Chapter 12 © Copyright 2007 Prentice-HallElectric Circuits Fundamentals - Floyd Chapter 12

© Copyright 2007 Prentice-HallElectric Circuits Fundamentals - Floyd 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.

Chapter 12 © Copyright 2007 Prentice-HallElectric Circuits Fundamentals - Floyd Summary Impedance of series RL circuits In a series RL circuit, the total impedance is the phasor sum of R and X L. R is plotted along the positive x-axis. R X L is plotted along the positive y-axis. XLXL Z It is convenient to reposition the phasors into the impedance triangle. R XLXL Z Z is the diagonal

Chapter 12 © Copyright 2007 Prentice-HallElectric Circuits Fundamentals - Floyd Summary Impedance of series RL circuits R = 1.2 k  X L = 960  Sketch the impedance triangle and show the values for R = 1.2 k  and X L = 960 . Z = 1.33 k  39 o

Chapter 12 © Copyright 2007 Prentice-HallElectric Circuits Fundamentals - Floyd Summary Analysis of series RL circuits Ohm’s law is applied to series RL circuits using quantities of Z, V, and I. Because I is the same everywhere in a series circuit, you can obtain the voltage phasors by simply multiplying the impedance phasors by the current.

Chapter 12 © Copyright 2007 Prentice-HallElectric Circuits Fundamentals - Floyd Summary Analysis of series RL circuits Assume the current in the previous example is 10 mA rms. Sketch the voltage phasors. The impedance triangle from the previous example is shown for reference. V R = 12 V V L = 9.6 V The voltage phasors can be found from Ohm’s law. Multiply each impedance phasor by 10 mA. x 10 mA = R = 1.2 k  X L = 960  Z = 1.33 k  39 o V S = 13.3 V 39 o

Chapter 12 © Copyright 2007 Prentice-HallElectric Circuits Fundamentals - Floyd Summary Variation of phase angle with 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 X L increases with increasing f. This determines the frequency response of RL circuits.

Chapter 12 © Copyright 2007 Prentice-HallElectric Circuits Fundamentals - Floyd Summary Phase shift R VRVR V out L V in V out V in For a given frequency, a series RL circuit can be used to produce a phase lead by a specific amount between an input voltage and an output by taking the output across the inductor. This circuit is also a basic high-pass filter, a circuit that passes high frequencies and rejects all others.

Chapter 12 © Copyright 2007 Prentice-HallElectric Circuits Fundamentals - Floyd Summary R VLVL V out L V in V out V in Reversing the components in the previous circuit produces a circuit that is a basic lag network. This circuit is also a basic low-pass filter, a circuit that passes low frequencies and rejects all others. Phase shift

Chapter 12 © Copyright 2007 Prentice-HallElectric Circuits Fundamentals - Floyd Summary Sinusoidal 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. Admittance is the reciprocal of impedance. Inductive susceptance is the reciprocal of inductive reactance.

Chapter 12 © Copyright 2007 Prentice-HallElectric Circuits Fundamentals - Floyd Summary Sinusoidal response of parallel RL circuits VSVS GBLBL Y G BLBL In a parallel RL circuit, the admittance phasor is the sum of the conductance and inductive susceptance phasors. The magnitude of the susceptance is The magnitude of the phase angle is

Chapter 12 © Copyright 2007 Prentice-HallElectric Circuits Fundamentals - Floyd Summary Sinusoidal response of parallel RL circuits Y G BLBL G is plotted along the positive x-axis. B L is plotted along the negative y-axis. Y is the diagonal VSVS GBLBL Some important points to notice are:

Chapter 12 © Copyright 2007 Prentice-HallElectric Circuits Fundamentals - Floyd Summary Sinusoidal response of parallel RL circuits VSVS L 25.3 mH Y = 1.18 mS G = 1.0 mS B L = mS Draw the admittance phasor diagram for the circuit. f = 10 kHz R 1.0 k  The magnitude of the conductance and susceptance are:

Chapter 12 © Copyright 2007 Prentice-HallElectric Circuits Fundamentals - Floyd Summary Analysis of parallel RL circuits Ohm’s law is applied to parallel RL circuits using quantities of Y, V, and I. 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.

Chapter 12 © Copyright 2007 Prentice-HallElectric Circuits Fundamentals - Floyd Summary Analysis of parallel RL circuits Assume the voltage in the previous example is 10 V. Sketch the current phasors. The admittance diagram from the previous example is shown for reference. The current phasors can be found from Ohm’s law. Multiply each admittance phasor by 10 V. x 10 V = I R = 10 mA I L = 6.29 mA I S = 11.8 mA Y = 1.18 mS G = 1.0 mS B L = mS

Chapter 12 © Copyright 2007 Prentice-HallElectric Circuits Fundamentals - Floyd Summary Phase angle of parallel RL circuits Notice that the formula for inductive susceptance is the reciprocal of inductive reactance. Thus B L and I L are inversely proportional to f: IRIR ILIL ISIS  As frequency increases, B L and I L decrease, so the angle between I R and I S must decrease as well.

Chapter 12 © Copyright 2007 Prentice-HallElectric Circuits Fundamentals - Floyd Summary The power triangle x 10 mA = Recall that in a series RC or RL circuit, you could multiply the impedance phasors by the current to obtain the voltage phasors. The earlier example from this chapter is shown for review: R = 1.2 k  X L = 960  Z = 1.33 k  39 o V R = 12 V V L = 9.6 V V S = 13.3 V 39 o

Chapter 12 © Copyright 2007 Prentice-HallElectric Circuits Fundamentals - Floyd Summary Multiplying the voltage phasors by I rms gives the power triangle (equivalent to multiplying the impedance phasors by I 2 ). 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. The rms current in the earlier example was 10 mA. Show the power triangle. x 10 mA = P true = 120 mW P r = 96 mVAR The power triangle 39 o P a = 133 mVA V R = 12 V V L = 9.6 V V S = 13.3 V 39 o

Chapter 12 © Copyright 2007 Prentice-HallElectric Circuits Fundamentals - Floyd Inductive susceptance (B L ) The ability of an inductor to permit current; the reciprocal of inductive reactance. The unit is the siemens (S). Key Terms

Chapter 12 © Copyright 2007 Prentice-HallElectric Circuits Fundamentals - Floyd Quiz 1. If the frequency is increased in a series RL circuit, the phase angle will a. increase b. decrease c. be unchanged

Chapter 12 © Copyright 2007 Prentice-HallElectric Circuits Fundamentals - Floyd 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

Chapter 12 © Copyright 2007 Prentice-HallElectric Circuits Fundamentals - Floyd 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. none of the above V in V out

Chapter 12 © Copyright 2007 Prentice-HallElectric Circuits Fundamentals - Floyd Quiz 4. In a series RL circuit, the phase angle can be found from the equation a. b. c. both of the above are correct d. none of the above is correct

Chapter 12 © Copyright 2007 Prentice-HallElectric Circuits Fundamentals - Floyd 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. 0 o b. 30 o c. 45 o d. 90 o

Chapter 12 © Copyright 2007 Prentice-HallElectric Circuits Fundamentals - Floyd Quiz 6. Susceptance is the reciprocal of a. resistance b. reactance c. admittance d. impedance

Chapter 12 © Copyright 2007 Prentice-HallElectric Circuits Fundamentals - Floyd Quiz 7. In a parallel RL circuit, the magnitude of the admittance can be expressed as a. b. c. Y = G + B L d.

Chapter 12 © Copyright 2007 Prentice-HallElectric Circuits Fundamentals - Floyd 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

Chapter 12 © Copyright 2007 Prentice-HallElectric Circuits Fundamentals - Floyd Quiz 9. The unit used for measuring true power is the a. volt-ampere b. watt c. volt-ampere-reactive (VAR) d. kilowatt-hour

Chapter 12 © Copyright 2007 Prentice-HallElectric Circuits Fundamentals - Floyd 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

Chapter 12 © Copyright 2007 Prentice-HallElectric Circuits Fundamentals - Floyd Quiz Answers: 1. a 2. a 3. c 4. a 5. c 6. b 7. d 8. d 9. b 10. a