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chapter 12 RL Circuits Electronics Fundamentals 8th edition Floyd/Buchla © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 12 1 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. VL VR V R lags V S VL lead s VS R L VS I I lags V S Electronics Fundamentals 8th edition Floyd/Buchla © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 12 1 Impedance of series RL circuits 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. XL R tan 1 Z Z is the diagonal Z XL XL R R It is convenient to reposition the phasors into the impedance triangle. Electronics Fundamentals 8th edition Floyd/Buchla © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 12 1 Impedance of series RL circuits Sketch the impedance triangle and show the values for R = 1.2 kW and XL = 960 W. Z 1.2 kW 2 + 0.96 kW 1.33 kW tan 1 0.96 kW 1.2 kW 39 Electronics Fundamentals 8th edition Floyd/Buchla 2 Z = 1.33 kW 39o XL = 960 W R = 1.2 kW © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 12 1 Analysis of series RL circuits Ohm’s law is applied to series RL circuits using quantities of Z, V, and I. V IZ I V Z Z V 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. Electronics Fundamentals 8th edition Floyd/Buchla © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 12 1 Analysis of series RL circuits Assume the current in the previous example is 10 mArms. Sketch the voltage phasors. The impedance triangle from the previous example is shown for reference. The voltage phasors can be found from Ohm’s law. Multiply each impedance phasor by 10 mA. Z = 1.33 kW 39o R = 1.2 kW Electronics Fundamentals 8th edition Floyd/Buchla x 10 mA = XL = 960 W VS = 13.3 V 39o VL = 9.6 V VR = 12 V © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 12 1 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. Increasing f As frequency changes, X Z the impedance triangle for an RL circuit changes Z X as illustrated here because XL increases with Z X increasing f. This determines the frequency R response of RL circuits. 3 2 1 1 Electronics Fundamentals 8th edition Floyd/Buchla 2 L3 L2 L1 3 © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 12 1 Phase shift 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. R Vout Vin Vout f Vin L Vout (phase lead) Vin f Electronics Fundamentals 8th edition Floyd/Buchla VR © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 12 1 Phase shift 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. L Vin VL R Vin Vin Vout f (phase lag) Vout f Vout Electronics Fundamentals 8th edition Floyd/Buchla © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 12 1 Sinusoidal response of parallel RL circuits For parallel circuits, it is useful to review conductance, susceptance and admittance, introduced in Chapter 10. G Conductance is the reciprocal of resistance. Inductive susceptance is the reciprocal of inductive reactance. BL 1 R 1 XL 1 Admittance is the reciprocal of impedance. Y Z Electronics Fundamentals 8th edition Floyd/Buchla © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 12 1 Sinusoidal response of parallel RL circuits In a parallel RL circuit, the admittance phasor is the sum of the conductance and inductive susceptance phasors. The magnitude of the susceptance is Y G 2 + BL 2 The magnitude of the phase angle is tan 1 G VS G BL BL Electronics Fundamentals 8th edition Floyd/Buchla BL G Y © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 12 1 Sinusoidal response of parallel RL circuits Some important points to notice are: G is plotted along the positive x-axis. BL is plotted along the negative y-axis. BL G tan 1 Y is the diagonal G VS G BL BL Electronics Fundamentals 8th edition Floyd/Buchla Y © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 12 1 Sinusoidal response of parallel RL circuits Draw the admittance phasor diagram for the circuit. The magnitude of the conductance and susceptance are: 1 1 1 B 0.629 mS G 1.0 mS L 2 10 kHz 25.3 mH R 1.0 kW Y G 2 + BL 2 1.0 mS + 0.629 mS 1.18 mS 2 2 G = 1.0 mS VS f = 10 kHz Electronics Fundamentals 8th edition Floyd/Buchla R 1.0 kW L 25.3 mH BL = 0.629 mS Y= 1.18 mS © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 12 1 Analysis of parallel RL circuits Ohm’s law is applied to parallel RL circuits using quantities of Y, V, and I. I Y V I V Y I VY 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. Electronics Fundamentals 8th edition Floyd/Buchla © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 12 1 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. G = 1.0 mS BL = 0.629 mS Y= 1.18 mS Electronics Fundamentals 8th edition Floyd/Buchla x 10 V = IL = 6.29 mA IR = 10 mA IS = 11.8 mA © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 12 1 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: BL 1 2 fL As frequency increases, BL and IL decrease, so the angle between IR and IS must decrease as well. IL Electronics Fundamentals 8th edition Floyd/Buchla IR IS © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 12 1 Series-Parallel RL circuits Series-parallel RL circuits are combinations of both series and parallel elements. The solution of these circuits is similar to resistive combinational circuits but you need to combine reactive elements using phasors. The components in the yellow box are in series and those in the green box are also in series. Z1 R12 X L21 and Z 2 R22 X L22 Electronics Fundamentals 8th edition Floyd/Buchla R2 R1 Z1 L1 Z2 L2 The two boxes are in parallel. You can find the branch currents by applying Ohm’s law to the source voltage and the branch impedance. © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 12 1 The power triangle 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: Z = 1.33 kW 39o R = 1.2 kW Electronics Fundamentals 8th edition Floyd/Buchla x 10 mA = XL = 960 W VS = 13.3 V 39o VL = 9.6 V VR = 12 V © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 12 1 The power triangle Multiplying the voltage phasors by Irms gives the power triangle (equivalent to multiplying the impedance phasors by I2). 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 = VS = 13.3 V 39o VR = 12 V Electronics Fundamentals 8th edition Floyd/Buchla VL = 9.6 V Pa = 133 mVA Pr = 96 mVAR 39o Ptrue = 120 mW © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 12 1 Power factor The power factor was discussed in Chapter 15 and applies to RL circuits as well as RC circuits. Recall that it is the relationship between the apparent power in volt-amperes and true power in watts. Volt-amperes multiplied by the power factor equals true power. Power factor is defined as PF = cos Electronics Fundamentals 8th edition Floyd/Buchla © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 12 1 Apparent power Apparent power consists of two components; a true power component, that does the work, and a reactive power component, that is simply power shuttled back and forth between source and load. Power factor corrections for an inductive load (motors, generators, etc.) are done by adding a parallel capacitor, which has a canceling effect. Electronics Fundamentals 8th edition Floyd/Buchla (W) PPatrue (VA) Pr (VAR) Pt (W) © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 12 1 Frequency Response of RL Circuits Series RL circuits have a frequency response similar to series RC circuits. 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 10 V rms 0 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 Vout (V) 9.98 Plotting the response: 8.46 1.57 0.79 9 8 7 6 5 4 3 2 1 0.1 Electronics Fundamentals 8th edition Floyd/Buchla 1 10 20 100 f (kHz) © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 12 1 Frequency Response of RL Circuits Reversing the position of the R and L components, produces the high-pass response. The output is taken across the inductor. Vin Vout 10VV rms rms 10 10 V dc 10 V rms 0 10 V dc 100 100 W W 100 W ƒ = 100 Hz 1 kHz ƒ = 10 kHz 9.87 V rms 5.32 V rms 0.63 V rms 10 10 mH 10mH mH 10 mH 0 V dc Vout (V) 9.87 Plotting the response: 5.32 0.63 Electronics Fundamentals 8th edition Floyd/Buchla 10 9 8 7 6 5 4 3 2 1 0 0.01 0.1 1 10 f (kHz) © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 12 1 Quiz 1. If the frequency is increased in a series RL circuit, the phase angle will a. increase b. decrease c. be unchanged Electronics Fundamentals 8th edition Floyd/Buchla © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 12 1 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 Electronics Fundamentals 8th edition Floyd/Buchla © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 12 1 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 Vin Vout d. none of the above Electronics Fundamentals 8th edition Floyd/Buchla © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 12 1 Quiz 4. In a series RL circuit, the phase angle can be found from the equation XL R a. tan 1 b. VL tan VR 1 c. both of the above are correct d. none of the above is correct Electronics Fundamentals 8th edition Floyd/Buchla © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 12 1 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 Electronics Fundamentals 8th edition Floyd/Buchla © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 12 1 Quiz 6. Susceptance is the reciprocal of a. resistance b. reactance c. admittance d. impedance Electronics Fundamentals 8th edition Floyd/Buchla © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 12 1 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 BL 2 c. Y = G + BL d. Y G 2 + BL 2 Electronics Fundamentals 8th edition Floyd/Buchla © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 12 1 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 Electronics Fundamentals 8th edition Floyd/Buchla © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 12 1 Quiz 9. The unit used for measuring true power is the a. volt-ampere b. watt c. volt-ampere-reactive (VAR) d. kilowatt-hour Electronics Fundamentals 8th edition Floyd/Buchla © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 12 1 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 Electronics Fundamentals 8th edition Floyd/Buchla © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 12 1 Quiz Answers: Electronics Fundamentals 8th edition Floyd/Buchla 1. a 6. b 2. a 7. d 3. c 8. d 4. a 9. b 5. c 10. a © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved.