 Transformers ....................................................................................................................................107 6.1 General ...............................................................................................................................................107 6.2 Coupling Factor ..................................................................................................................................107 6.3 Transformation Ratio ..........................................................................................................................109 6.4 Resistance Transformation ................................................................................................................111 www.hps-systemtechnik.com V 0102 Alternating Current Technology 7 Establishing and Displaying Characteristics in AC Technology 1 Establishing and Displaying Characteristics in AC Technology 1.1 Fundamentals +U sinusoidal voltage Alternating current changes its direction of 0 t flow continuously, in contrast to direct cur- -U rent which always flows in the same directi- +U square-wave voltage on. The wave (function) of the current or re- 0 quired voltage respectively may adopt diffe- t -U rent shapes. Figure 1.1.1 shows some of the +U delta voltage common functions in electrical and electro- 0 nics engineering. In AC technology a distinc- t -U tion is made additionally between single- phase and multiphase AC voltages and cur- +U sawtooth voltage rents. The public power supply, for example, 0 t supplies a three-phase alternating current as -U a rule. +U trapezoidal voltage 0 The following experiments are restricted to t the sine-wave and square-wave voltages -U which are the most frequently occuring forms Figure 1.1.1 in electrical and electronic engineering. They deal with the characteristics frequency, root mean square value, peak value, peak-to-peak value, phase shift and power. Experiments in three-phase alternating current are conducted separately in the next chapter. 1.2 Characteristics of the Sine-Wave Voltage 1.2.1 General If a sine-wave AC voltage is applied to an ohmic resistor, it generates an AC current in it with the same function. Explanation of variables: ip = peak value of current (positive or negative) in A ipp = peak-to-peak value of current in A Irms = root mean square value of current in A V 0102 www.hps-systemtechnik.com 8 Alternating Current Technology Establishing and Displaying Characteristics in AC Technology +U T = period duration in s up = peak value of voltage in V, amplitude, peak value (positive or negative) + up u pp upp = peak-to-peak value of voltage in V U rms Urms= root mean square value of voltage in V 0 t -up T -U Figure 1.2.1.1 Figure 1.2.1.1 shows the function in time of a sinewave AC voltage. During one period the sine-wave voltage reaches its positive maximum value after every quarter period, crosses the zero point, reaches its negative maximum value and then returns to zero. Other characteristic values and the formulae for calculating them are specified below. Frequency f in Hz The frequency f gives the number of periods per second. 1 1 f = 1 Hz = T 1s Radian frequency ω in 1/s The radian frequency ω is the product of 2 . π and the frequency. ω= 2 . π. f Momentary values u, i u = up . sin (ω t) i = ip . sin (ω t) Wavelength λ in m (km) The wavelength results from the propagation speed (v in m/s) and the frequency f. v λ= f Root mean square values Urms, Irms This is the root mean square value of a sinusoidal shaped voltage or current. 1 1 U rms = u p ⋅ I rms = i p ⋅ 2 2 www.hps-systemtechnik.com V 0102 Alternating Current Technology 9 Establishing and Displaying Characteristics in AC Technology 1.2.2 Experiments Experiment Display the sine-wave voltage on an ohmic resistor with an oscilloscope and establish the following values from the screen image. A Y1 up = peak voltage ip = peak current R G U upp = peak-to-peak voltage 1 kΩ Urms= root mean square voltage B ⊥ Irms = root mean square current 0V T = period duration oscilloscope f = frequency Figure 1.2.2.1 ω = radian frequency λ = wavelength u = momentary value (after a third of a period) Experiment procedure • Set up the experiment according to circuit (figure 1.2.2.1), connect the function generator and connect the oscilloscope to test points A and B. • Set the oscilloscope according to the specifications made in the grid (figure 1.2.2.2). • Then the sine-wave AC voltage shown in the grid should be set on the function generator and the values specified under experiment determined. Settings: X = 0.1 ms / division Y1 = 1 V / division - 0 (Y 1 ) Figure 1.2.2.2 V 0102 www.hps-systemtechnik.com 10 Alternating Current Technology Establishing and Displaying Characteristics in AC Technology Peak voltage up Period duration T up = T= Peak current ip Frequency f up 1 ip = = f = = R T Peak-to-peak voltage upp Radian frequency ω u pp = 2 ⋅ u p ω = 2 ⋅π ⋅ f Root mean square voltage Urms Wavelength λ 6 (at a propagation speed of 300 · 10 m/s) 1 U rms = u p ⋅ 2 v λ= = f Root mean square current Irms Momentary voltage u after a third of a period U rms I rms = = Note of solution: ωt is the angle with the unit „rad“ R u = up . sin (ω t) = www.hps-systemtechnik.com V 0102 Alternating Current Technology 11 Establishing and Displaying Characteristics in AC Technology 1.3 Active Power with Sine-Wave Voltage 1.3.1 General U I u i 0 t Figure 1.3.1.1 If a sine-wave voltage is applied to a purely ohmic load a sinusoidal current flows through the load. AC current and AC voltage are in phase, i. e. they both achieve their positive and negative maximum value at the same time. The power which is converted in the purely ohmic load is known as the active power. It results from the product of current and voltage. The power curve can be displayed by multiplying the momentary values of voltage and current in a line diagram. P=U.I p = u . i (momentary values) P = I2 ⋅R U2 P= R (root mean square values) P; p = active power in W U; u = voltage in V I; i = current in A R = ohmic resistance in Ω V 0102 www.hps-systemtechnik.com 12 Alternating Current Technology Establishing and Displaying Characteristics in AC Technology 1.3.2 Experiments Experiment Scan the voltage and current on an ohmic resistor with an oscilloscope, transcribe the curves onto the given diagram and construct the power curve by multiplying the momentary values. Experiment procedure • Set up the experiment according to circuit (figure 1.3.2.1), connect the function generator and set the fol- lowing voltage with the oscilloscope: up = 7 V (sinusoidal); f = 1 kHz • Connect the oscilloscope: A Y1 - test point A to channel 1 (Y1) U test point B to channel 2 (Y2), inverse R 1 kΩ - test point C to ground • Settings on the oscilloscope: up = 7 V G (sinusoidal) C ⊥ - Sweep X: 0,1 ms / division - channel 1 (Y1): 2 V / division (voltage U) UM RM - channel 2 (Y2): 0.5 V / division (invert) 100 Ω (measuring voltage UM ⇒ current IA) - Center the zero line of channels 1 and 2. B Y2 0V - Triggering: Y1 oscilloscope Figure 1.3.2.1 • Draw the scanned voltages U and UM in the given diagram (figure 1.3.2.2). • Determine the momentary values u, uM, i and p at the times specified in table 1.3.2.2 and then construct the power curve in the diagram (figure 1.3.2.2). Attention: Make sure that the test point C is not connected to test points B or A through the ground of the devices used (function generator, oscilloscope). Use an isolating transformer if necessary.The 100 Ω resistor in the circuit ser- ves as a measuring resistor. The voltage UM at this resistor is proportional to the current flowing through both re- sistors. The current is calculated with the following formula: UM U I= = M (Ohm’s Law) RM 100 Ω The reference point of the two voltages has been put between the resistors 1 k Ω and 100 Ω for simultaneous representation of the voltage U and the measuring voltage UM. It must be taken into account that the two vol- tages are displayed with a 180° phase shift. The correct representation of the voltage curves is achieved by inverting one of the two voltages with the oscilloscope (in the experiment voltage UM; channel 2; Y2). www.hps-systemtechnik.com V 0102 Alternating Current Technology 13 Establishing and Displaying Characteristics in AC Technology Time t [ms] Voltage u [V] Voltage uM [V] Current i [mA] Active power p [mW] 0 0.1 0.15 0.25 0.35 0.4 0.5 0.6 0.65 0.75 0.85 0.9 1.0 Table 1.3.2.1 P [ mW ] I [ mA] U [V] 40 20 8 30 6 20 10 4 10 2 0 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 t [ ms] -10 -2 -20 -10 -4 -30 -6 -40 -20 -8 -50 -10 -60 -30 -12 Figure 1.3.2.2 V 0102 www.hps-systemtechnik.com 14 Alternating Current Technology Establishing and Displaying Characteristics in AC Technology Question 1: What does the power curve tell us? Answer: Question 2: What is the root mean square value of the power? Answer: P=U.I P= or up ⋅ ip P= = 2 1 U = up ⋅ = 2 1 I = ip ⋅ = 2 www.hps-systemtechnik.com V 0102 Alternating Current Technology 15 Establishing and Displaying Characteristics in AC Technology 1.4 Characteristics of the Square-Wave AC Voltage 1.4.1 General Figure 1.4.1.1 shows the function in time of a square-wave AC voltage. If such a voltage is applied to an oh- mic load a DC current flows through it which changes its direction at certain intervals. T = period duration in s +U up = peak voltage in V (positive or negative) upp = peak-to-peak voltagein V u pp +u p Correspondingly for the current: 0 ip = peak currentin A (positive or negative) t ipp = peak-to-peak current in A -u p -U T Figure 1.4.1.1 If the time intervals of the change in direction are equal and if the voltages acting in both directions are also equal one refers to a symmetrical square-wave voltage. Other characteristic values and the formulas for their calculation are specified below. Root mean square values Urms und Irms In a symmetrical square-wave voltage or current respectively the root mean square value for the voltage and current is equal to the peak values. Urms = up Irms = ip Power P in W (only on ohmic loads) This is calculated using the rms values of the voltage and current. 2 2 U rms P = U rms ⋅ I rms P = I rms ⋅R P= R Frequency f in Hz The frequency f gives the number of periods per second. 1 1 f = 1 Hz = T s V 0102 www.hps-systemtechnik.com 16 Alternating Current Technology Establishing and Displaying Characteristics in AC Technology 1.4.2 Experiments Experiment Display the square-wave AC voltage on an ohmic resistor with the oscilloscope and establish the following va- lues from the screen image. A Y1 up = peak voltage ip = peak current R Urms, Irms = root mean square voltage and current G U (square wave) 1 kΩ P = power T = period duration B ⊥ f = frequency 0V oscilloscope Figure 1.4.2.1 Experiment procedure • Set up the experiment according to circuit (figure 1.4.2.1), connect the function generator and connect the oscilloscope to test points A and B. • Set the oscilloscope according to the specifications made in the grid (figure 1.4.2.2). • Then the square-wave AC voltage shown in the grid (figure 1.4.2.2) should be set on the function genera- tor in connection with the oscilloscope and the values of the characteristics specified under experiment de- termined. Settings: X = 0.1 ms / division Y1 = 1 V / division - 0 (Y 1 ) Figure 1.4.2.2 www.hps-systemtechnik.com V 0102 Alternating Current Technology 17 Establishing and Displaying Characteristics in AC Technology Peak voltage up Peak-to-peak voltage upp up = upp = 2 . up = Peak current ip Power P up ip = = P = Urms . Irms = R p = up . ip = Root mean square voltage Urms Period duration T Urms = up = T= Root mean square current Irms Frequency f 1 Irms = ip = f = = T V 0102 www.hps-systemtechnik.com 18 Alternating Current Technology Establishing and Displaying Characteristics in AC Technology Notes: www.hps-systemtechnik.com V 0102 Alternating Current Technology 19 Three-Phase Alternating Current 2 Three-Phase Alternating Current 2.1 Fundamentals U1 U2 U3 U1 U 0 t 12 0° 12 0° 120° 120° U3 U2 Figure 2.1.1 Figure 2.1.2 Three-phase current systems result when several phase-shifted voltages are connected together. Generation of the phase-shifted, sinusoidal AC voltages is achieved by an appropriate arrangement of coils in the three-phase current generator. The most common three-phase generator which is usually used in the public power supply network supplies three sinusoidal AC voltages which are linked to each other and phase-shifted by 120°. Figure 2.1.1 shows the three voltages as a line diagram and figure 2.1.2 shows them as a pointer diagram. V 0102 www.hps-systemtechnik.com 20 Alternating Current Technology Three-Phase Alternating Current Two basic circuit types are common in the coils of the three-phase current generator and on the load side when connecting motors, for example; the star circuit and the delta circuit (see also figure 2.1.3 and figure 2.1.4). L1 L1 phase star point UL1, L3 U L1, L2 U L3, L1 U L1, L2 L2 rotation direction U L2, L3 U L2, N U L1, N L2 L3 U L2, L3 U L3, N N L3 Figure 2.1.3 Star circuit of the generator coils Figure 2.1.4 Delta circuit of the generator coils L1, L2, L3 = conductor (phase) N = zero conductor (neutral wire) UL1,L2 / UL2,L3 / UL3,L1 = conductor voltages between the conductors specified in the index (in V) The conductor voltages in the delta circuit are also known as the delta voltage. UL1,N / UL2,N / UL3,N = phase voltages (star voltages) between the lines specified in the index (in V) In the star circuit three conductor voltages (400 V) are available between the conductors L1, L2 and L3 as well as three phase voltages (230 V) between the conductors L1, L2 and L3 and the zero conductor (N). In the delta circuit three voltages are available between the conductors L1, L2 and L3 with 400 V each. The specified voltages are rms values. The frequency of the sinusoidal AC voltages is 50 Hz. In the following experiments the behaviour of voltages, currents and powers in star and delta circuits is in- vestigated. The three necessary sinusoidal AC voltages are not taken directly from the mains, they are generated e- lectronically by phase shifters. For riskless experimentation the voltage 230 V is limited to 7 V and the voltage 400 V is limited to 12 V. www.hps-systemtechnik.com V 0102 Alternating Current Technology 21 Three-Phase Alternating Current 2.2 Potential Gradient in Three-Phase Current Systems 2.2.1 General L1 U L1, N U L1 ,L 1 ,L 2 L3 U 12 0° 12 0° U N L2 U L3, 120° ,N L3 U L2, L3 L2 Figure 12.2.1.1 Voltage pointer diagram for the three-phase current system The three sinusoidal, 120° phase-shifted AC voltages generated by a three-phase current generator are inter- linked in such a way that three conductor voltages and three phase voltages can be taken off at conductors L1, L2, L3 and N in starcircuited coils. The voltage pointer diagram (figure 2.2.1.1) shows the relationships between the conductor voltages and the phase voltages with respect to the phase relation and the voltage ratios. The conductor voltage UL and the phase voltage Uphase are linked with factor, also called the linkage factor. In the public power supply network therefore, at a phase voltage of 230 V there is a conductor voltage of: U L = 3 ⋅ U phase = 1.73 ⋅ 230 V = 400 V V 0102 www.hps-systemtechnik.com 22 Alternating Current Technology Three-Phase Alternating Current 2.2.2 Experiments Experiment Display the phase voltages of a three-phase current system on the oscilloscope, draw the displayed voltage curves in a diagram (figure 2.2.2.2) and determine the angle of phase shift between the individual voltages. Then measure the phase and conductor voltages with a multimeter and establish the linkage factor. What is the peak value of the phase and conductor voltages? L1 Experiment procedure channel 1 (Y1) Settings on the oscilloscope: L2 channel 2 (Y2) - Center zero conductor of channel 1 and channel 2 G - Sweep X: 2 ms / division L3 3 - Channel 1 (Y1): 5 V / division N - Channel 2 (Y2): 5 V / division ground ⊥ - Triggering: Channel 1 oscilloscope 7 V / 12 V Figure 2.2.2.1 • Apply voltage UL1,N to channel 1 and voltage UL2,N to channel 2 of the oscilloscope. Draw the displayed vol- tage curves in the provided (figure 2.2.2.2). Apply voltage UL3,N to channel 2 and draw this curve in the dia- gram, too. 12 U [V] 10 8 6 4 2 0 5 10 15 20 t [ms] -2 -4 -6 -8 -10 Figure 12.2.2.2 -12 www.hps-systemtechnik.com V 0102 Alternating Current Technology 23 Three-Phase Alternating Current Angle of phase shift ϕ Linkage factor Period duration T = 20 ms =ˆ 360° The linkage factor results from the ratio of the con- Time shift of individual voltages according to the dia- ductor voltage UL to the phase voltage Uphase. gram (figure 2.2.2.2) UL = U phase t = Frequency f ϕ = 1 f = = T Phase voltages Uphase Peak value of the phase voltage us UL1,N = u p = U phase ⋅ 2 = UL2,N = UL3,N = Conductor voltages UL Peak value of the conductor voltage us UL1,L2 = up =UL ⋅ 2 = UL2,L3 = UL3,L1 = V 0102 www.hps-systemtechnik.com 24 Alternating Current Technology Three-Phase Alternating Current Notes: www.hps-systemtechnik.com V 0102 Alternating Current Technology 25 Three-Phase Alternating Current 2.3 Loads in Star Circuit 2.3.1 General If loads are connected in star circuit to a three-phase current network as shown in figure 2.3.1.1, the same current flows in lines L1, L2 and L3 at symmetrical load (R1 = R2 = R3), the zero line is currentless. L1 I L1 L1, L2, L3 = conductor (phase) N = zero conductor R1 IL1, IL2, IL3 = conductor currents in A IN = zero conductor current in A N IN R1, R2, R3 = loads R3 R2 L2 I L2 L3 I L3 Figure 2.3.1.1 Conductor current IL The individual conductor currents IL which are also referred to as phase currents (Iphase) are calculated accor- ding to Ohm's Law with the formula: UR U phase IL = I phase = R R On an unsymmetrical load a compensation current IN flows through the zero conductor. Power Ptot The power consumption Ptot on a load in star circuit consists of the individual phase powers. In the experi- ment the individual phase powers are designated PR1, PR2 and PR3 corresponding to the ohmic resistors. Ptot = PR1 + PR2 + PR3 A symmetrical load gives the following relationships: Ptot = 3 . Pphase Pphase = Uphase . Iphase V 0102 www.hps-systemtechnik.com 26 Alternating Current Technology Three-Phase Alternating Current On inductive and capacitive load: Active power Ptot Ptot = 3 ⋅ U phase ⋅ I phase ⋅ cos ϕ Ptot = 3 ⋅ U L ⋅ I L ⋅ cos ϕ Apparent power Stot S tot = 3 ⋅ U phase ⋅ I phase S tot = 3 ⋅ U L ⋅ I L Reactive power Qtot Qtot = 3 ⋅ U phase ⋅ I phase ⋅ sin ϕ Qtot = 3 ⋅ U L ⋅ I L ⋅ sin ϕ www.hps-systemtechnik.com V 0102 Alternating Current Technology 27 Three-Phase Alternating Current 2.3.2 Experiments Experiment Measure the currents IL and IN and the voltages UL and Uphase (rms values) with a multimeter in a three-phase AC current network on a star circuited, symmetrical and unsymmetrical ohmic load and then calculate the powers Pphase and Ptot. L1 R1 = R2 = R3 = 1 kΩ G L2 (symmetrical load) 3 L3 R1 = 1 k Ω N R2 = 680 Ω I L3 I L1 IN I L2 7 V / 12 V R3 = 330 Ω (unsymmetrical load) R1 R3 R2 Figure 2.3.2.1 Experiment procedure • Connect symmetrical load according to the circuit in figure 2.3.2.1. • Measure voltages and currents according to the following table 2.3.2.1 with the multimeter and calculate the powers. • Then repeat these measurements and calculations on an unsymmetrical load. • Enter all the measured values in the table. Star circuit Symmetrical load Unsymmetrical load IL1 IL2 Conductor currents IL, IN (Iphase) IL3 IN UL1, L2 Conductor voltages UL UL2, L3 UL3, L1 UL1, N Phase voltages Uphase UL2, N UL3, N PR1 Phase powers Pphase PR2 PR3 Total power Ptot Ptot Table 2.3.2.1 V 0102 www.hps-systemtechnik.com 28 Alternating Current Technology Three-Phase Alternating Current Calculating the phase power and total power: The phase powers correspond to the powers PR 1, PR 2 and PR 3. Symmetrical load PR1 = UL1,N . IL1 = PR2 = UL2,N . IL2 = PR3 = UL3,N . IL3 = Ptot = 3 . Pphase = Unsymmetrical load PR1 = UL1,N . IL1 = PR2 = UL2,N . IL2 = PR3 = UL3,N . IL3 = Ptot = PR1 + PR2 + PR3 = www.hps-systemtechnik.com V 0102 Alternating Current Technology 29 Three-Phase Alternating Current 2.4 Loads in Delta Circuit 2.4.1 General If loads are connected in delta circuit to three-phase AC mains (figure 2.4.1.1) every load is applied directly to the conductor voltage. The conductor voltage is equal to the phase voltage: UL = Uphase The conductor currents IL1, IL2 and IL3 are divided in the branching points into the phase currents IR1; IR2; IR3. I L1 L1 L1, L2, L3 = conductor (phase) N = zero conductor I R1 IL1, IL2, IL3 = conductor currents in A R3 R1 R1, R2, R3 = loads N IR1, IR2, IR3 = phase currents Iphase in A I R3 R2 I L2 I R2 L2 L3 I L3 generator loads Figure 2.4.1.1 Phase current Iphase On symmetrical load (R1 = R2 = R3) the phase currents are 3 smaller than the conductor currents: IL I phase = 3 Power Ptot The power consumption of a load in delta circuit consists of the individual phase powers Pphase. In the experi- ment the individual phase powers are designated PR1, PR2 and PR3 corresponding to the ohmic resistors: Ptot = PR1 + PR2 + PR3 A symmetrical load gives the following relationships: Ptot = 3 . Pphase Phase power Pphase The phase power Pphase results respectively from the phase current Iphase and the phase voltage Uphase: Pphase = Iphase . Uphase V 0102 www.hps-systemtechnik.com 30 Alternating Current Technology Three-Phase Alternating Current Below are some more formulae which are used for calculating the powers on an inductive and capacitive symmetrical load. Active power Ptot Ptot = 3 . Uphase . Iphase . cos ϕ Ptot = 3 . UL . IL . cos ϕ Apparent power Stot Stot = 3 . Uphase . Iphase Stot = 3 . UL . IL Reactive power Qtot Qtot = 3 . Uphase . Iphase . sin ϕ Qtot = 3 . UL . IL . sin ϕ 2.4.2 Experiments Experiment Measure the phase currents and conductor currents as well as the conductor voltages in a three-phase AC mains on a delta circuited symmetrical and unsymmetrical load and then calculate the phase power and total power. L1 R1 = R2 = R3 = 1 kΩ G L2 (symmetrical load) 3 L3 R1 = 1 k Ω N R2 = 680 Ω I L3 I L1 I L2 R3 = 330 Ω 7 V / 12 V (unsymmetrical load) I R1 R3 R1 I R3 R2 I R2 Figure 2.4.2.1 Experiment procedure • Connect symmetrical load according to the circuit in figure 2.4.2.1. • Measure voltages and currents according to the following table 2.4.2.1 with the multimeter and calculate the powers. Then repeat these measurements and calculations on an unsymmetrical load. • Enter all the measured values in the table. www.hps-systemtechnik.com V 0102 Alternating Current Technology 31 Three-Phase Alternating Current Delta circuit Symmetrical load Unsymmetrical load IL1 Conductor currents IL IL2 IL3 IR1 Phase currents Iphase IR2 IR3 UL1, L2 Conductor voltages UL (Uphase) UL2, L3 UL3, L1 PR1 Phase powers Pphase PR2 PR3 Total power Ptot Ptot Table 2.4.2.1 Calculating the phase power and total power: Symmetrical load PR1 = IR1 . UL1,L2 = PR2 = IR2 . UL2,L3 = PR3 = IR3 . UL3,L1 = Ptot = 3 . Pphase = Unsymmetrical load PR1 = IR1 . UL1,L2 = PR2 = IR2 . UL2,L3 = PR3 = IR3 . UL3,L1 = Ptot = PR1 + PR2 + PR3 = V 0102 www.hps-systemtechnik.com 32 Alternating Current Technology Three-Phase Alternating Current Notes: www.hps-systemtechnik.com V 0102 Alternating Current Technology 33 Three-Phase Alternating Current 2.5 Measurements on a Defective Star Circuit 2.5.1 General If an outer conductor (L1, L2 or L3) fails in a star circuit connected to the three-phase AC mains, the originally consumed power on symmetrical load is reduced by a third. If the zero conductor also fails, the power consump- tion is reduced by half as a result of two loads in series being applied to the remaining conductor voltage. If two outer conductors are broken in a star circuit with symmetrical load, the power consumed is only a third of the original power consumption. 2.5.2 Experiments Experiment Measure the conductor currents, the conductor voltages and the phase voltages on a defective star circuit and then calculate the phase and total powers. L1 I L1 R1 = R2 = R3 = 1 kΩ L2 I L2 G L3 I L3 3 N IN 7 V / 12 V A R1 R3 R2 D C B Figure 2.5.2.1 Experiment procedure • Connect load according to the circuit (figure 2.5.2.1). • Make the respective break in the conductor at the generator output in accordance with table 2.5.2.1 and make the necessary measurements. Measure the phase and conductor voltages directly at test points A ... D. • Enter all the measured values in the table 2.5.2.1. V 0102 www.hps-systemtechnik.com 34 Alternating Current Technology Three-Phase Alternating Current Use the following formulae to calculate the powers: Phase powers Pphase Pphase = Uphase . Iphase (e.g. PR 1 = UL1,N . IL1) Total power Ptot Ptot = PR 1 + PR 2 + PR 3 Conductor breaks Star circuit none L1 L1 / L2 L1 / N IL1 7.0 IL2 7.0 Conductor currents IL (Iphase) in mA IL3 7.0 IN 0 UL1, L2 12.1 Conductor voltages UL in V UL2, L3 12.1 UL3, L1 12.1 UL1, N 7.0 Phase voltages Uphase in V UL2, N 7.0 UL3, N 7.0 PR 1 49.0 Phase powers Pphase in mW PR 2 49.0 PR 3 49.0 Total power Ptot in mW Ptot 147.0 Table 2.5.2.1 www.hps-systemtechnik.com V 0102 Alternating Current Technology 35 Three-Phase Alternating Current Calculation of the phase and total powers: Conductor break L1 PR1 = UL1,N . IL1 = PR2 = UL2,N . IL2 = PR3 = UL3,N . IL3 = Ptot = PR1 + PR2 + PR3 = Conductor breaks L1 and L2 PR1 = UL1,N . IL1 = PR2 = UL2,N . IL2 = PR3 = UL3,N . IL3 = Ptot = PR1 + PR2 + PR3 = Conductor breaks L1 and N PR1 = UL1,N . IL1 = PR2 = UL2,N . IL2 = PR3 = UL3,N . IL3 = Ptot = PR1 + PR2 + PR3 = V 0102 www.hps-systemtechnik.com 36 Alternating Current Technology Three-Phase Alternating Current Notes: www.hps-systemtechnik.com V 0102 Alternating Current Technology 37 Three-Phase Alternating Current 2.6 Measurements on a Defective Delta Circuit 2.6.1 General If outer conductors or phases are broken in the case of a delta circuited load connected to the three-phase AC mains, the consumed power can be at the worst only a sixth of the maximum power. The following table 2.6.1.1 lists the power consumptions in different defective delta circuits. Break one phase two phases one outer conductor one phase and one outer conductor P = 2 ⋅ Pmax P = 1 ⋅ Pmax P = 1 ⋅ Pmax 3 3 2 P = 1 ⋅ Pmax P = 1 ⋅ Pmax 3 6 Table 2.6.1.1 2.6.2 Experiments Experiment Measure the conductor currents, the phase currents and the conductor voltages (phase voltages) on a defective delta circuit and then calculate the phase and total powers. Experiment procedure L1 • Connect load according to the circuit (figure G L2 2.6.2.1). Make the respective break in the L3 3 conductor and phase in accordance with N table 2.6.2.1 and make the necessary mea- surements. 7 V / 12 V A • Make the breaks in the phase by pulling out I R1 the appropriate resistors (R1, R2, R3). Measu- R3 R1 re the conductor voltages (phase voltages) at I R3 test points A ... C. C B • Enter all the measured values in the table R2 I R2 2.6.2.1. Figure 2.6.2.1 V 0102 www.hps-systemtechnik.com 38 Alternating Current Technology Three-Phase Alternating Current Use the following formulas to calculate the powers: Phase powers Pphase Pphase = Uphase . Iphase (e.g. PR 1 = UL1,L2 . IR 1) Total power Ptot Ptot = PR 1 + PR 2 + PR 3 Conductor and phase breaks Delta circuit none R2 L1 R1 / R3 L2 / R2 L1 / R2 IL1 21.3 Conductor currents IL in mA IL2 21.3 IL3 21.3 IR 1 12.3 Phase currents Iphase in mA IR 2 12.3 IR 3 12.3 UL1, L2 12.1 Conductor voltages UL (Uphase) in V UL2, L3 12.1 UL3, L1 12.1 PR 1 148.8 Phase powers Pphase in mW PR 2 148.8 PR 3 148.8 Total power Ptot in mW Ptot 446.4 Table 2.6.2.1 www.hps-systemtechnik.com V 0102 Alternating Current Technology 39 Three-Phase Alternating Current Calculation of the phase and total powers: Phase breaks R2 Phase breaks R1 and R3 PR1 = UL1,L2 . IR1 = PR1 = UL1,L2 . IR1 = PR2 = UL2,L3 . IR2 = PR2 = UL2,L3 . IR2 = PR3 = UL3,L1 . IR3 = PR3 = UL3,L1 . IR3 = Ptot = PR1 + PR2 + PR3 = Ptot = PR1 + PR2 + PR3 = Conductor break L1 Conductor and phase breaks L2 and R2 PR1 = UL1,L2 . IR1 = PR1 = UL1,L2 . IR1 = PR2 = UL2,L3 . IR2 = PR2 = UL2,L3 . IR2 = PR3 = UL3,L1 . IR3 = PR3 = UL3,L1 . IR3 = Ptot = PR1 + PR2 + PR3 = Ptot = PR1 + PR2 + PR3 = Conductor and phase breaks L1 and R2 PR1 = UL1,L2 . IR1 = PR2 = UL2,L3 . IR2 = PR3 = UL3,L1 . IR3 = Ptot = PR1 + PR2 + PR3 = V 0102 www.hps-systemtechnik.com 40 Alternating Current Technology Three-Phase Alternating Current Notes: www.hps-systemtechnik.com V 0102 Alternating Current Technology 41 Capacitor in the AC Circuit 3 Capacitor in the AC Circuit 3.1 Fundamentals Next to the resistors capacitors are the most frequently used components in electronic circuits. Their applications are very versatile: Symbol • isolation of DC and AC current • phase shifting of current and voltage Figure 3.1.1 • short-circuiting of AC voltages • as reactive resistor, timing delay element and • energy store • smoothing of DC voltages • setting up filters and resonant circuits etc. Capacitors have a variety of different designs, the major ones are: - wirewound capacitors - ceramic capacitors - electrolytic capacitors - variable capacitors The most important characteristics of these components are: Capacitance C This is the capacity for the load factor Q and is given by the dielectric constant (ε0 . εr) and the effective plate surface divided by the distance apart of the plates. ε ⋅ε ⋅ A C= 0 r d C = capacitance, unit: farad (F) or As / V ε0 = electric field constant (8.85 . 10 -12 As / Vm) εr = dielectric constant 2 A = plate surface in m d = distance between plates in m Rated voltage The rated voltage is the highest permissible continuous voltage which may be applied to the capacitor. V 0102 www.hps-systemtechnik.com 42 Alternating Current Technology Capacitor in the AC Circuit Peak voltage The peak voltage is the briefly permissible maximum voltage (peak value or peak-to-peak value) Insulating resistance Rp The insulating resistance is the specific resistance of the insulator used. The insulating resistance should be as large as possible ( > 1 GΩ ) so that the residual current flowing through the charged capacitor remains as small as possible. Capacitive reactance XC The capacitive reactance is determined by the value of capacitance and the frequency of the applied AC voltage. 1 XC = ω ⋅C XC = capacitive reactancein Ω ω = radian frequency in Hz (2 . π . f ) Dielectric loss factor tan δ The dielectric loss factor gives the ratio between the capacitive reactance XC and the effective resistance (in- sulating resistance) RP of a capacitor. XC tan δ = RP tan δ = dielectric loss factor Rp = insulating resistance in Ω Charge Q The charge Q stored in a capacitor is dependent on the charging current and the charging time. Q=I.t Q = charge in As = C (Coulomb) t = time in s I = charging current in A (constant) In a loaded capacitor: Q=C.U C = capacitance in F U = maximum charging voltage in V www.hps-systemtechnik.com V 0102 Alternating Current Technology 43 Capacitor in the AC Circuit 3.2 Charging and Discharging Process of a Capacitor 3.2.1 General When charging and discharging a capacitor on DC voltage, current and voltage run according to an e-func- tion whereby the voltage has risen to 63 % of the final voltage after 1 τ during charging and has sunk to 37 % of the initial voltage during discharging. τ is the time constant given by resistance x capacitance. During charging or discharging, the current sinks to 37 % of its initial value after 1 τ . After about 5 τ the char- ging or discharging process is ended. τ=R.C τ = time constant in s C = capacitance in F R = resistance to current limiting in Ω The momentary current and the momentary voltage when charging and discharging a capacitor are calcula- ted with the following formulas: Momentary voltage value uC when charging ( uC = U ⋅ 1 − e − t / τ ) U = charging voltage e = base number 2.718 t = charging time in s Momentary voltage value uC when discharging u C = U ⋅ e −t / τ U = voltage on the charged capacitor t = charging time in s Momentary current value iC when charging U iC = ⋅ e − t / τ R R = resistance to the current limiting in Ω Momentary current value iC when discharging U iC = − ⋅ e − t / τ R V 0102 www.hps-systemtechnik.com 44 Alternating Current Technology Capacitor in the AC Circuit 3.2.2 Experiments Experiment Display the charge and discharge curve of voltage and current of a capacitor on the oscilloscope and deter- mine the following values from the curve: - time constant τ - capacitance C - momentary voltage uC after a charging time of 2 ms - momentary current iC at a discharging time of 2.5 ms - Then check the values determined from the curves by calculation. - How great is the charge Q after a charging time of 5 ms? Experiment procedure • Set up the experiment according to circuit (figure 3.2.2.1) and connect the function generator with a positive square-wave voltage: U = 6 V; f = 100 Hz; V=2 A Y1 4.7 k Ω G up = 6 V B Y2 f = 100 Hz 0.22 µF C ⊥ 0V oscilloscope Figure 3.2.2.1 • Connect the oscilloscope: - Test point A to channel 1 (Y1), for recording the input voltage - Test point C to ground - Test point B to channel 2 (Y2), for recording the capacitor voltage or displaying the capacitor currents • To display the capacitor current switch over the resistor • 4.7 kΩ and the capacitor 0.22 µF in the circuit; the voltage applied to the resistor, which is proportional to the capacitor current, is then displayed on the oscilloscope. Make other settings on the oscilloscope in ac- cordance with the specifications below the grids (figure 3.2.2.2 and figure 3.2.2.3). • Draw the displayed voltage curves into the grids (figure 3.2.2.2 and figure 3.2.2.3) respectively. • Determine the values specified in the experiment from the recorded voltage curves. www.hps-systemtechnik.com V 0102 Alternating Current Technology 45 Capacitor in the AC Circuit Settings: X = 1 ms / division Y1 = 2 V / division Y2 = 2 V / division Triggering: Y1 - 0 (Y 1 ) Remarks: Y1: Input voltage Y2: Capacitor voltage - 0 (Y 2 ) Figure 3.2.2.2 Settings: X = 1 ms / division Y1 = 2 V / division Y2 = 2 V / division Triggering: Y1 - 0 (Y1 ) - 0 (Y2 ) Remarks: Y1: Input voltage Y2: Voltage on the resistor (proportional to the capacitor current) Figure 3.2.2.3 Time constant τ From screen image: Calculated: V 0102 www.hps-systemtechnik.com 46 Alternating Current Technology Capacitor in the AC Circuit Capacitance C From screen image: Calculated: Momentary voltage value uC after a charging time of 2 ms From screen image: Calculated: Momentary current value iC after a discharging time of 2.5 ms From screen image: Calculated: Charge Q after a charging time of 5 ms Only calculated: www.hps-systemtechnik.com V 0102 Alternating Current Technology 47 Capacitor in the AC Circuit 3.3 Phase Shift between Current and Voltage on the Capacitor 3.3.1 General If a sinusoidal AC voltage is applied to a capacitor, UC UC IC it is periodically charged and discharged. IC Because of the AC voltage the polarity of the ca- pacitor’s charge also changes periodically. 0 t The current IC reaches its maximum value every time the voltage UC crosses zero. 90° The current IC in a capacitor and the voltage UC have a 90° phase shift. Figure 3.3.1.1 3.3.2 Experiments Experiment Display the current and voltage curves of a capacitor on the oscilloscope and determine the phase shift bet- ween the current IC and the voltage UC from the screen image. A Y1 Experiment procedure • Set up the experiment according to the cir- 1 kΩ cuit (figure 3.3.2.1) and connect the functi- on generator: u pp = 3 V G C ⊥ upp = 3 V (sinusoidal); f = 1 kHz (sinusoidal) Attention: 0.22 µ F Make sure that test point C is not connected with test points A or B through ground of the B Y2 0V equipment used (function generator, oscilloscope oscilloscope). Use an isolating transformer if Figure 3.3.2.1 necessary. The resistor (1 kΩ) in the circuit serves as a measuring resistor. The voltage UR applied to it is proportional to the capacitor current IC. For simultaneously displaying the capacitor voltage UC and the capacitor current IC (UR) the reference point of the voltages to be measured has been placed between the capacitor and the measuring resistor (1 kΩ) (test point C). It must be taken into account that the two voltages are displayed with a 180° shift to each other. The actual display of the voltage curves is made by inverting one of the two voltages with the oscilloscope (in the expe- riment UC, channel 2, Y2). V 0102 www.hps-systemtechnik.com 48 Alternating Current Technology Capacitor in the AC Circuit • Connect the oscilloscope: - Test point A to channel 1 (Y1) - Test point B to channel (Y2), inverse - Test point C to ground Make other settings on the oscilloscope in accordance with the specifications below the grid (figure 3.3.2.2). • Draw the displayed voltage curves into the grid (figure 3.3.2.2) and determine the phase shift between the capacitor voltage UC and the capacitor current IC (UR). Settings: X = 0.1 ms / division - 0 (Y 1 ) Y1 = 1 V / division Y2 = 1 V / division (inverse) Triggering: Y1 Remarks: Y1: Voltage UR (capacitor current IC) Y2: Capacitor voltage UC - 0 (Y 2 ) Figure 3.3.2.2 Phase shift between the capacitor current and the capacitor voltage: Period duration T T= Phase shift ϕ ϕ= www.hps-systemtechnik.com V 0102 Alternating Current Technology 49 Capacitor in the AC Circuit 3.4 Capacitive Reactance of a Capacitor 3.4.1 General A capacitor has a current limiting effect in the AC circuit which results from the countervoltage during charge reversal. This current limiting effect is referred to as the capacitive reactance XC. The value of the capacitive reactance depends on the capacitance of a capacitor and the frequency of the applied AC voltage. The capa- citive reactance is calculated on a sinusoidal AC voltage with the following formula: 1 XC = XC = capacitive reactance of the capacitor in Ω 2 ⋅π ⋅ f ⋅ C 2 ⋅π ⋅ f = radian frequency ω in 1/s C = capacitance in F UC If capacitor current and capacitor voltage are known the capacitive reactance can be cal- XC = IC culated with Ohm's Law: 3.4.2 Experiments Experiment Display the current and voltage curve of different capacitors at different frequencies on the oscilloscope. The respective capacitive reactance XC is to be determined from the screen image by means of the peak-to-peak values and then checked by calculation. Experiment procedure A Y1 • Set up the experiment according to the cir- cuit (figure 3.4.2.1), set a voltage upp = 8 V 1 kΩ (sinusoidal); f = 0.1 kHz on the function ge- nerator and connect it with the circuit. u pp = 8 V G C ⊥ (sinusoidal) • Connect the oscilloscope: - Test point A to channel 1 (Y1) C - Test point B to channel 2 (Y2) - Test point C to ground B Y2 0V • Read off the peak-to-peak values of UR and oscilloscope UC for the frequencies and capacitors Figure 3.4.2.1 C = 0.22 / 0.47 / 1 µF specified in table 3.4.2.1 and enter them in the table. • Calculate values IC and XC and enter these in table 3.4.2.1 also. • Transfer the XC values to diagram (figure 3.4.2.2) for constructing the characteristic curve XC = f (f). Attention: Make sure that test point C is not connected with test points A or B through ground of the equipment used (function generator, oscilloscope). Use an isolating transformer if necessary. The resistor (1 kΩ) in the cir- cuit serves as a measuring resistor. The voltage UR applied to it is proportional to the capacitor current IC. The respective capacitor current is calculated with the formula IC = UR / R. V 0102 www.hps-systemtechnik.com 50 Alternating Current Technology Capacitor in the AC Circuit f [kHz] 0.1 0.2 0.3 0.4 0.5 0.6 0.7 1.0 µF UC [V] 0.47 µF 0.22 µF 1.0 µF UR [V] 0.47 µF 0.22 µF 1.0 µF IC [mA] 0.47 µF 0.22 µF 1.0 µF XC [kΩ] 0.47 µF 0.22 µF Table 3.4.2.1 Figure 3.4.2.2 Question: What do the curves tell us? Answer: Check the capacitive reactance XC of the capacitor C = 0.47 µF at f = 600 Hz by calculation. 1 XC = = 2 ⋅π ⋅ f ⋅ C www.hps-systemtechnik.com V 0102 Alternating Current Technology 51 Capacitor in the AC Circuit 3.5 Series Circuiting of Capacitors 3.5.1 General C1 C2 C3 When capacitors are circuited in series the total I + capacitance is less than the smallest individual capacitance. It is calculated with the following UC 1 UC 2 UC 3 formula: 1 U C tot Ctot = U 1 1 1 + + ⋅⋅⋅ C1 C 2 C 3 - When only two capacitors are circuited in series Figure 3.5.1.1 the following results: C1 ⋅ C 2 Ctot = C1 + C 2 The partial voltages on the individual capacitors behave inversely proportional to the respective capacitance, their sum gives the total voltage UC tot. The capacitor current in a series circuit of capacitors is the same at all points. The same applies to the electrical charge shift. 3.5.2 Experiments Experiment Prove, by measuring the current and voltage and on the assumption that the capacitive reactance of a capa- citor is XC = 1/ω . C, that in a series circuit of capacitors the total capacitance is less than the smallest indivi- dual capacitance. R1 C1 C2 C3 Experiment procedure A B C D E • Set up the experiment according to the circuit (figure 3.5.2.1) and connect the function ge- 1 kΩ 0.22 µ F 0.47 µF 1µF nerator: Urms = 3 V G (sinusoidal) • Urms = 3 V (sinusoidal); f = 1 kHz f = 1kHz • Measure the voltages at test points A ... E with a multimeter and enter the values in the 0V table 3.5.2.1. Figure 3.5.2.1 • The voltage at the measuring resistor R1 (1 kΩ) is proportional to the capacitor current IC and is used for its calculation. V 0102 www.hps-systemtechnik.com 52 Alternating Current Technology Capacitor in the AC Circuit Partial and total voltage [V] Test points A-B (UR 1) B-C (UC 1) C-D (UC 2) D-E (UC 3) B-E (UC tot) Table 3.5.2.1 • Calculate the capacitor current, capacitive reactances and capacitances. • Check the total capacitance Ctot by calculation. Calculation of the capacitor current across the measuring resistor 1 kΩ: U R1 IC = = RM Calculation of the capacitive reactances Calculation of capacitances U C1 1 X C1 = = C1 = = IC ω ⋅ X C1 UC2 1 XC2 = = C2 = = IC ω ⋅ X C2 UC3 1 X C3 = = C3 = = IC ω ⋅ X C3 U C tot 1 X C tot = = Ctot = = IC ω ⋅ X C tot Checking the total capacitance by calculation 1 Ctot = = 1 1 1 + + C1 C 2 C 3 www.hps-systemtechnik.com V 0102 Alternating Current Technology 53 Capacitor in the AC Circuit 3.6 Parallel Circuiting of Capacitors 3.6.1 General I C tot + In a parallel circuit of capacitors the total capacitance is IC 1 IC 2 IC 3 equal to the sum of the individual capacitances. Ctot = C1 + C2 + C3 ... C1 C2 C3 U The partial currents behave proportionally to the respective - capacitance, their sum gives the total current IC tot. Figure 3.6.2.1 The voltage UC is the same in a parallel circuit of all capacitors. 3.6.2 Experiments Experiment Prove, by measuring the current and voltage and on the assumption that the capacitive reactance of a capa- citor is XC = 1 / ω . C, that the total capacitance in a parallel circuit of capacitors is equal to the sum of the in- dividual capacitances. A I C tot B Experiment procedure C E G • Set up the experiment according to the circuit IC 1 IC 2 IC 3 (figure 3.6.2.1), connect the function generator U rms = 3 V G (sinusoidal) D F H and set the following voltage: f = 1 kHz C1 C2 C3 Urms = 3 V (sinusoidal); f = 1 kHz 0.22 µ F 0.47 µ F 1µF I 0V • Measure the total current IC tot and the partial Figure 3.6.2.1 currents IC 1, IC 2, IC 3 with a multimeter and enter the values in the table 3.6.2.1. • Calculate the capacitive reactances XC tot, XC 1, XC 2, XC 3 with the formula XC = UC / IC. • Establish the individual capacitances and total capacitance with the formula C = 1 / ω . XC. • Check the total capacitance Ctot established in the experiment by calculation. V 0102 www.hps-systemtechnik.com 54 Alternating Current Technology Capacitor in the AC Circuit Partial and total voltage [mA] Capacitor voltage [V] Test points Test points A-B (IC tot) C-D (IC 1) E-F (IC 2) G-H (IC 3) D-I (UC 1) F-I (UC 2) H-I (UC 3) Table 3.6.2.1 Calculation of the capacitive reactances Calculation of capacitances U C1 1 X C1 = = C1 = = IC1 ω ⋅ X C1 UC2 1 X C2 = = C2 = = IC 2 ω ⋅ X C2 UC3 1 X C3 = = C3 = = IC3 ω ⋅ X C3 U C tot 1 X C tot = = C tot = = I C tot ω ⋅ X C tot Checking the total capacitance by calculation Ctot = C1 + C2 + C3 = www.hps-systemtechnik.com V 0102 Alternating Current Technology 55 Capacitor in the AC Circuit 3.7 Reactive Power of a Capacitor 3.7.1 General If a capacitor is applied to a sinusoidal AC voltage a phase shift results between the current and the voltage caused by the capacitive reactance. The reactive power Q which occurs in the capacitor can be represented in the line diagram (figure 3.7.1.1) by multiplication of the current and voltage values. Energy is fed to the capacitor during the positive half-wave. During the negative half-wave the capacitor radia- tes this stored energy. Q U I U I Q 0 t Figure 3.7.1.1 No active power is converted in the ideal capacitor. The following definitions are valid: QC = UC . IC (rms values) qC = uC . iC (momentary values) QC; qC = capacitive reactancein var UC; uC = capacitor voltagein V IC; iC = capacitor currentin A V 0102 www.hps-systemtechnik.com 56 Alternating Current Technology Capacitor in the AC Circuit 3.7.2 Experiments Experiment Display the current and voltage curves of a capacitor on the oscilloscope, transfer the curves to a diagram and construct the power curve by multiplication of the momentary values of current and voltage. Experiment procedure A Y1 • Set up the experiment according to the circuit (fi- UR gure 3.7.2.1), connect the function generator and 1 kΩ up= 4 V set the following voltage: G (sinusoidal) C ⊥ up = 4 V (sinusoidal); f = 1 kHz f = 1 kHz UC 0.22 µ F • Connect the oscilloscope: - Test point A to channel 1 (Y1) B Y2 - Test point B to channel 2 (Y2), inverse 0V oscilloscope - Test point C to ground Figure 3.7.2.1 • Settings on the oscilloscope: - Sweep: X = 0.1 ms / division - Channel 1: Y1 = 1 V / division (voltage UR ~ capacitor current IC) - Channel 2: Y2 = 1 V / division (capacitor voltage UC) - Center zero line of Y1 and Y2 - Triggering: Y1 • Draw the screen image on the diagram (figure 3.7.2.2) and establish the values for construction of the po- wer curve by multiplication of the voltage and current values (e. g. every 0.1 ms). Attention: Make sure that test point C is not connected with test points A or B through ground of the equipment used (function generator, oscilloscope). Use an isolating transformer if necessary. The resistor (1 kΩ) in the circuit serves as a measuring resistor. The voltage UR applied to it is proportional to the capacitor current IC (in mA). For simultaneously displaying the capacitor voltage UC and the capacitor current IC (UR) the reference point of the voltages to be measured has been placed between the capacitor and the measuring resistor (1 kΩ) (test point C). It must be taken into account that the two voltages are displayed with a 180° shift to each other. The actual display of the voltage curves is made by inverting one of the two voltages with the oscilloscope (in the experiment UC, channel 2, Y2). www.hps-systemtechnik.com V 0102 Alternating Current Technology 57 Capacitor in the AC Circuit Time t [ms] Current iC [mA] Voltage uC [V] Reactive power qC [mvar] 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 Table 3.7.2.1 5 I C [mA] U C [V] 4 QC [mvar] 3 2 1 0 0.2 0.4 0.6 0.8 1.0 t [ms] -1 -2 -3 -4 -5 Figiure 3.7.2.2 V 0102 www.hps-systemtechnik.com 58 Alternating Current Technology Capacitor in the AC Circuit Notes: www.hps-systemtechnik.com V 0102 Alternating Current Technology 59 Coil in the AC Circuit 4 Coil in the AC Circuit 4.1 Fundamentals Coils are generally wirewound (usually copper) whereby the number of windings and the gauge of the wire can be extremely varied. Coils are widely used in electrical en- Symbol gineering and electronics, e. g.: - generating induction voltages (transformer) Figure 4.1.1 - generating magnetic forces (electric motor, relay, contactors) - filtering out AC currents - storing magnetic energy - function as reactance Coils are manufactured in a variety of designs and sizes depending on their applications. The two major ca- tegories are air coils and coils with a magnetic core. The choice of the magnetic core material and the design of the core are greatly influential in determining the characteristics of a coil. The most important characteristics of a coil are: Inductance L The inductance specifies how great the generated countervoltage is at a predefined change in current. The inductance depends on the number of windings and the coil constant. L = N 2 ⋅ AL L = inductance, unit: Henry (H) N = number of windings AL = coil constant in Henry (H) Coil constant AL µ ⋅µ ⋅ A AL = 0 r d µ0 = magnetic field constant (1.257 . 10 Vs / Am) -6 µr = relative permeability 2 A = core cross section in m d = length of coil in m V 0102 www.hps-systemtechnik.com 60 Alternating Current Technology Coil in the AC Circuit Inductive reactance XL The inductive reactance is determined by the value of inductance and the frequency of the applied AC voltage. XL = ω . L XL = inductive reactance in Ω ω = radian frequency in Hz (2 . π . f ) Loss factor tan δ The loss factor tan δ gives the ratio between the equivalent resistance of a coil and the inductive reactance XL. R tan δ = V XL δ = loss angle tan δ = loss factor RV = equivalent resistance in Ω The equivalent resistance consists mainly of the ohmic resistance of the coil, hysteresis and capacitance los- ses as well as losses due to eddy currents. Coil quality Q The coil quality, as the name implies, specifies the quality of a coil and is equal to the reciprocal of the loss factor tan δ. XL Q= RV Q = coil quality www.hps-systemtechnik.com V 0102 Alternating Current Technology 61 Coil in the AC Circuit 4.2 Switch-On and Switch-Off Process of a Coil 4.2.1 General Current and voltage run according to an e-function when switching the DC voltage of a coil on and off. If a DC voltage is applied, the current increases to 63 % of its final value within 1 τ and sinks to 37 % of its ini- tial value when switching off. After 5 τ the current reaches approximately its final value or zero. τ is the time constant. It results from the inductance of the coil and the ohmic resistance effective in the circuit. L τ = time constant in s τ= R L = inductance in H R = ohmic resistance in Ω (coil, line, internal resistance of the source) The voltage on the coil is reduced to 37 % of its maximum value (on applying the C voltage) after 1 τ and reaches its smallest value, which is dependent on the ohmic coil resistance, after about 5 τ. If the DC voltage on the coil is switched off the self-induction voltage is active in the circuit. This has reverse polarity to the originally applied DC voltage and reaches zero after about 5 τ. The momentary current value iL and the momentary voltage value uL when switching on and off a DC voltage are calculated for the coil according to the following formulas: Momentary value of current iL in the Momentary value of voltage uL in switch-on process the switch-on process iL = U R ( 1 − e−t / τ ) ( i L = I max 1 − e −t / τ ) u L = U ⋅ e −t / τ U = applied DC voltage in V U = maximum coil voltage R = ohmic resistance active in the circuit in Ω t = switch-on duration in s e = base number 2.718 Momentary value of current iL in the Momentary value of voltage uL in the switch-off process switch-off process iL = ( U −t / τ R e ) ( i L = I max e −t / τ ) ( u L = − U e −t / τ ) U = max. self-induction voltage of the coil V 0102 www.hps-systemtechnik.com 62 Alternating Current Technology Coil in the AC Circuit 4.2.2 Experiments Experiment Display the current and voltage curve on the oscilloscope when switching a DC voltage on and off on the coil and establish the following values from the curves: - time constant τ - inductance L - momentary current iL at a switch-on duration of 0.2 ms. Experiment procedure Set up the experiment according to the circuit (figure 4.2.2.1) and connect the function generator with a posi- tive square-wave voltage: U = 6 V; f = 1 kHz; V=2 A Y1 1 kΩ G up = 6 V B Y2 f = 1kHz 100 mH C ⊥ 0V oscilloscope Figure 4.2.2.1 • Connect the oscilloscope: - Test point A to channel 1 (Y1), for oscilloscoping the input voltage - Test point B to channel 2 (Y2), for oscilloscoping the coil voltage or displaying the coil current - Test point C to ground • To display the coil current, switch over the resistor 1 kΩ and the coil 100 mH in the circuit; the voltage at the resistor is then displayed on the oscilloscope. This voltage is proportional to the coil current. • Make other settings on the oscilloscope in accordance with the specifications below the grids (figure 4.2.2.2 and figure 4.2.2.3). • Establish the values specified in the experiment from the recorded voltage curves. • The values established from the curves should then be checked by calculation. www.hps-systemtechnik.com V 0102 Alternating Current Technology 63 Coil in the AC Circuit Settings: X = 0.1 ms / division Y1 = 2 V / division Y2 = 2 V / division Triggering: Y1 - 0 (Y1 ) Remarks: - 0 (Y2 ) Y1: Input voltage Y2: Coil voltage Figure 4.2.2.2 Settings: X = 0.1 ms / division Y1 = 2 V / division Y2 = 2 V / division Triggering: Y1 - 0 (Y 1 ) Remarks: Y1: Input voltage Y2: Voltage on the resistor (proportional to the coil current) - 0 (Y 2 ) Figure 4.2.2.3 Time constant τ Momentary value iL From screen image: at a switch-on duration of 0.2 ms From screen image: Calculated: Calculated: Inductance L From screen image: Calculated: V 0102 www.hps-systemtechnik.com 64 Alternating Current Technology Coil in the AC Circuit Notes: www.hps-systemtechnik.com V 0102 Alternating Current Technology 65 Coil in the AC Circuit 4.3 Phase Shift between Current and Voltage on a Coil 4.3.1 General If an AC voltage is applied to a coil, the coil current UL IL cannot follow the rapid voltage changes on account UL IL of the inductance. The current IL in a coil follows the voltage UL by 90°. 0 In the case of a sinusoidal AC voltage, for example, t the current IL reaches the negative peak value a 90° quarter of a period (90°) later than the voltage UL (figure 4.3.1.1). Figure 4.3.1.1 The equivalent resistance of the coil is ignored in this consideration. 4.3.2 Experiments Experiment Display the current and voltage curves of a coil on the oscilloscope and determine the phase shift between the current IL and the voltage UL from the screen image. A Y1 Experiment procedure • Set up the experiment according to the circuit 1 kΩ (figure 4.3.2.1), connect the function generator and set the following voltage: u pp = 3 V G C ⊥ upp = 3 V (sinusoidal); f = 1 kHz (sinusoidal) 100 mH Attention: Make sure that test point C is not connected with B Y2 test points A or B through ground of the 0V oscilloscope equipment used (function generator, oscilloscope). Use an isolating transformer if Figure 4.3.2.1 necessary. The resistor (1 kΩ) in the circuit serves as a measuring resistor. The voltage UR applied to it is proportional to the capacitor current IL. For simultaneously displaying the coil voltage UL and the coil current IL (UR) the reference point of the voltages to be measured has been placed between the coil and the measuring resistor (1 kΩ) (test point C). V 0102 www.hps-systemtechnik.com 66 Alternating Current Technology Coil in the AC Circuit It must be taken into account that the two voltages are displayed with a 180° shift to each other.The actual display of the voltage curves is made by inverting one of the two voltages with the oscilloscope (in the experiment UL, channel 2, Y2). Settings: X = 0.1 ms / division Y1 = 1 V / division - 0 (Y1 ) Y2 = 1 V / division (inverse) Triggering: Y1 Remarks: Y1: Voltage UR (coil current IL) Y2: Coil voltage UL - 0 (Y2 ) Figure 4.3.2.2 • Connect the oscilloscope: - Test point A to channel 1 (Y1) - Test point B to channel 2 (Y2), inverse - Test point C to ground • Make other settings on the oscilloscope in accordance with the specifications below the grid (figure 4.3.2.2). • Draw the displayed voltage curves into the grid (figure 4.3.2.2) and determine the phase shift between the coil voltage UL and the capacitor current IL (UR). Phase shift between the coil current IL and the coil voltage UL: Period duration T T= Phase shift ϕ ϕ= www.hps-systemtechnik.com V 0102 Alternating Current Technology 67 Coil in the AC Circuit 4.4 Inductive Reactance of a Coil 4.4.1 General A coil has a current limiting effect in the AC circuit which results from the counter voltage generated in it. This current limiting effect is referred to as the inductive reactance XL. The value of the inductive reactance depends on the inductance of a coil and the frequency of the applied AC voltage. The inductive reactance is calculated on a sinusoidal AC voltage with the following formula: XL = 2 . π . f . L If coil current and coil voltage are known the inductive reactance can be calculated with the formula:: XL = inductive reactance of the capacitor in Ω UL XL = 2.π.f = radian frequency ω in 1/s IL L = inductance in H 4.4.2 Experiments Experiment Display the current and voltage curves of different coils at different frequencies on the oscilloscope and re- cord the curve XL = f (f). The respective inductive reactance is determined from the screen image by means of the peak-to-peak values. Experiment procedure A Y1 • Set up the experiment according to the cir- cuit (figure 4.4.2.1), set a voltage upp = 8 V 470 Ω (sinusoidal); f = 1 kHz on the function ge- u pp = 8 V nerator and connect it with the circuit. G 1…6 kHz C ⊥ (sinusoidal) • Connect the oscilloscope: - Test point A to channel 1 (Y1) L - Test point B to channel 2 (Y2), inverse - Test point C to ground B Y2 0V • Read off the peak-to-peak values of UR and oscilloscope UL for the frequencies and capacitors Figure 4.4.2.1 L = 8 mH / 100 mH specified in table 4.4.2.1 and enter them in 8 mH = transformer coil N = 900 the table. • Calculate values IL and XL and enter these in table 4.4.2.1 also. • Transfer the XL values to diagram 4.4.2.2 for constructing the characteristic curve XL = f (f). Attention: Make sure that test point C is not connected with test points A or B through ground of the equip- ment used (function generator, oscilloscope). Use an isolating transformer if necessary. The resistor (470 Ω) in the circuit serves as a measuring resistor. The voltage UR applied to it is proportional to the coil current IL. The respective coil current IL is calculated with the formula: IL = UR / R V 0102 www.hps-systemtechnik.com 68 Alternating Current Technology Coil in the AC Circuit f [kHz] 1 2 3 4 5 6 UL [V] 8 mH 100 mH UR [V] 8 mH 100 mH IL [mA] 8 mH 100 mH XL [kΩ] 8 mH 100 mH Table 4.4.2.1 5 4 3 XL / kΩ 2 1 0 0 1 2 3 4 5 6 7 f / kHz Figure 4.4.2.2 Question: What do the curves tell us? Answer: Calculate the inductive reactance XL of the coil 100 mH with the formula XL = ω . L (at f = 2kHz) and compare it with the value established in the experiment. XL = ω . L = Value established in the experiment: www.hps-systemtechnik.com V 0102 Alternating Current Technology 69 Coil in the AC Circuit 4.5 Series Circuiting of Coils 4.5.1 General L1 L2 L3 When coils are circuited in series the total IL + inductance is equal to the sum of the individual inductances. UL1 UL 2 UL 3 Ltot = L1 + L2 + L3 ... The partial voltages on the individual coils U U L tot behave proportionally to the respective inductance, their sum gives the total voltage - UL tot. Figure 4.5.1.1 The coil current in a series circuit of coils is the same at all points. 4.5.2 Experiments Experiment Prove, by measuring the current and voltage, that in a series circuit of coils the total inductance is equal to the sum of the individual inductances. Between the inductive reactance and the inductance of a coil there is a relationship: XL = ω . L L1 L2 L3 Experiment procedure A I1 B C D E • Set up the experiment according to the circuit (fi- 100 mH 8 mH 8 mH gure 4.5.2.1) and set a voltage of U rms = 5 V G Urms = 5 V (sinusoidal); f = 1 kHz f = 1kHz (sinusoidal) on the function generator and connect it with the circuit. 0V Figiure 4.5.2.1 L = 8 mH / 100 mH 8 mH = transformer coil N = 900 • Measure the current IL between test points A and B as well as the voltages at the test points B ... E with the multimeter and enter the values in table 4.5.2.1. • Calculate the individual inductive reactances, inductances and the total inductance in that order from the measured values. V 0102 www.hps-systemtechnik.com 70 Alternating Current Technology Coil in the AC Circuit Coil current [mA] Partial and total voltage [V] Test points Test points A-B (IL) B-C (UL 1) C-D (UL 2) D-E (UL 3) B-E (UL tot) Table 4.5.2.1 Calculation of the inductive reactances Calculation of inductances U L1 X L1 X L1 = = L1 = = IL ω U L2 X L2 X L2 = = L2 = = IL ω U L3 X L3 X L3 = = L3 = = IL ω U L tot X L tot X L tot = = Ltot = = IL ω www.hps-systemtechnik.com V 0102 Alternating Current Technology 71 Coil in the AC Circuit 4.6 Parallel Circuiting of Coils 4.6.1 General I L tot In a parallel circuit of coils the total inductance is less + than the smallest individual inductance. It is calculated IL 1 IL 2 IL 3 with the following formula: 1 L1 L2 L3 Ltot = 1 1 1 U + + ⋅⋅⋅ L1 L2 L3 - With only two parallel-circuited coils the following Figure 4.6.1.1 formula is used: L1 ⋅ L2 Ltot = L1 + L2 The current through the individual coils behaves inversely proportional to the respective inductance. The vol- tage UL is the same in a parallel circuit on all coils. 4.6.2 Experiments Experiment Prove, by measuring the current and voltage, that the total inductance in a parallel circuit of coils is less than the smallest individual inductance. Between the measured inductive reactances and the inductance there is a relationship: XL = ω . L A B I L tot C E G IL 1 IL 2 IL 3 U rms = 5 V G f = 5 kHz D F H (sinusoidal) L1 L2 L3 100 mH 8 mH 8 mH I 0V Figure 4.6.2.1 8 mH = transformer coil N = 900 Experiment procedure • Set up the experiment according to the circuit (figure 4.6.2.1), connect the function generator and set the following voltage: Urms = 5 V (sinusoidal); f = 5 kHz • Measure the total current IL tot and the partial currents IL 1, IL 2, IL 3 with a multimeter and enter the values in the table 4.6.2.1. V 0102 www.hps-systemtechnik.com 72 Alternating Current Technology Coil in the AC Circuit Partial and total current [mA] Partial voltage UL [V] Test points Test points A-B (IL tot) C-D (IL 1) E-F (IL 2) G-H (IL 3) D-I (UL 1) F-I (UL 2) H-I (UL 3) Table 4.6.2.1 • Calculate the inductive reactances XL tot, XL 1, XL 2 and XL 3 with the formula XL = UL / IL. • Calculate the individual inductances and the total inductance with the formula L = XL / ω. • Check the value of total inductances Ltot established in the experiment by calculation. Calculation of the inductive reactances Calculation of inductances UL1 X L1 X L1 = = L1 = = I L1 ω UL2 XL2 XL2 = = L2 = = IL2 ω UL3 X L3 XL3 = = L3 = = IL3 ω U L tot X L tot X L tot = = Ltot = = I L tot ω Checking the total inductance by calculation 1 Ltot = = 1 1 1 + + L1 L2 L3 www.hps-systemtechnik.com V 0102 Alternating Current Technology 73 Coil in the AC Circuit 4.7 Reactive Power of a Coil 4.7.1 General If a coil is applied to a sinusoidal AC voltage a phase shift of 90° results between the current and the voltage caused by the inductive reactance. The reactive power Q which occurs in the coil can be represented in the line diagram (figure 4.7.1.1) by mul- tiplication of the current and voltage values. Energy is fed to the coil during the positive half-wave. During the negative half-wave the coil radiates this sto- red energy. No active power is converted in the ideal coil (no ohmic resistance). Q U U I I Q 0 t Figure 4.7.1.1 The following definitions are valid: QL = UL . IL (rms values) qL = uL . iL (momentary values) QL; qL = inductive reactance in var UL; uL = coil voltage in V IL; iL = coil current in A V 0102 www.hps-systemtechnik.com 74 Alternating Current Technology Coil in the AC Circuit 4.7.2 Experiments Experiment Display the current and voltage curves of a coil on the oscilloscope, transfer the curves to a diagram and construct the power curve by multiplication of the momentary values of current and voltage. A Y1 Experiment procedure • Set up the experiment according to the cir- UR cuit (figure 4.7.2.1), connect the function 1 kΩ generator and set the following voltage: up = 4 V G C ⊥ up = 4 V (sinusoidal); f = 1 kHz f = 1 kHz (sinusoidal) • Connect the oscilloscope: UL - Test point A to channel 1 (Y1) 100 mH - Test point B to channel 2 (Y2), inverse B Y2 - Test point C to ground 0V Figure 4.7.2.1 oscilloscope • Settings on the oscilloscope: - Sweep X: 0,1 ms / division - Channel 1, Y1: 1 V / division ((voltage UR ~ coil current IL) - Channel 2, Y2: 1 V / division (coil voltage UL) - Center zero line of Y1 and Y2. - Triggering: Y1 • Draw the screen image on the diagram (figure 4.7.2.2) and establish the values for construction of the po- wer curve by multiplication of the voltage and current values (e. g. every 0.1 ms). Enter the values into the diagram (figure 4.7.2.2). Attention: Make sure that test point C is not connected with test points A or B through ground of the equipment used (function generator, oscilloscope). Use an isolating transformer if necessary. The resistor (1 kΩ) in the circuit serves as a measuring resistor. The voltage UR applied to it is proportional to the coil current IL (in mA). The actual display of the voltage curves is made by inverting one of the two voltages with the oscilloscope (in the experiment UL, channel 2, Y2). For simultaneously displaying the coil voltage UL and the coil current IL (UR) the reference point of the volta- ges to be measured has been placed between the coil and the measuring resistor (1 kΩ) (test point C). It must be taken into account that the two voltages are displayed with a 180° shift to each other. www.hps-systemtechnik.com V 0102 Alternating Current Technology 75 Coil in the AC Circuit Time t [ms] Current iL [mA] Voltage uL [V] Reactive power qL [mvar] 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 Table 4.7.2.1 5 I L [mA] U L [V] 4 Q L [mvar] 3 2 1 0 0.2 0.4 0.6 0.8 1.0 t [ ms] -1 -2 -3 -4 -5 Figure 4.7.2.2 V 0102 www.hps-systemtechnik.com 76 Alternating Current Technology Coil in the AC Circuit Notes: www.hps-systemtechnik.com V 0102 Alternating Current Technology 77 Interconnecting Resistor, Capacitor and Coil 5 Interconnecting Resistor, Capacitor and Coil 5.1 Fundamentals The following experiments deal with the interaction of resistor, capacitor and coil on a sinusoidal AC voltage. The aim is to measure and calculate the phase relation of current and voltage as well as the resulting values in parallel and series circuits of resistor, capacitor and coil. The values in accordance with phase and amount (rms values) can be shown either by a pointer diagram or with the oscilloscope. The experiments should be carried out with a constant frequency of 1 kHz. Experiments dealing with the fre- quency behaviour of resistor, capacitor and coil (e. g. high-pass, low-pass and band-pass) have not been ta- ken into consideration here. 15.2 Series Circuiting of Resistor and Capacitor 5.2.1 General If a sinusoidal AC voltage is applied to a series circuit of resistor and capacitor the same current flows through both components. R C I UR UC U Figure 5.2.1.1 Phase shifts occur between the voltages UR, UC and U due to the capacitive reactance XC of the capacitor. These can be shown by a line diagram or pointer diagram (see figure 5.2.1.2). V 0102 www.hps-systemtechnik.com 78 Alternating Current Technology Interconnecting Resistor, Capacitor and Coil The phase shift between the active voltage UR and the reactive voltage UC is constantly 90°, the same applies to the phase shift between the current and the reactive voltage UC. The phase shift between the apparent voltage U and the voltages UR and UC is determined by the voltage ra- tio UC to UR or the resistor ratio XC to R. UR = active voltage in V I = current in A U = apparent voltage, total voltage in V UC = reactive voltage, capacitor voltage in V ϕ = phase angle in ° (degrees) Figure 5.2.1.2 Pointer diagram for voltages The resistances can be represented by a pointer diagram as follows: R = active resistance in Ω XC = capacitive reactance in Ω Z = apparent resistance in Ω Figure 5.2.1.3 Pointer diagram for resistors www.hps-systemtechnik.com V 0102 Alternating Current Technology 79 Interconnecting Resistor, Capacitor and Coil Due to the phase shift between current and voltage, a simple numeric addition of the partial voltages and par- tial resistances in the series circuit of resistor and capacitor is not possible. They are calculated with the fol- lowing formulas: Apparent voltage U Capacitive reactance XC XC = Z . sin ϕ U = U R2 + U C2 U=Z.I Apparent resistance Z Tangent of the phase angle U UC X C Z = R 2 + X C2 Z= tan ϕ = = I UR R Active resistance R R = Z . cos ϕ 5.2.2 Experiments Experiment Measure and calculate the active voltage UR, the reactive voltage UC, the current I, the phase angle ϕ, the apparent resistance Z and the capacitive reactance XC in a series circuit of resistor and capacitor. Then draw the pointer diagrams for voltage and resistance from the measured values. A B C 1 kΩ U rms = 5 V G (sinusoidal) D ~ f = 1 kHz 0.22 µF 0V E Figure 5.2.2.1 Experiment procedure • Set up the experiment according to the circuit (figure 5.2.2.1), connect the function generator and set the following voltage in connection with the multimeter:: Urms = 5 V (sinusoidal); f = 1 kHz V 0102 www.hps-systemtechnik.com 80 Alternating Current Technology Interconnecting Resistor, Capacitor and Coil Current I Active voltage UR (test points A - B) (test points C - D) I= UR = Reactive voltage UC (test points D - E) UC = • Calculate the phase angle ϕ, apparent resistance Z and capacitive reactance XC. Phase angle ϕ Apparent resistance Z UC U tan δ = = Z= = UR I Capacitive reactance XC UC XC = Z . sin ϕ (sin ϕ = = ) XC = U • Draw pointer diagrams from the measured values. 0.5 V / division 100 Ω / division Figure 5.2.2.2 Pointer diagram for voltages Figure 5.2.2.3 Pointer diagram for resistances www.hps-systemtechnik.com V 0102 Alternating Current Technology 81 Interconnecting Resistor, Capacitor and Coil 5.3 Parallel Circuiting of Resistor and Capacitor 5.3.1 General If a sinusoidal AC voltage is connected to the parallel circuit of resistor and capacitor, the same voltage is ap- plied to both components. The current I is divided into the capacitor current IC and the active current IR. Phase shifts occur between the currents I, IC and IR due to the capacitive reactance XC of the capacitor. These can be represented in a line diagram or a pointer diagram (see figure 5.3.1.2). Figure 5.3.1.1 The voltage U is in phase with the active current IR whereas the reactive current IC of the capacitor, the volta- ge U and the active current are constantly phase shifted by 90°. The phase shift between the apparent current I and the currents IR and IC is determined by the current ratio IC to IR or the conductance ratio BC to G. IR = active current in A IC = reactive current, capacitor current in A I = apparent current, total current in A U = voltage in V ϕ = phase angle in ° (degrees) Figure 5.3.1.2 Pointer diagram for currents V 0102 www.hps-systemtechnik.com 82 Alternating Current Technology Interconnecting Resistor, Capacitor and Coil The conductances can also be shown on a pointer diagram. Y = apparent conductance in S (Siemens) G = active conductance in S BC = reactive conductance (capacitive) in S Figure 5.3.1.3 Pointer diagram for conductances Direct addition of the partial currents IC and IR and the conductances G and BC for determining the apparent conductance Y is not possible on account of the phase shifts between current and voltage on the capacitor; the following formulas must be used. Apparent current I Cosine, sine and tan of the angle ϕ IR G I = I R2 + I C2 cos ϕ = = I Y U I C BC I= sin ϕ = = Z I Y I C BC tan ϕ = = IR G Apparent conductance Y Y = G 2 + BC2 1 Y= Z Z = apparent resistance www.hps-systemtechnik.com V 0102 Alternating Current Technology 83 Interconnecting Resistor, Capacitor and Coil 5.3.2 Experiments Experiment Measure the apparent current I, the reactive current IC, the active current IR in a parallel circuit of resistor and capacitor and calculate the reactive conductance Y, the apparent resistance Z and the phase angle ϕ from the measured values. Then draw the pointer diagrams for current and conductance from the measured values. Experiment procedure • Set up the experiment according to the circuit (figure 5.3.2.1), connect the function generator and set the following voltage in connection with the multimeter: Urms = 5 V (sinusoidal); f = 1 kHz I A B IC IR U rms = 5 V C E (sinusoidal) G f = 1 kHz D F ~ 0.22 µF 1 kΩ 0V Figure 5.3.2.1 Apparent current I Reactive current IC Active current IR (test points A - B) (test points C - D) (test points E - F) I= IC = IR = • Calculate the reactive conductance Y, the apparent resistance Z and the phase angle ϕ. Reactive conductance Y Apparent resistance Z 1 Y = G 2 + BC2 = Z= = Y 1 Phase angle ϕ G= = R IC tan ϕ = = 1 1 IR BC = = =ω ⋅C = XC 1 ω ⋅C V 0102 www.hps-systemtechnik.com 84 Alternating Current Technology Interconnecting Resistor, Capacitor and Coil • Construct the pointer diagrams from the measured values. 1 mA / division 0.2 mS / division Figure 5.3.2.2 Pointer diagram for currents Figure 5.3.2.3 Pointer diagram for conductances www.hps-systemtechnik.com V 0102 Alternating Current Technology 85 Interconnecting Resistor, Capacitor and Coil 5.4 Series Circuiting of Resistor and Coil 5.4.1 General R L I If a sinusoidal AC voltage is applied to a series circuit of resistor and coil the same current flows through both UR UL components. U Figure 5.4.1.1 Phase shifts occur between the voltages UR, UL and U due to the inductive reactance XL of the coil. These can be shown by a line diagram or pointer diagram (figure 5.4.1.2). The resistances can also be represented by a pointer diagram (figure 5.4.1.3) as follows: Figure 5.4.1.2 Pointer diagram for voltages Figure 5.4.1.3 Pointer diagram for resistances UR = active voltage in V R = active resistance in Ω I = current in A XL = inductive reactance in Ω U = apparent voltage, total voltage in V Z = apparent resistance in Ω UL = reactive voltage, capacitor voltage in V ϕ = phase angle in ° (degrees) V 0102 www.hps-systemtechnik.com 86 Alternating Current Technology Interconnecting Resistor, Capacitor and Coil Due to the phase shift between current and voltage, a simple numeric addition of the partial voltages and par- tial in the series circuit of resistor and coil is not possible. They are calculated with the following formulas: Apparent voltage U Inductive reactance XL XL = Z . sin ϕ U = U R2 + U L2 U=Z.I Apparent resistance Z Tan and cosine of the phase angle ϕ UL X L Z = R 2 + X L2 tan ϕ = = UR R U UL XL Z= cos ϕ = = I U Z Active resistance R R = Z . cos ϕ www.hps-systemtechnik.com V 0102 Alternating Current Technology 87 Interconnecting Resistor, Capacitor and Coil 5.4.2 Experiments Experiment Measure the coil voltage UL, the active voltage UR, the current I and the phase shift between the apparent vol- tage U and the active voltage UR with the oscilloscope on a series circuit of resistor and coil. Calculate the inductive reactance XL of the coil, the apparent resistance Z and the phase shift between the apparent voltage U and the coil voltage UL. The equivalent resistance of the coil can be ignored in the calculations on account of its low value (13 Ω). Experiment procedure • Set up the experiment according to the circuit (figure 5.4.2.1), connect the function generator and set the following voltage in connection with the multimeter: upp = 6 V (sinusoidal); f = 1 kHz I • Measure the peak-to-peak voltage on the coil (test A points A - B) and the resistor (test points B - C) with the oscilloscope. 100 mH u pp = 6 V G (sinusoidal) B Attention: f = 1 kHz When measuring, make sure that the oscilloscope and 1 kΩ generator grounds are not connected through mains; use an isolating transformer if necessary. C 0V Figure 5.4.2.1 Coil voltage uL test points A - B uL = Active voltage uR test points B - C uR = The current i can be calculated by way of the voltage drop across the 1 kΩ resistor. Current i uR i= = R V 0102 www.hps-systemtechnik.com 88 Alternating Current Technology Interconnecting Resistor, Capacitor and Coil • Connect the oscilloscope as follows to determine the phase angle between the apparent voltage U and the active voltage UR: Test point A to channel 1 (Y1) Test point B to channel 2 (Y2) Test point C to ground The other settings on the oscilloscope should be made according to the specifications below grid (figure 5.4.2.2). Settings: X = 0.1 ms / division Y1 = 1 V / division Y2 = 1 V / division Triggering: Y1 - 0 (Y1 ,Y2 ) Remarks: Y1: Apparent voltage U Y2: Active voltage UR Figure 5.4.2.2 • Draw the displayed curves in the grid (figure 5.4.2.2) and determine the phase angle ϕ. Period duration T Apparent resistance Z T= Z = R 2 + X L2 = Phase angle ϕ Phase angle cos ϕ ϕ= between the apparent voltage U and the coil voltage UL UL cos ϕ = = U Inductive reactance XL XL = ω . L = www.hps-systemtechnik.com V 0102 Alternating Current Technology 89 Interconnecting Resistor, Capacitor and Coil 5.5 Parallel Circuiting of Resistor and Coil 5.5.1 General I If a sinusoidal AC voltage is applied to a parallel circuit of resistor and coil the same voltage is applied to both components. The current I is divided into the coil current IL and the active current IR. IL IR Phase shifts result between the currents I, IL and IR due to the inductive reactance XL of the coil. These can be represented by a line diagram L R or a pointer diagram (figure 5.5.1.2). U Figure 5.5.1.1 IR = active current in A U = voltage in V I = apparent current, total current in A IL = reactive current, coil current in A ϕ = phase angle in ° (degrees) Figure 5.5.1.2 Pointer diagram for currents The voltage U is in phase with the active current UR whereas the reactive current IL is constantly 90° later than the voltage U and the active current IR. The phase shift between the apparent current I and the currents IR and IC is determined by the current ratio IL to IR or the conductance ratio BL to G. V 0102 www.hps-systemtechnik.com 90 Alternating Current Technology Interconnecting Resistor, Capacitor and Coil The conductances can also be represented in a pointer diagram which has the following form: Y = apparent conductance in S (Siemens) G = active conductance in S BL = reactive inductance in S Figure 5.5.1.3 Pointer diagram for conductances Direct addition of the partial currents IL and IR and the conductances G and BL for determining the apparent current would lead to an incorrect result due to the phase shift between current and voltage. They are calculated by way of a geometric addition with the following formulas: Apparent current I U I = I R2 + I L2 I= Z Apparent conductance Y 1 Y = G 2 + BL2 Y= Z (Z = apparent resistance) Cosine, sine and tan of the angle ϕ IR G cos ϕ = = I Y I L BL sin ϕ = = I Y I L BL tan ϕ = = IR G www.hps-systemtechnik.com V 0102 Alternating Current Technology 91 Interconnecting Resistor, Capacitor and Coil 5.5.2 Experiments Experiment Measure the apparent current I, the reactive current IL, the active current IR on a parallel circuit of resistor and coil and calculate the active conductance G, the reactive conductance BL, the apparent conductance Y and the phase angle ϕ. Then construct the pointer diagrams for current and conductance. Experiment procedure I A B • Set up the experiment according to circuit (figure 5.5.2.1), connect the function genera- IL IR tor and set the following voltage in connecti- U rms = 5 V on with the multimeter: C E (sinusoidal) G Urms = 5 V (sinusoidal); f = 1 kHz f = 1 kHz D F ~ 100 mH 1 kΩ 0V Figure 5.5.2.1 • Measuring with the multimeter: Apparent current I (test points A - B) I= Reactive current IL (test points C - D) IL = Active current IR (test points E - F) IR = V 0102 www.hps-systemtechnik.com 92 Alternating Current Technology Interconnecting Resistor, Capacitor and Coil • Calculate the reactive conductance BL, apparent conductance Y, phase angle ϕ and active conductance G and construct the pointer diagrams from the measured values. Active conductance G Apparent conductance Y 1 G= = Y = G 2 + BC2 = R Reactive conductance BL Phase angle ϕ 1 I BL = = sin ϕ = L = ω⋅L I 1 mA / division 0.2 mS / division Figure 5.5.2.2 Pointer diagram for currents Fiure 5.5.2.3 Pointer diagram for conductances www.hps-systemtechnik.com V 0102 Alternating Current Technology 93 Interconnecting Resistor, Capacitor and Coil 5.6 Series Circuiting of Capacitor and Coil 5.6.1 General C L I If a sinusoidal AC voltage is applied to a series circuit of coil and capacitor the partial voltage UC follows the current I by 90° UC UL and the partial voltage UL precedes the current I by 90°, i. e. both partial voltages (also called reactive voltages) are in opposite phase (180°). U Figure 5.6.1.1 If both voltages have the same value, they equalize each other UL (resonance case). If one of the partial voltages is greater, the total voltage U is either inductive or capacitive, i. e. the current I 90 ° is 90° later. I The losses of the coil and the capacitor are ignored in this 90 ° consideration. UC Figure 5.6.1.2 Pointer diagram for voltages (UL = UC) UL UC UL I UL - UC UC - U L I UC UC UL Figure 5.6.1.3 Figure 5.6.1.4 Pointer diagram for voltages (UL > UC) Pointer diagram for voltages (UL < UC) V 0102 www.hps-systemtechnik.com 94 Alternating Current Technology Interconnecting Resistor, Capacitor and Coil 5.6.2 Experiments Experiment Measure the partial voltages UC and UL in the series circuits (coil / capacitor) and calculate whether the total voltage U precedes or follows the current I. Then construct the pointer diagrams. Experiment procedure A • Set up the experiment according to the circuit (fi- gure 5.6.2.1), connect the function generator and 1 µF set the following voltage: U rms = 3 V G (sinusoidal) Urms = 3 V (sinusoidal); f = 1 kHz B ~ f = 1 kHz 100 mH (8 mH) 0V C Figure 5.6.2.1 L = 8 mH / 100 mH 8 mH = transformer coil N = 900 • Measure voltages UC and UL. Then construct the pointer diagrams. Series circuit of C = 1 µF; L = 100 mH: Series circuit of C = 1 µF; L = 8 mH: Capacitor voltage UC Capacitor voltage UC (test points A - B) (test points A - B) UC = UC = Coil voltage UL Coil voltage UL (test points B - C) (test points B - C) UL = UL = Total voltage U Total voltage U U = UL - UC = U = UC - UL = www.hps-systemtechnik.com V 0102 Alternating Current Technology 95 Interconnecting Resistor, Capacitor and Coil 1 V / division 1 V / division I I Figure 5.6.2.2 Figure 5.6.2.3 Pointer diagram: Pointer diagram: C = 1 µF; L = 100 mH; f = 1 kHz C = 1 µF; L = 8 mH; f = 1 kHz V 0102 www.hps-systemtechnik.com 96 Alternating Current Technology Interconnecting Resistor, Capacitor and Coil Notes: www.hps-systemtechnik.com V 0102 Alternating Current Technology 97 Interconnecting Resistor, Capacitor and Coil 5.7 Parallel Circuiting of Capacitor and Coil 5.7.1 General If capacitor and coil are circuited in parallel and supplied with a sinusoidal I AC voltage, the same voltage is applied to both components. The total cur- rent I is divided into the coil current IL and the capacitor current IC, whereby IL IC IL precedes the voltage U by 90°. L C The currents IL and IC are at opposite phase (180°) and, depending on their size, equalize each other wholly or partly (see pointer diagrams). U Figure 5.7.1.1 When IC = IL the total current I is zero (resonance case). IC 90° U 90° IL Figure 5.7.1.2 Pointer diagram for currents (IL = IC) IC IC IL U IC - IL IL - IC U IL IL IC Figure 5.7.1.3 Pointer diagram for currents (IC > IL) Figure 5.7.1.4 Pointer diagram for currents (IC < IL) When IC > IL a capacitive residual current is left over, i. e. the total current I is capacitive and precedes volta- ge U by 90°. When IC < IL an inductive residual current is left over, i.e. the total current I is inductive and follows the voltage U by 90°. Coil and capacitor losses have been ignored in these considerations. V 0102 www.hps-systemtechnik.com 98 Alternating Current Technology Interconnecting Resistor, Capacitor and Coil 5.7.2 Experiments Experiment Measure the total current I, the coil current IL and the capacitor current IC on two parallel circuits (coil / capa- citor). Using the measured values, determine whether the total current I precedes or follows the voltage U. Then draw the corresponding pointer diagrams. I A B Experiment procedure • Set up the experiment according to the circuit IL IC (figure 5.7.2.1), connect the function generator U rms = 3 V and set the following voltage: C E (sinusoidal) G Urms = 3 V (sinusoidal); f = 1 kHz f = 1 kHz D F ~ L C 0V Figure 5.7.2.1 L / C-Kombination: 100 mH / 0.47 µF; 100 mH / 0.1 µF • Measure the currents I, IL and IC Then construct the pointer diagrams. Parallel of L = 100 mH; C = 0.47 µF Parallel of L = 100 mH; C = 0.1 µF Capacitor current IC Capacitor current IC (test points E - F) (test points E - F) IC = IC = Coil current IL Coil current IL (test points C - D) (test points C - D) IL = IL = Total current I Total current I (test points A - B) (test points A - B) I= I= www.hps-systemtechnik.com V 0102 Alternating Current Technology 99 Interconnecting Resistor, Capacitor and Coil 2 mA / division 1 mA / division U U Figure 5.7.2.2 Figure 5.7.2.3 Pointer diagram: Pointer diagram: L = 100 mH; C = 0.47 µF; f = 1 kHz L = 100 mH; C = 0.1 µF; f = 1 kHz V 0102 www.hps-systemtechnik.com 100 Alternating Current Technology Interconnecting Resistor, Capacitor and Coil Notes: www.hps-systemtechnik.com V 0102 Alternating Current Technology 101 Interconnecting Resistor, Capacitor and Coil 5.8 Series Circuiting of Resistor, Capacitor and Coil 5.8.1 General When connecting a sinusoidal AC voltage to a I R C L series circuit of resistor, capacitor and coil, the same current flows through all three components. UR UC UL The voltage UR is phase equal with the current I. The voltages UR, UC, UL and U are phase-shifted. The resistances behave in accordance with the voltages (see also pointer diagrams figure 5.8.1.2 U Figure 15.8.1.1 and figure 5.8.1.3). UL XL UR R ϕ I ϕ UC - U L XC - X L U Z UC UL XC XL Figure 5.8.1.2 Figure 5.8.1.3 Pointer diagram voltages (UC > UL) Pointer diagram resistances (XC > XL) UR = active voltage in V XL = inductive reactance, UL = reactive voltage (inductive), coil resistance in Ω coil voltage in V XC = capacitive reactance, UC = reactive voltage (capacitive), capacitor resistance in Ω capacitor voltage in V R = active resistance in Ω U = phase voltage, total voltage in V Z = apparent resistance in Ω I = current in A ϕ = phase angle in ° (degrees) The pointer diagrams show the case in which the voltage UC or the resistance XC is greater than the voltage UL or the resistance XL respectively, i. e. the capacitive part is superior. The voltage U follows the active voltage UR. If the inductive part is superior (UL > UC) the relationships are ac- cordingly reversed. If the inductive and capacitive parts are equal they equalize each other due to the phase shift of 180°; in this case the voltage U is equal to the active voltage UR and the apparent resistance Z is equal to the active re- sistance R. V 0102 www.hps-systemtechnik.com 102 Alternating Current Technology Interconnecting Resistor, Capacitor and Coil Below are some formulas for calculating values in a series circuit of resistor, coil and capacitor. Apparent voltage U U = U R2 + (U L − U C ) 2 U=Z.I Apparent resistance Z U Z = R 2 + (X L − X C ) 2 Z= I Tan of the phase angle ϕ U L − UC X L − X C tan ϕ = = UR R www.hps-systemtechnik.com V 0102 Alternating Current Technology 103 Interconnecting Resistor, Capacitor and Coil 5.8.2 Experiments Experiment Measure the voltages UR, UC and UL in a series circuit of resistor, capacitor and coil, determine whether the total voltage U precedes or follows the voltage UR and measure the phase angle ϕ with the oscilloscope. A Experiment procedure L 100 mH • Set up the experiment according to the circuit (figure 5.8.2.1), connect the function U rms = 3 V B generator and set the following voltage: G (sinusoidal) C 0.47 µ F ~ f = 1 kHz Urms = 3 V (sinusoidal); f = 1 kHz C R 470 Ω 0V D Figure 5.8.2.1 • Measuring with the multimeter: Coil voltage UL (test points A - B) UL = Capacitor voltage UC (test points B - C) UC = Active voltage UR (test points C - D) UR = • Determination of phase relation of U to UR: ............................................................................... ............................................................................... V 0102 www.hps-systemtechnik.com 104 Alternating Current Technology Interconnecting Resistor, Capacitor and Coil • Connect the oscilloscope as follows to determine the phase angle ϕ between the total voltage U and the active voltage UR: Test point C to channel 1 (Y1) Test point A to channel 2 (Y2) Test point D to ground • Make the other settings on the oscilloscope according to the specifications below the grid (figure 5.8.2.2). • Draw the displayed voltage curves in the grid (figure 5.8.2.2) and determine the phase angle ϕ. Settings: X = 0.1 ms / division Y1 = 2 V / division Y2 = 2 V / division - 0 (Y1) Triggering: Y1 Remarks: - 0 (Y2) Y1: Active voltage UR Y2: Total voltage, apparent voltage U Figure 5.8.2.2 Period duration T T= Phase angle ϕ ϕ= www.hps-systemtechnik.com V 0102 Alternating Current Technology 105 Interconnecting Resistor, Capacitor and Coil 5.9 Parallel Circuiting of Resistor, Capacitor and Coil 5.9.1 General If a sinusoidal AC voltage is connected to a parallel circuit of I resistor, capacitor and coil the voltage on all components is the same. The total current I is divided into active current IR, capacitor IC IL current IC and coil current IL. A phase shift occurs between the IR currents IL, IC, IR and I due to the reactances XL of the coil and XC of the capacitor (see pointer diagram figure 5.9.1.2). U C L R Figure 5.9.1.1 I = total current, apparent current in A IC IC = reactive current (capacitive), capacitor current in A IL IL = reactive current (inductive), coil current in A I IR = active current in A IC - I L ϕ U = voltage in V U ϕ = phase angle in ° (degrees) IR IL Figure 5.9.1.2 Pointer diagram for currents (IC > IL) The current IC precedes current IR constantly by 90°, whilst the current IL follows the active current IR constantly by 90°. Currents IC and IL are therefore in opposite phase (180°) and equalize each other wholly or partly depending on their size. When IC = IL the two currents equalize each other, the total current I is equal to the active current in phase and value (resonance). When IC > IL a capacitive residual current is left over, the total current I precedes the active current IR. When IC < IL an inductive residual current is left over, the total current I follows the active current IR. V 0102 www.hps-systemtechnik.com 106 Alternating Current Technology Interconnecting Resistor, Capacitor and Coil The pointer diagram figure 5.9.1.3 shows how the conductances BL, BC, G and Y behave in a parallel circuit of resistor, capacitor and coil. BC = reactive conductance (capacitive) in S BC BL BL = reactive conductance (inductive) in S Y = apparent conductance in S Y G = active conductance in S B C- B L ϕ G BL Figure 5.9.1.3 Pointer diagram for conductances (BC > BL) The pointer diagrams are prepared for the case IC > IL or BC > BL. Formulas for calculating the values in a parallel circuit of R, L and C: Apparent current I I = I R2 + (I C − I L ) 2 Apparent conductance Y Y = G 2 + (BC − BL ) 2 Tan of the phase angle ϕ I − IL tan ϕ = C IR B − BL tan ϕ = C G www.hps-systemtechnik.com V 0102 Alternating Current Technology 107 Interconnecting Resistor, Capacitor and Coil 5.9.2 Experiments Experiment Measure the current I, IL, IC and IR in a parallel circuit of resistor, capacitor and and calculate the phase angle ϕ . Then construct the pointer diagrams for currents and conductances. Experiment procedure I A B • Set up the experiment according to the circuit (figure 5.9.2.1), connect the func- IR IL tion generator and set the following vol- IC tage: C E G Urms = 3 V Urms = 3 V (sinusoidal); f = 1 kHz G (sinusoidal) D F H ~ f = 1 kHz R C L 470 Ω 0.47 µF 100 mH 0V Figure 5.9.2.1 • Measuring with the multimeter: Total current I Coil current IL (test points A - B) (test points G - H) I= IL = Active current IR Phase angle ϕ (test points C - D) I − IL tan ϕ = C = IR IR = Capacitor current IC (test points E - F) IC = V 0102 www.hps-systemtechnik.com 108 Alternating Current Technology Interconnecting Resistor, Capacitor and Coil Calculations for constructing the pointer diagrams: Reactive conductance (capacitive) BC Active conductance G BC = ω . C = 1 G= = R Reactive conductance (inductive) BL Apparent conductance Y 1 BL = = Y = G 2 + (BC − BL ) 2 = ω⋅L 2 mA / division 1 mS / division Figure 5.9.2.2 Pointer diagram for currents Figure 5.9.2.3 Pointer diagram for conductances www.hps-systemtechnik.com V 0102 Alternating Current Technology 109 Interconnecting Resistor, Capacitor and Coil 5.10 Active, Reactive and Apparent Power 5.10.1 General The previous experiments treated exclusively the behaviour I of voltage, current and resistance connecting capacitors, coils and ohmic resistors. The object of the experiment are the resulting powers. IC IL IR Like the voltages and currents the powers are also mutually U UC UL UR phase-shifted due to the reactances. (sinusoidal) Figure 5.10.1.1 In figures 5.10.1.2 to 5.10.1.4 the power ratios in a parallel circuit of resistor, capacitor and coil (figure 5.10.1.1) are shown as pointer diagrams. QC QL QC P S ϕ QC - QL Q L- Q C ϕ S P QL QL QC Figure 5.10.1.2 Figure 5.10.1.3 Pointer diagram for powers (QC > QL) Pointer diagram for powers QL > QC S = apparent power in VA P = active power in W QL = inductive reactive power in var QC = capacitive reactive power in var ϕ = phase angle in ° (degrees) If the capacitive reactive power QC is greater than the inductive reactive power QL (QC > QL) the apparent power S precedes the active power P (figure 5.10.1.2). In the reverse case (QL > Q) the apparent power S follows the active power P (figure 5.10.1.3). V 0102 www.hps-systemtechnik.com 110 Alternating Current Technology Interconnecting Resistor, Capacitor and Coil If both reactive powers are the same (QL = QC) they neutralize each other and the apparent power S is equal to the active power P (figure 5.10.1.4). QC The representation of the power ratios in a series circuit of resistor, ca- P (S) pacitor and coil are similar to those in a parallel circuit except that the power pointers QC and QL are reversed. QL Figure 5.10.1.4 Pointer diagram for powers QL = QC The individual powers are calculated with the following formulas: Apparent power S S = P 2 + (QL − QC ) 2 S=U.I Active power P P = U . I . cos ϕ P = S . cos ϕ Reactive power QL, QC Q Q = U . I . sin ϕ sin ϕ = S www.hps-systemtechnik.com V 0102 Alternating Current Technology 111 Interconnecting Resistor, Capacitor and Coil 5.10.2 Experiments Experiment Measure the apparent power S, the active power P, the reactive powers QC and QL and calculate the phase angle ϕ in a parallel circuit of resistor, capacitor and coil and then construct the corresponding pointer diagram. Experiment procedure I A B • Set up the experiment according to the circuit (figure 5.10.2.1), connect the func- IR IL tion generator and set the following vol- IC tage: C E G Urms = 3 V Urms = 3 V (sinusoidal); f = 1 kHz G (sinusoidal) D F H ~ f = 1 kHz R C L 470 Ω 0.47 µF 100 mH 0V Figure 5.10.2.1 • Measure the individual currents with the multimeter in order to calculate the powers. Total current I (test points A - B) I= Active current IR (test points C - D) IR = Capacitor current IC (test points E - F) IC = Coil current IL (test points G - H) IL = V 0102 www.hps-systemtechnik.com 112 Alternating Current Technology Interconnecting Resistor, Capacitor and Coil Calculate the powers and the phase angle with the given formulae and then draw the pointer diagram. Active power P Apparent power S P =U . IR = S=U.I= Reactive power (capacitive) QC Phase angle ϕ QC =U . IC = P cos ϕ = = S Reactive power (inductive) QL QL =U . IL = 6 mW (mvar, mVA) / division Figure 5.10.2.2 Pointer diagram for power www.hps-systemtechnik.com V 0102 Alternating Current Technology 113 Transformers 6 Transformers 6.1 Fundamentals Transformers consist of two or more windings (coils) which are magnetically coupled by a core. They are u- sed for voltage, current and resistance matching and transformation as well as for isolation of electrical circu- its (galvanic isolation). In the ideal, i. e. lossless, transformer, the power consumed is equal to the power produced. In practice los- ses occur in the winding and in the core of the transformer so that only a part of the consumed power is pas- sed on to the load resistor RL. The losses consist mainly of the copper losses (ohmic resistance of the coil windings), the core losses and losses caused by introduction of an air gap into the cross section of the core to improve the transformation properties. softmagnetic core I1 I2 input side G U1 U2 RL output side primary coil secondary coil Figure 6.1.1 6.2 Coupling Factor 6.2.1 General To get a fixed magnetic coupling between the primary and secondary coils of a transformer, they are connec- ted through a core. Depending on the coupling factor the magnetic flux in the core is greater or smaller and an accordingly higher or lower voltage is induced in the secondary coil. To avoid distortion of the signals to be transformed and premature magnetic saturation of the core material by DC current, the coupling factor is reduced by a break in the core (air gap). V 0102 www.hps-systemtechnik.com 114 Alternating Current Technology Transformers 6.2.2 Experiments Experiment Urms = 4 V Determine the size of the magnetic coupling bet- f = 1 kHz ween two coils by doing voltage measurements for: U1 N1 N2 G ~ 900 900 U2 - coils with core - coils with core and air gap - coils without core 0V Experiment procedure Figure 6.2.2.1 • Put together a primary and secondary coil with 900 windings each with the split tape core as Primary Secondary coil coil shown in the diagram (figure 6.2.2.2). • Connect the function generator to the primary coil as shown in the circuit diagram (figure 6.2.2.1) and set a sinusoidal voltage Urms = 4 V ; f = 1 kHz. • Measure the secondary voltage U2 with the mul- timeter and enter the value in the table 6.2.2.1 under „Coils with core“. N1 N2 900 900 • To measure the secondary voltage on coils with core and air gap, place a piece of paper between Figure 6.2.2.2 the upper and lower halves of the split tape core to simulate the air gap. • To determine the secondary voltage in „Coils N1 N2 without core“ arrange the coils as shown in figure 900 900 6.2.2.3. Figure 6.2.2.3 Question: Why do the secondary voltages differ? Coupling factor Answer: U1 [rms] U2 [rms] Coils with core 4V Coils with core and air gap 4V Coils without core 4V Table 6.2.2.1 www.hps-systemtechnik.com V 0102 Alternating Current Technology 115 Transformers 6.3 Transformation Ratio 6.3.1 General In a transformer the ratio of the number of windings on the primary side to the number of windings on the se- condary side is referred to as the transformation ratio. U1, I1 = primary voltage and current I1 I2 U2, I2 = secondary voltage and current N1, N2 = primary and secondary windings N1 N2 U1 U2 Figure 6.3.1.1 This ratio corresponds to the ratio of primary voltage U1 to secondary voltage U2 when the transformer is on no-load. The transformation ratio of the current is inverse to the voltage or winding ratio. N1 U 1 I 2 Transformation ratio r: r= = = N2 U2 I1 6.3.2 Experiments Experiment Determine the transformation ratio on the primary and secondary sides of a transformer with different num- bers of windings by measuring voltage and current. I1 I2 A Urms = 6 V N1 = N2 = Urms = 6 V N1 = N2 = G 300 V G 300 A ~ f = 1 kHz U 1 900 U2 ~ f = 1 kHz 900 900 900 0V 0V Figure 6.3.2.1 Figure 6.3.2.2 Experiment procedure • Put together a primary and secondary coil with 900 and 300 windings each with the split tape core as shown in the diagram (figure 6.3.2.3). • Connect the function generator to the secondary coil as shown in the circuit diagram (figure 6.3.2.1) and set a sinusoidal voltage Urms = 6 V; f = 1 kHz. V 0102 www.hps-systemtechnik.com 116 Alternating Current Technology Transformers Primary Secondary • Measure the secondary voltage U2 at the secondary coil with 300 coil coil and 900 windings and enter the values in the table 6.3.2.1. • Calculate the transformation ratio of the voltages with the formula r = U1 / U2. • Then measure the primary and secondary currents on the secondary coils with 300 and 900 windings as shown in the circuit (figure 6.3.2.2) and enter the values in table 6.3.2.2. The voltage U1 of N1 N2 6 V (rms) should be kept constant. 900 300 900 Figure 6.3.2.3 • Calculate the transformation ratio r of the currents with the formula: I2 r= = I1 N1 N2 U1 [V] U2[V] r N1 N2 I1 [mA] I2[mA] r 900 300 6 900 300 900 900 6 900 900 Table 6.3.2.1 Table 6.3.2.2 Transformation ratios of the voltages: Transformation ratios of the currents: When N1 = 900; N2 = 300 When N1 = 900; N2 = 300 U1 I2 r= = r= = U2 I1 When N1 = 900; N2 = 900 When N1 = 900; N2 = 900 U1 I2 r= = r= = U2 I1 www.hps-systemtechnik.com V 0102 Alternating Current Technology 117 Transformers 6.4 Resistance Transformation 6.4.1 General If a transformer is not loaded (no-load) the transformation ratio between primary and secondary voltages cor- responds approximately to the ratio between the number of windings on the primary and secondary coils (transformation ratio). U1 N r= = 1 U2 N2 When there is a load on the secondary coil, a current I2 flows through the load resistor RL. This current is ac- tive, multiplied by the reciprocal of the transformation ratio, in the primary coil of the transformer. 1 I2 I1 = I 2 ⋅ r= r I1 Primary current and primary voltage together form the input resistance of the transformer. U1 R1 = I1 The output or load resistance is given by: U2 R2 = R L = I2 If both resistances are put in relationship, the following resistance transformation is achieved: U1 R1 I U ⋅I = 1 = 1 2 = r ⋅ r = r2 or: R1 = R L ⋅ r 2 RL U 2 U 2 ⋅ I1 I2 This means the secondary load RL is transformed to the primary side of the transformer with the square of the transformation ratio N1 / N2. 6.4.2 Experiments Experiment Establish the resistances R1 and R2 by measuring the current and voltage on the primary and secondary si- des of a transformer. Do this for different transformation ratios of the number of primary windings to the num- ber of secondary windings and for different loads RL. Then form the ratio between R1 and R2 and compare with the square of the respective transformation ratio. V 0102 www.hps-systemtechnik.com 118 Alternating Current Technology Transformers I1 I2 Experiment procedure mA mA • Put together the transformer with the split tape core and the coils (N1 = 900; N2 = 300) as shown in the Urms = 6 V diagram (figure 6.4.2.2). G N1 N2 U2 RL ~ f = 1 kHz • Connect the function generator as shown in the cir- cuit diagram (figure 6.4.2.1) and set a sinusoidal 0V voltage of Figure 6.4.2.1 RL = 100 Ω / 1 kΩ, Urms = 6 V; f = 1 kHz (at RL = 10 Ω). N1 = 900, N2 = 300 / 900 • Measure current and voltage at the winding ratios and loads RL specified in table 6.4.2.3. Primary Secondary coil coil • Calculate the resistances R1 and R2 according to Ohm’s Law with the formula R = U / I. • Then calculate the transformation ratios of the re- sistances according to the specifications and com- pare with the winding ratios. N1 N2 300 300 900 Figure 6.4.2.2 N1 N2 r RL [Ω] U1 [V] U2 [V] I1 [mA] I2 [mA] R1 [Ω] R2 [Ω] 900 300 0.33 100 900 900 1 1000 Table 6.4.2.3 Ratio between the resistances: When N1 = 900; N2 = 300; RL = 100 Ω When N1 = 900; N2 = 900; RL = 1000 Ω R1 R1 r2 = = r2 = = R2 R2 r = r2 = = r = r2 = = www.hps-systemtechnik.com V 0102