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DC/AC Fundamentals: A Systems Approach Thomas L. Floyd David M. Buchla RL Circuits Chapter 12 Ch.12 Summary Sinusoidal Response of RL Circuits When both resistance and inductance are in a series circuit, the phase angle between the applied voltage and total current is between 0 and 90, depending on the values of resistance and reactance. VS VR VS VL VR lags VS VL leads VS R L VS VS I I lags VS DC/AC Fundamentals: A Systems Approach Thomas L. Floyd © 2013 by Pearson Higher Education, Inc Upper Saddle River, New Jersey 07458 • All Rights Reserved Ch.12 Summary Series RL Circuit Impedance In a series RL circuit, the total impedance is the phasor sum of R and XL. R is plotted along the positive x-axis. XL is plotted along the positive y-axis. Z is the diagonal. It is convenient to reposition the phasors into an impedance triangle. XL DC/AC Fundamentals: A Systems Approach Thomas L. Floyd Z q tan 1 XL R XL Z q q R R © 2013 by Pearson Higher Education, Inc Upper Saddle River, New Jersey 07458 • All Rights Reserved Ch.12 Summary Series RL Circuit Impedance Sketch the impedance triangle and show the values for R = 1.2 kW and XL = 960 W. Z R 2 X L2 (1.2 kW)2 (0.96 kW)2 1.33 kΩ X q tan 1 L R 1 0.96 kΩ tan 1.2 kΩ 39 DC/AC Fundamentals: A Systems Approach Thomas L. Floyd Z = 1.33 kW XL = 0.96 kW q = 39o R = 1.2 kW © 2013 by Pearson Higher Education, Inc Upper Saddle River, New Jersey 07458 • All Rights Reserved Ch.12 Summary Series RL Circuit Analysis Ohm’s law is applied to series RL circuits using quantities of Z, V, and I. V IZ V I Z V Z I Because I is the same everywhere in a series circuit, you can multiply the impedance phasor values by the circuit current to obtain the voltage phasor values. DC/AC Fundamentals: A Systems Approach Thomas L. Floyd © 2013 by Pearson Higher Education, Inc Upper Saddle River, New Jersey 07458 • All Rights Reserved Ch.12 Summary Series RL Circuits Analysis Assume the current in the previous example is 10 mA. Sketch the voltage phasors. (The impedance triangle from the previous example is shown for reference.) The voltage phasor diagram can be found using Ohm’s law. Multiply each impedance phasor by 10 mA (as shown below): Z = 1.33 kW XL = 0.96 kW x 10 mA = VS = 13.3 V VL = 9.6 V q = 39o q = 39o R = 1.2 kW VR = 12 V DC/AC Fundamentals: A Systems Approach Thomas L. Floyd © 2013 by Pearson Higher Education, Inc Upper Saddle River, New Jersey 07458 • All Rights Reserved Ch.12 Summary Phase Angle Vs. Frequency Phasor diagrams that have reactance phasors can only be drawn for a single frequency because X is a function of frequency. As frequency changes, the impedance triangle for an RL circuit changes (as illustrated here) because XL is directly proportional to f. This determines the frequency response of RL circuits. DC/AC Fundamentals: A Systems Approach Thomas L. Floyd R R q1 q3 q2 q2 Increasing f q1 Z Z33 q3 X XC3 ff 3 3 X XC2 f 22 X XC1 ff1 Z Z22 Z Z11 © 2013 by Pearson Higher Education, Inc Upper Saddle River, New Jersey 07458 • All Rights Reserved Ch.12 Summary Phase Shift A series RL circuit can be used to produce a specific phase lead between an input voltage and an output by taking the output across the inductor. This circuit is a basic high-pass filter, a circuit that passes high frequencies and rejects all others. This filter passes frequencies that are above a specific frequency, called the cutoff frequency. V Vin R Vout (phase lead) Vin L Vin Vout Vout VR DC/AC Fundamentals: A Systems Approach Thomas L. Floyd © 2013 by Pearson Higher Education, Inc Upper Saddle River, New Jersey 07458 • All Rights Reserved Ch.12 Summary Phase Shift Reversing the components in the previous circuit produces a circuit that is a basic lag network. This circuit is a lowpass filter, a circuit that passes low frequencies and rejects all others. This filter passes low frequencies up to a frequency called the cutoff frequency. L VL Vin Vin Vout Vin R Vout (phase lag) f Vout DC/AC Fundamentals: A Systems Approach Thomas L. Floyd © 2013 by Pearson Higher Education, Inc Upper Saddle River, New Jersey 07458 • All Rights Reserved Ch.12 Summary AC Response of Parallel RL Circuits For parallel circuits, it is useful to review conductance, susceptance and admittance, introduced in Chapter 10. Conductance is the reciprocal of resistance. 1 G R 1 XL Inductive susceptance is the reciprocal of inductive reactance. BL Admittance is the reciprocal of impedance. 1 Y Z DC/AC Fundamentals: A Systems Approach Thomas L. Floyd © 2013 by Pearson Higher Education, Inc Upper Saddle River, New Jersey 07458 • All Rights Reserved Ch.12 Summary AC Response of Parallel RL Circuits In a parallel RL circuit, the admittance phasor is the sum of the conductance and capacitive susceptance phasors: Y G 2 BL2 From the diagram, the phase angle is: 1 BL q VS q tan G DC/AC Fundamentals: A Systems Approach Thomas L. Floyd G G BL BL Y © 2013 by Pearson Higher Education, Inc Upper Saddle River, New Jersey 07458 • All Rights Reserved Ch.12 Summary AC Response of Parallel RL Circuits Draw the admittance phasor diagram for the circuit below. The magnitudes of conductance, susceptance, and admittance are: 1 1 1 1 B .629 mS G 1 mS C X L 2(10 kHz)(25.3 mH) R 1 kΩ Y G 2 BL2 (1 mS) 2 (0.629 mS) 2 1.18 mS q VS f =10 kHz R 1 kW L 25.3 mH G= 1.0 mS Y= 1.18 mS BL = 0.629 mS DC/AC Fundamentals: A Systems Approach Thomas L. Floyd © 2013 by Pearson Higher Education, Inc Upper Saddle River, New Jersey 07458 • All Rights Reserved Ch.12 Summary Analysis of Parallel RL Circuits Ohm’s law is applied to parallel RL circuits using quantities of Y, V, and I. I V Y I VY I Y V Because V is the same across all components in a parallel circuit, you can obtain the current in a given component by simply multiplying the admittance of the component by the voltage as illustrated in the following example. DC/AC Fundamentals: A Systems Approach Thomas L. Floyd © 2013 by Pearson Higher Education, Inc Upper Saddle River, New Jersey 07458 • All Rights Reserved Ch.12 Summary Analysis of Parallel RL Circuits If the voltage in the previous example is 10 V, sketch the current phasor diagram. The admittance diagram from the previous example is shown for reference. The current phasor diagram can be found from Ohm’s law. Multiply each admittance phasor by 10 V. G = 1.0 mS BL = 0.629 mS Y= 1.18 mS DC/AC Fundamentals: A Systems Approach Thomas L. Floyd IR = 10 mA x 10 V = IL = 6.29 mA IS = 11.8 mA © 2013 by Pearson Higher Education, Inc Upper Saddle River, New Jersey 07458 • All Rights Reserved Ch.12 Summary Phase Angle of Parallel RL Circuits Notice that the formula for inductive susceptance is the reciprocal of inductive reactance. Thus BL and IL are inversely proportional to f: 1 BL 2fL As frequency increases, BL and IL decrease, so the angle between IR and IS must decrease as well. DC/AC Fundamentals: A Systems Approach Thomas L. Floyd q IL IR IS © 2013 by Pearson Higher Education, Inc Upper Saddle River, New Jersey 07458 • All Rights Reserved Ch.12 Summary Series-Parallel RL Circuits Series-parallel RL circuits are combinations of both series and parallel elements. These circuits can be solved by methods from Z1 series and parallel circuits. Z2 For example, the components in the green box are in series: RR1 L1 C RR22 L 2C2 Z1 R12 X L2 The components in the yellow box are in parallel: R2 X L 2 Z2 R22 X L22 DC/AC Fundamentals: A Systems Approach Thomas L. Floyd The total impedance can be found by converting the parallel components to an equivalent series combination, then adding the result to R1 and XL1 to get the total reactance. © 2013 by Pearson Higher Education, Inc Upper Saddle River, New Jersey 07458 • All Rights Reserved Ch.12 Summary The Power Triangle As shown earlier, you can multiply the impedance phasors for a series RL circuit by the current to obtain the voltage phasors. The earlier example is shown below for review: Z = 1.33 kW XL = 0.96 kW x 10 mA = VS = 13.3 V VL = 9.6 V q = 39o q = 39o R = 1.2 kW VR = 12 V Multiplying each value in the left-hand triangle gives you the corresponding value in the right-hand triangle. DC/AC Fundamentals: A Systems Approach Thomas L. Floyd © 2013 by Pearson Higher Education, Inc Upper Saddle River, New Jersey 07458 • All Rights Reserved Ch.12 Summary The Power Triangle (Cont’d) Multiplying the voltage phasors by Irms (10 mA) gives the power triangle values (because P = V I ). Apparent power is the product of the magnitude of the current and magnitude of the voltage and is plotted along the hypotenuse of the power triangle. x 10 mA = VS = 13.3 V VL = 9.6 V q = 39o VR = 12 V DC/AC Fundamentals: A Systems Approach Thomas L. Floyd Pa = 133 mVA q 39o Pr = 96 mVAR Ptrue = 120 mW © 2013 by Pearson Higher Education, Inc Upper Saddle River, New Jersey 07458 • All Rights Reserved Ch.12 Summary Power Factor The power factor was introduced in Chapter 10 and applies to RL circuits (as well as RC circuits). Recall that it is the relationship between the apparent power (in VA) and true power (in W). True power equals the product of Voltamperes and power factor. Power factor can be determined using: PF cos q Power factor can vary from 0 (for a purely reactive circuit) to 1 (for a purely resistive circuit). DC/AC Fundamentals: A Systems Approach Thomas L. Floyd © 2013 by Pearson Higher Education, Inc Upper Saddle River, New Jersey 07458 • All Rights Reserved Ch.12 Summary Apparent Power Apparent power consists of two components; the true power component, which does the work, and a reactive power component, that is simply power shuttled back and forth between source and load. Components such as transformers, motors, and generators are rated in VA rather than watts. Pa (VA) Pr ( VAR) q Ptrue (W) DC/AC Fundamentals: A Systems Approach Thomas L. Floyd © 2013 by Pearson Higher Education, Inc Upper Saddle River, New Jersey 07458 • All Rights Reserved Ch.12 Summary Frequency Response of RL Circuits The response of a series RL circuit is similar to that of a series RC circuit. In the case of the low-pass response shown here, the output is taken across the resistor. Vout Vin 10 10V V rms rms 10 V dc 0 10 V rms 10 V dc 10 mH 10mH mH 10 ƒ == 110kHz ƒƒ kHz = 20 kHz 8.46 1.57 V V rms rms V rms 10 V dc 0.79 100 W 100W W 100 100 W 0 V out (V) Plotting the response: 9.98 9.98 8.46 8.46 9 8 7 6 5 4 3 1.57 1.57 0.79 0.79 2 1 0.1 0.1 DC/AC Fundamentals: A Systems Approach Thomas L. Floyd 11 10 10 20 100 f (kHz) © 2013 by Pearson Higher Education, Inc Upper Saddle River, New Jersey 07458 • All Rights Reserved Ch.12 Summary Frequency Response of RL Circuits Reversing the position of the R and L components produces a high-pass response. In this case, the output is taken across the inductor. Vin Vout 10VV rms rms 10 10 V dc 0 10 V rms 10 V dc 100 100 W W 100 W ƒ = 100 Hz 1 kHz ƒ = 10 kHz 10 10 mH 10mH mH 10 mH 9.87 V rms 5.32 V rms 0.63 V rms 0 V dc Vout (V) 9.87 9.87 Plotting the response: 5.32 5.32 10 9 8 7 6 5 4 3 0.63 0.63 DC/AC Fundamentals: A Systems Approach Thomas L. Floyd 2 1 0 0.01 0.1 0.1 11 10 10 f (kHz) © 2013 by Pearson Higher Education, Inc Upper Saddle River, New Jersey 07458 • All Rights Reserved Ch.12 Summary Key Terms Inductive The ability of an inductor to permit susceptance (BL) current; the reciprocal of inductive reactance, measured in siemens (S). DC/AC Fundamentals: A Systems Approach Thomas L. Floyd © 2013 by Pearson Higher Education, Inc Upper Saddle River, New Jersey 07458 • All Rights Reserved Ch.12 Summary Quiz 1. If the frequency is increased in a series RL circuit, the phase angle will a. increase b. decrease c. remain unchanged DC/AC Fundamentals: A Systems Approach Thomas L. Floyd © 2013 by Pearson Higher Education, Inc Upper Saddle River, New Jersey 07458 • All Rights Reserved Ch.12 Summary Quiz 2. If you multiply each of the impedance phasors in a series RL circuit by the current, the result is the a. voltage phasors b. power phasors c. admittance phasors d. none of the above DC/AC Fundamentals: A Systems Approach Thomas L. Floyd © 2013 by Pearson Higher Education, Inc Upper Saddle River, New Jersey 07458 • All Rights Reserved Ch.12 Summary Quiz 3. For the circuit shown, the output voltage a. is in phase with the input voltage b. leads the input voltage c. lags the input voltage d. leads the resistor voltage Vin DC/AC Fundamentals: A Systems Approach Thomas L. Floyd Vout © 2013 by Pearson Higher Education, Inc Upper Saddle River, New Jersey 07458 • All Rights Reserved Ch.12 Summary Quiz 4. Which of the equations below can be used to calculate the phase angle in a series RL circuit? a. q tan 1 X L R b. q tan 1VL V R c. both of the above are correct d. none of the above is correct DC/AC Fundamentals: A Systems Approach Thomas L. Floyd © 2013 by Pearson Higher Education, Inc Upper Saddle River, New Jersey 07458 • All Rights Reserved Ch.12 Summary Quiz 5. In a series RL circuit, if the inductive reactance is equal to the resistance, the source current will lag the source voltage by a. 0o b. 30o c. 45o d. 90o DC/AC Fundamentals: A Systems Approach Thomas L. Floyd © 2013 by Pearson Higher Education, Inc Upper Saddle River, New Jersey 07458 • All Rights Reserved Ch.12 Summary Quiz 6. Susceptance is the reciprocal of a. resistance b. reactance c. admittance d. impedance DC/AC Fundamentals: A Systems Approach Thomas L. Floyd © 2013 by Pearson Higher Education, Inc Upper Saddle River, New Jersey 07458 • All Rights Reserved Ch.12 Summary Quiz 7. In a parallel RL circuit, the magnitude of the admittance can be expressed as a. Y 1 1 1 G BL b. Y G 2 BL2 c. Y G B L d. Y G 2 BL2 DC/AC Fundamentals: A Systems Approach Thomas L. Floyd © 2013 by Pearson Higher Education, Inc Upper Saddle River, New Jersey 07458 • All Rights Reserved Ch.12 Summary Quiz 8. If you increase the frequency in a parallel RL circuit, a. the total admittance will increase b. the total current will increase c. both a and b d. none of the above DC/AC Fundamentals: A Systems Approach Thomas L. Floyd © 2013 by Pearson Higher Education, Inc Upper Saddle River, New Jersey 07458 • All Rights Reserved Ch.12 Summary Quiz 9. The unit used for measuring true power is the a. volt-ampere b. watt c. volt-ampere-reactive (VAR) d. kilowatt-hour DC/AC Fundamentals: A Systems Approach Thomas L. Floyd © 2013 by Pearson Higher Education, Inc Upper Saddle River, New Jersey 07458 • All Rights Reserved Ch.12 Summary Quiz 10. A power factor of zero implies that the a. circuit is entirely reactive b. reactive and true power are equal c. circuit is entirely resistive d. maximum power is delivered to the load DC/AC Fundamentals: A Systems Approach Thomas L. Floyd © 2013 by Pearson Higher Education, Inc Upper Saddle River, New Jersey 07458 • All Rights Reserved Ch.12 Summary Answers 1. a 6. b 2. a 7. d 3. c 8. d 4. a 9. b 5. c 10. a DC/AC Fundamentals: A Systems Approach Thomas L. Floyd © 2013 by Pearson Higher Education, Inc Upper Saddle River, New Jersey 07458 • All Rights Reserved