Alexander_2

Review Questions 201Review Questions5.1 The two input terminals of an op amp are labeled as: 5.6 If vs ϭ 8 mV in the circuit of Fig. 5.41, the output voltage is: (a) high and low. (b) positive and negative. (a) Ϫ44 mV (b) Ϫ8 mV (c) inverting and noninverting. (c) 4 mV (d) 7 mV (d) differential and nondifferential. 5.7 Refer to Fig. 5.41. If vs ϭ 8 mV, voltage va is:5.2 For an ideal op amp, which of the following statements (a) Ϫ8 mV (b) 0 mV are not true? (c) 10͞3 mV (d) 8 mV (a) The differential voltage across the input terminals is zero. 5.8 The power absorbed by the 4-k⍀ resistor in Fig. 5.42 is: (b) The current into the input terminals is zero. (c) The current from the output terminal is zero. (a) 9 mW (b) 4 mW (d) The input resistance is zero. (c) 2 mW (d) 1 mW (e) The output resistance is zero.5.3 For the circuit in Fig. 5.40, voltage vo is:(a) Ϫ6 V (b) Ϫ5 V + 4 kΩ(c) Ϫ1.2 V (d) Ϫ0.2 V − 2 kΩ 10 kΩ ix 6 V +− + vo − − + 2 kΩ + Figure 5.421 V +− 3 kΩ For Review Questions 5.8. vo −Figure 5.40For Review Questions 5.3 and 5.4.5.4 For the circuit in Fig. 5.40, current ix is: 5.9 Which of these amplifiers is used in a digital-to-analog converter?(a) 0.6 mA (b) 0.5 mA (a) noninverter(c) 0.2 mA (d) 1͞12 mA (b) voltage follower (c) summer5.5 If vs ϭ 0 in the circuit of Fig. 5.41, current io is: (d) difference amplifier(a) Ϫ10 mA (b) Ϫ2.5 mA 5.10 Difference amplifiers are used in (please check all that apply):(c) 10͞12 mA (d) 10͞14 mA (a) instrumentation amplifiers 4 kΩ 8 kΩ (b) voltage followers (c) voltage regulators a − (d) buffers + io (e) summing amplifiers10 mV +− vs +− (f ) subtracting amplifiers + 2 kΩ vo −Figure 5.41 Answers: 5.1c, 5.2c,d, 5.3b, 5.4b, 5.5a, 5.6c, 5.7d, 5.8b,For Review Questions 5.5, 5.6, and 5.7. 5.9c, 5.10a,f.

202 Chapter 5 Operational AmplifiersProblemsSection 5.2 Operational Amplifiers + vo 741 5.1 The equivalent model of a certain op amp is shown − in Fig. 5.43. Determine: (a) the input resistance +− (b) the output resistance 1 mV (c) the voltage gain in dB Figure 5.45 For Prob. 5.6. 60 Ω 5.7 The op amp in Fig. 5.46 has Ri ϭ 100 k⍀, Ro ϭ 100 ⍀, A ϭ 100,000. Find the differential − 1.5 MΩ + 8 × 104vd voltage vd and the output voltage vo. vd − + −+ vdFigure 5.43 +−For Prob. 5.1.5.2 The open-loop gain of an op amp is 100,000. Calculate 10 kΩ 100 kΩ the output voltage when there are inputs of ϩ10 ␮V on the inverting terminal and ϩ20 ␮V on the 1 mV +− + noninverting terminal. vo5.3 Determine the output voltage when Ϫ20 ␮V is − applied to the inverting terminal of an op amp and ϩ30 ␮V to its noninverting terminal. Assume that Figure 5.46 the op amp has an open-loop gain of 200,000. For Prob. 5.7.5.4 The output voltage of an op amp is Ϫ4 V when the Section 5.3 Ideal Op Amp noninverting input is 1 mV. If the open-loop gain 5.8 Obtain vo for each of the op amp circuits in Fig. 5.47. of the op amp is 2 ϫ 106, what is the inverting input? 10 kΩ5.5 For the op amp circuit of Fig. 5.44, the op amp has an open-loop gain of 100,000, an input resistance of 10 k⍀, and an output resistance of 100 ⍀. Find the voltage gain vo͞vi using the nonideal model of the op amp. 2 kΩ 2V − + − + − − + + ++ + vo 1 mA vo 1 V +− 2 kΩ vo − − − vi +−Figure 5.44 (a) (b)For Prob. 5.5. Figure 5.47 For Prob. 5.8.5.6 Using the same parameters for the 741 op amp in 5.9 Determine vo for each of the op amp circuits in Example 5.1, find vo in the op amp circuit of Fig. 5.48. Fig. 5.45.

Problems 2031 mA 2 kΩ + 5 kΩ 25 kΩ + vo vs +− − − − vo + + − +− 4 V 10 kΩ + Figure 5.51 − For Prob. 5.12. +− 1 V 5.13 Find vo and io in the circuit of Fig. 5.52.3 V +− 2 kΩ + 10 kΩ 90 kΩ vo + io − − 1 V +− 100 kΩ + 50 kΩFigure 5.48 10 kΩ voFor Prob. 5.9. −5.10 Find the gain vo͞vs of the circuit in Fig. 5.49. 20 kΩ + + Figure 5.52vs +− − vo For Prob. 5.13. − 10 kΩ 5.14 Determine the output voltage vo in the circuit of Fig. 5.53. 10 kΩ 10 kΩ 20 kΩ 10 kΩ + 5 kΩ − vo + −Figure 5.49 2 mAFor Prob. 5.10.5.11 Using Fig. 5.50, design a problem to help other Figure 5.53 students better understand how ideal op amps work. For Prob. 5.14. Section 5.4 Inverting Amplifier R2 5.15 (a) Determine the ratio vo͞is in the op amp circuit of R1 − io Fig. 5.54. + (b) Evaluate the ratio for R1 ϭ 20 k⍀, R2 ϭ 25 k⍀, R3 ϭ 40 k⍀. R3 + R1 R3V +− R4 R5 vo R2 −Figure 5.50 − +For Prob. 5.11. + vo − 5.12 Calculate the voltage ratio vo͞vs for the op amp is circuit of Fig. 5.51. Assume that the op amp is ideal. Figure 5.54 For Prob. 5.15.

204 Chapter 5 Operational Amplifiers5.16 Using Fig. 5.55, design a problem to help students 5.19 Determine io in the circuit of Fig. 5.58. better understand inverting op amps. 2 kΩ 4 kΩ 10 kΩ R1 ix R3 750 mV +− 4 kΩ io − iy + – 2 kΩ +V −+ R4 Figure 5.58 R2 For Prob. 5.19. 5.20 In the circuit of Fig. 5.59, calculate vo of vs ϭ 2 V. 8 kΩFigure 5.55 4 kΩ 4 kΩ 2 kΩFor Prob. 5.16. 9 V +− − + 5.17 Calculate the gain vo͞vi when the switch in Fig. 5.56 + vo is in: − (a) position 1 (b) position 2 (c) position 3. vs +− 12 kΩ 1 Figure 5.59 80 kΩ 2 For Prob. 5.20. 2 MΩ 3 5.21 Calculate vo in the op amp circuit of Fig. 5.60. 10 kΩ 5 kΩ − 4 kΩvi +− ++ − 10 kΩ vo − + + vo 3 V +− 1 V +− −Figure 5.56 Figure 5.60For Prob. 5.17. For Prob. 5.21.*5.18 For the circuit shown in Figure 5.57, solve for the 5.22 Design an inverting amplifier with a gain of Ϫ15. Thevenin equivalent circuit looking into terminals A and B. 5.23 For the op amp circuit in Fig. 5.61, find the voltage gain vo͞vs. 10 kΩ Rf 10 kΩ −7.5 V +− + 2.5 Ω R1 – R2 + + vs +− vo −Figure 5.57For Prob. 5.18. Figure 5.61 For Prob. 5.23.* An asterisk indicates a challenging problem.

Problems 2055.24 In the circuit shown in Fig. 5.62, find k in the voltage 5.28 Find io in the op amp circuit of Fig. 5.66. transfer function vo ϭ kvs. 50 kΩR1 vs R2 Rf + 10 kΩ − io −+ vo + 20 kΩ − −R3 R4 + +− 0.4 V Figure 5.66 For Prob. 5.28.Figure 5.62 5.29 Determine the voltage gain vo͞vi of the op ampFor Prob. 5.24. circuit in Fig. 5.67.Section 5.5 Noninverting Amplifier R1 5.25 Calculate vo in the op amp circuit of Fig. 5.63. + − 12 kΩ vi + R2 − + + − vo R2 – + R1 3.7 V +− 20 kΩ vo −Figure 5.63 Figure 5.67For Prob. 5.25. For Prob. 5.29. 5.26 Using Fig. 5.64, design a problem to help other 5.30 In the circuit shown in Fig. 5.68, find ix and the students better understand noninverting op amps. power absorbed by the 20-k⍀ resistor. + − 60 kΩ ix − + 30 kΩ 20 kΩV +− R2 io 1.2 V +− R1 R3Figure 5.64 Figure 5.68For Prob. 5.26. For Prob. 5.30. 5.27 Find vo in the op amp circuit of Fig. 5.65. 5.31 For the circuit in Fig. 5.69, find ix. 12 kΩ 16 Ω v1 − v2 8 Ω 6 kΩ + + 3 kΩ ix − +7.5 V +− 12 Ω + 4 mA 6 kΩ vo 24 Ω vo − −Figure 5.65 Figure 5.69For Prob. 5.27. For Prob. 5.31.

206 Chapter 5 Operational Amplifiers5.32 Calculate ix and vo in the circuit of Fig. 5.70. Find + the power dissipated by the 60-k⍀ resistor. −a vs −+ R2 R1 + ix 20 kΩ − b4 mV +− 50 kΩ 60 kΩ 30 kΩ + Figure 5.73 10 kΩ For Prob. 5.36. vo − Section 5.6 Summing Amplifier 5.37 Determine the output of the summing amplifier in Fig. 5.74.Figure 5.70 2 V 10 kΩ 30 kΩFor Prob. 5.32. −+ 5.33 Refer to the op amp circuit in Fig. 5.71. Calculate ix −2 V 20 kΩ and the power absorbed by the 3-k⍀ resistor. −+ − 1 kΩ 4.5 V 30 kΩ ++ +− vo −1 mA 4 kΩ + ix Figure 5.74 − 3 kΩ For Prob. 5.37. 2 kΩ 5.38 Using Fig. 5.75, design a problem to help other students better understand summing amplifiers.Figure 5.71 V1 R1 +For Prob. 5.33. −+ V2 R2 − 5.34 Given the op amp circuit shown in Fig. 5.72, express +− + vo in terms of v1 and v2. V3 R3 vo −+ − R5 R1 vin + V4 R4 v1 − +− v2 R4 + Figure 5.75 R2 R3 vo For Prob. 5.38. –Figure 5.72 5.39 For the op amp circuit in Fig. 5.76, determine theFor Prob. 5.34. value of v2 in order to make vo ϭ Ϫ16.5 V.5.35 Design a noninverting amplifier with a gain of 7.5. +2 V 10 kΩ 50 kΩ vo v2 20 kΩ5.36 For the circuit shown in Fig. 5.73, find the Thevenin 50 kΩ − equivalent at terminals a-b. (Hint: To find RTh, apply –1 V + a current source io and calculate vo.) Figure 5.76 For Prob. 5.39.

Problems 2075.40 Referring to the circuit shown in Fig. 5.77, 5.46 Using only two op amps, design a circuit to solve determine Vo in terms of V1 and V2. v1 Ϫ v2 v3 Ϫvout ϭ 3 ϩ 2 100 kΩ 200 kΩ 100 kΩ − 10 Ω Section 5.7 Difference AmplifierV1 +− V2 +− + 5.47 The circuit in Fig. 5.79 is for a difference amplifier. + Find vo given that v1 ϭ 1 V and v2 ϭ 2 V. 40 Ω Vo − 30 kΩFigure 5.77 2 kΩFor Prob. 5.40.5.41 An averaging amplifier is a summer that provides 2 kΩ − + an output equal to the average of the inputs. By + vo using proper input and feedback resistor values, v1 +− v2 +− − one can get 20 kΩ Ϫvout ϭ 1 (v1 ϩ v2 ϩ v3 ϩ v4) 4 Using a feedback resistor of 10 k⍀, design an Figure 5.79 averaging amplifier with four inputs. For Prob. 5.47.5.42 A three-input summing amplifier has input resistors 5.48 The circuit in Fig. 5.80 is a differential amplifier with R1 ϭ R2 ϭ R3 ϭ 75 k⍀. To produce an driven by a bridge. Find vo. averaging amplifier, what value of feedback resistor is needed?5.43 A four-input summing amplifier has R1 ϭ R2 ϭ 20 kΩ 80 kΩ R3 ϭ R4 ϭ 80 k⍀. What value of feedback resistor 30 kΩ is needed to make it an averaging amplifier? − 60 kΩ +5.44 Show that the output voltage vo of the circuit in 10 kΩ 20 kΩ Fig. 5.78 is + 10 mV 80 kΩ vo ϭ (R3 ϩ R4) (R2v1 ϩ R1v2) 40 kΩ vo R3(R1 ϩ R2) Figure 5.80 R3 R4 vo For Prob. 5.48. R1 − v1 R2 + v2 5.49 Design a difference amplifier to have a gain of 4 and a common-mode input resistance of 20 k⍀ at eachFigure 5.78 input.For Prob. 5.44. 5.50 Design a circuit to amplify the difference between5.45 Design an op amp circuit to perform the following two inputs by 2.5. operation: (a) Use only one op amp. vo ϭ 3v1 Ϫ 2v2 All resistances must be Յ 100 k⍀. (b) Use two op amps.

208 Chapter 5 Operational Amplifiers5.51 Using two op amps, design a subtractor. R2 R2 22*5.52 Design an op amp circuit such that R1 − RG + vo ϭ 4v1 ϩ 6v2 Ϫ 3v3 Ϫ 5v4 + vo vi + R2 − Let all the resistors be in the range of 20 to 200 k⍀. − 2 R2 R1 2*5.53 The ordinary difference amplifier for fixed-gain (c) operation is shown in Fig. 5.81(a). It is simple and reliable unless gain is made variable. One way of Figure 5.81 providing gain adjustment without losing simplicity For Prob. 5.53. and accuracy is to use the circuit in Fig. 5.81(b). Another way is to use the circuit in Fig. 5.81(c). Show that:(a) for the circuit in Fig. 5.81(a), vo ϭ R2 Section 5.8 Cascaded Op Amp Circuits vi R1(b) for the circuit in Fig. 5.81(b), 5.54 Determine the voltage transfer ratio vo͞vs in the op amp circuit of Fig. 5.82, where R ϭ 10 k⍀. vo ϭ R2 1 R1 vi R1 1 ϩ R 2RG +(c) for the circuit in Fig. 5.81(c), R − vo ϭ R2 a1 ϩ R2 b R − R vi R1 2RG + + R vs + vo R2 − − R1 − + Figure 5.82 + vo For Prob. 5.54.− −vi+ R1 5.55 In a certain electronic device, a three-stage amplifier R2 is desired, whose overall voltage gain is 42 dB. The individual voltage gains of the first two stages are to (a) be equal, while the gain of the third is to be one- fourth of each of the first two. Calculate the voltage R1 R1 R2 gain of each. 2 2 − 5.56 Using Fig. 5.83, design a problem to help other− RG + students better understand cascaded op amps.vi R2+ R1 + R2 R4 2 vo R1 − R1 − R3 2 + + − vi + − (b) Figure 5.83 For Prob. 5.56.

Problems 2095.57 Find vo in the op amp circuit of Fig. 5.84. 5.61 Determine vo in the circuit of Fig. 5.88. 25 kΩ 50 kΩ 100 kΩ 100 kΩ + vo 20 kΩ − 0.2 V 10 kΩ 40 kΩvs1 − − 20 kΩ − − − 0.4 V 10 kΩ + + vo + + 100 kΩ 50 kΩ 50 kΩ vs2 Figure 5.88 For Prob. 5.61.Figure 5.84For Prob. 5.57. 5.62 Obtain the closed-loop voltage gain vo͞vi of the circuit in Fig. 5.89. 5.58 Calculate io in the op amp circuit of Fig. 5.85. 1 kΩ − 2 kΩ 10 kΩ R1 R2 Rf0.6 V +− + vi +− − − R3 − 5 kΩ + + ++ 3 kΩ io R4 vo 4 kΩ −Figure 5.85 Figure 5.89For Prob. 5.58. For Prob. 5.62. 5.59 In the op amp circuit of Fig. 5.86, determine the 5.63 Determine the gain vo͞vi of the circuit in Fig. 5.90. voltage gain vo͞vs. Take R ϭ 10 k⍀. 2R 4RR R R3 R4 R2 − − − + ++ ++ vs +− vo R1 R5 vo − − + − vi +− R6Figure 5.86 Figure 5.90For Prob. 5.59. For Prob. 5.63. 5.60 Calculate vo͞vi in the op amp circuit of Fig. 5.87. 5.64 For the op amp circuit shown in Fig. 5.91, find vo͞vs. 4 kΩ 5 kΩ 10 kΩ + G4 G3 − G+ − –vi + +− + G1 G vo – 2 kΩ − 10 kΩ + + vo vs −+ G2 –Figure 5.87 Figure 5.91For Prob. 5.60. For Prob. 5.64.

210 Chapter 5 Operational Amplifiers5.65 Find vo in the op amp circuit of Fig. 5.92. 5.68 Find vo in the circuit of Fig. 5.95, assuming that Rf ϭ ϱ (open circuit). 30 kΩ 50 kΩ – 10 kΩ 20 kΩ Rf + + 15 kΩ – −6 mV −+ + + 5 kΩ vo 8 kΩ – 15 mV +− − + + − 40 kΩ + 6 kΩ vo 1 kΩ − 2 kΩFigure 5.92 Figure 5.95For Prob. 5.65. For Probs. 5.68 and 5.69. 5.66 For the circuit in Fig. 5.93, find vo. 25 kΩ 40 kΩ 100 kΩ 5.69 Repeat the previous problem if Rf ϭ 10 k⍀. 5.70 Determine vo in the op amp circuit of Fig. 5.96. 20 kΩ − 20 kΩ − + ++6 V +− 4 V +− 10 kΩ vo 2 V +− −Figure 5.93 30 kΩ 40 kΩFor Prob. 5.66. −A + 10 kΩ 20 kΩ 20 kΩ 60 kΩ5.67 Obtain the output vo in the circuit of Fig. 5.94. 1 V +− −B C vo + − 80 kΩ 80 kΩ 10 kΩ + − 40 kΩ − 2 V +− 10 kΩ + + vo 10 kΩ − 20 kΩ +0.3 V +− 20 kΩ 3 V +− − 10 kΩ + 4 V +−0.7 V +−Figure 5.94 Figure 5.96For Prob. 5.67. For Prob. 5.70.

Problems 211 5.74 Find io in the op amp circuit of Fig. 5.100.5.71 Determine vo in the op amp circuit of Fig. 5.97. 100 kΩ 32 kΩ 5 kΩ 20 kΩ 40 kΩ 100 kΩ 10 kΩ − io 1.6 kΩ1.5 V −+ 80 kΩ + 20 kΩ − − – 0.9 V +− + + + + 10 kΩ +− 0.6 V 20 kΩ vo − 10 kΩ – Figure 5.100 + + For Prob. 5.74. −2.25 V −+ Section 5.9 Op Amp Circuit Analysis with 30 kΩ 50 kΩ PSpiceFigure 5.97 5.75 Rework Example 5.11 using the nonideal op ampFor Prob. 5.71. LM324 instead of uA741. 5.76 Solve Prob. 5.19 using PSpice or MultiSim and op amp uA741.5.72 Find the load voltage vL in the circuit of Fig. 5.98. 5.77 Solve Prob. 5.48 using PSpice or MultiSim and op amp LM324. 5.78 Use PSpice or MultiSim to obtain vo in the circuit of Fig. 5.101. 100 kΩ 250 kΩ 10 kΩ 20 kΩ 30 kΩ 40 kΩ 20 kΩ − − + − − +1.8 V +− + + vL + + vo − − 2 kΩ 1 V +− 2 V +−Figure 5.98 Figure 5.101For Prob. 5.72. For Prob. 5.78. 5.79 Determine vo in the op amp circuit of Fig. 5.102, using PSpice or MultiSim.5.73 Determine the load voltage vL in the circuit of Fig. 5.99. 20 kΩ 10 kΩ 5 V +− 50 kΩ + 20 kΩ −10 kΩ − − 100 kΩ + + + 10 kΩ 5 kΩ vo+− 1.8 V 4 kΩ − 40 kΩ + 1 V +− − vL + −Figure 5.99 Figure 5.102For Prob. 5.73. For Prob. 5.79.

212 Chapter 5 Operational Amplifiers5.80 Use PSpice or MultiSim to solve Prob. 5.70. 5.86 Design a voltage controlled ideal current source (within the operating limits of the op amp) where the5.81 Use PSpice or MultiSim to verify the results in output current is equal to 200 vs(t) mA. Example 5.9. Assume nonideal op amps LM324.Section 5.10 Applications5.82 A five-bit DAC covers a voltage range of 0 to 7.75 V. Calculate how much voltage each bit is worth. 5.83 Design a six-bit digital-to-analog converter. 5.87 Figure 5.105 displays a two-op-amp instrumentation (a) If |Vo| ϭ 1.1875 V is desired, what should amplifier. Derive an expression for vo in terms of v1 [V1V2V3V4V5V6] be? and v2. How can this amplifier be used as a (b) Calculate |Vo| if [V1V2V3V4V5V6] ϭ [011011]. subtractor? (c) What is the maximum value |Vo| can assume?*5.84 A four-bit R-2R ladder DAC is presented in Fig. 5.103.(a) Show that the output voltage is given by v1 + R4 −ϪVo ϭ Rf a V1 ϩ V2 ϩ V3 ϩ V4 b R2 R3 2R 4R 8R 16R − R1(b) If Rf ϭ 12 k⍀ and R ϭ 10 k⍀, find |Vo| for v2 + vo [V1V2V3V4] ϭ [1011] and [V1V2V3V4] ϭ [0101]. Rf Figure 5.105 For Prob. 5.87. 2RV1 − Vo + R 2RV2 R *5.88 Figure 5.106 shows an instrumentation amplifier 2R driven by a bridge. Obtain the gain vo͞vi of theV3 amplifier. R 2RV4 2RFigure 5.103 20 kΩ 30 kΩ + 25 kΩ 500 kΩFor Prob. 5.84. vi 80 kΩ − 5.85 In the op amp circuit of Fig. 5.104, find the value of 40 kΩ 10 kΩ − vo R so that the power absorbed by the 10-k⍀ resistor is + 10 mW. Take vs ϭ 2 V. 2 kΩ + 10 kΩ − − 25 kΩ−+ vs 10 kΩ + R 500 kΩ 40 kΩFigure 5.104 Figure 5.106For Prob. 5.85. For Prob. 5.88.

Comprehensive Problems 213Comprehensive Problems5.89 Design a circuit that provides a relationship between 5.92 Refer to the bridge amplifier shown in Fig. 5.109. output voltage vo and input voltage vs such that Determine the voltage gain vo͞vi. vo ϭ 12vs Ϫ 10. Two op amps, a 6-V battery, and several resistors are available. 30 kΩ 60 kΩ + 20 kΩ5.90 The op amp circuit in Fig. 5.107 is a current − RL vo amplifier. Find the current gain io͞is of the amplifier. + − 20 kΩ 50 kΩ − vi +− − + + 4 kΩ io Figure 5.109 For Prob. 5.92.is 5 kΩ 2 kΩ *5.93 A voltage-to-current converter is shown in Fig. 5.110, which means that iL ϭ Avi if R1R2 ϭ R3R4. Find the constant term A.Figure 5.107 R1 R3For Prob. 5.90. + − 5.91 A noninverting current amplifier is portrayed in + Fig. 5.108. Calculate the gain io͞is. Take R1 ϭ 8 k⍀ and R2 ϭ 1 k⍀. vi R4 − R2 iL + − RL R1 R2 Figure 5.110 io For Prob. 5.93. is R2Figure 5.108For Prob. 5.91.



Capacitors and chapterInductors 6But in science the credit goes to the man who convinces the world, notto the man to whom the idea first occurs. —Francis DarwinEnhancing Your Skills and Your Career Photo by Charles AlexanderABET EC 2000 criteria (3.c), “an ability to design a system,component, or process to meet desired needs.”The “ability to design a system, component, or process to meetdesired needs” is why engineers are hired. That is why this is themost important technical skill that an engineer has. Interestingly, yoursuccess as an engineer is directly proportional to your ability to com-municate but your being able to design is why you will be hired inthe first place. Design takes place when you have what is termed an open-endedproblem that eventually is defined by the solution. Within the contextof this course or textbook, we can only explore some of the elementsof design. Pursuing all of the steps of our problem-solving techniqueteaches you several of the most important elements of the designprocess. Probably the most important part of design is clearly defining whatthe system, component, process, or, in our case, problem is. Rarely isan engineer given a perfectly clear assignment. Therefore, as a student,you can develop and enhance this skill by asking yourself, your col-leagues, or your professors questions designed to clarify the problemstatement. Exploring alternative solutions is another important part of thedesign process. Again, as a student, you can practice this part of thedesign process on almost every problem you work. Evaluating your solutions is critical to any engineering assignment.Again, this is a skill that you as a student can practice on every prob-lem you work. 215

216 Chapter 6 Capacitors and InductorsIn contrast to a resistor, which spends 6.1 Introductionor dissipates energy irreversibly, aninductor or capacitor stores or releases So far we have limited our study to resistive circuits. In this chapter,energy (i.e., has a memory). we shall introduce two new and important passive linear circuit ele- ments: the capacitor and the inductor. Unlike resistors, which dissipate Dielectric with permittivity ⑀ energy, capacitors and inductors do not dissipate but store energy, Metal plates, which can be retrieved at a later time. For this reason, capacitors and each with area A inductors are called storage elements. The application of resistive circuits is quite limited. With the intro- duction of capacitors and inductors in this chapter, we will be able to analyze more important and practical circuits. Be assured that the cir- cuit analysis techniques covered in Chapters 3 and 4 are equally appli- cable to circuits with capacitors and inductors. We begin by introducing capacitors and describing how to com- bine them in series or in parallel. Later, we do the same for inductors. As typical applications, we explore how capacitors are combined with op amps to form integrators, differentiators, and analog computers. d 6.2 CapacitorsFigure 6.1 A capacitor is a passive element designed to store energy in its elec-A typical capacitor. tric field. Besides resistors, capacitors are the most common electrical components. Capacitors are used extensively in electronics, communi-+− cations, computers, and power systems. For example, they are used in + the tuning circuits of radio receivers and as dynamic memory elements in computer systems.+q + − −q A capacitor is typically constructed as depicted in Fig. 6.1. + A capacitor consists of two conducting plates separated by an insu-+− lator (or dielectric). + In many practical applications, the plates may be aluminum foil while +− the dielectric may be air, ceramic, paper, or mica. v When a voltage source v is connected to the capacitor, as inFigure 6.2 Fig. 6.2, the source deposits a positive charge q on one plate and a neg-A capacitor with applied voltage v. ative charge Ϫq on the other. The capacitor is said to store the electric charge. The amount of charge stored, represented by q, is directly pro- portional to the applied voltage v so that q ϭ Cv (6.1)Alternatively, capacitance is the amount where C, the constant of proportionality, is known as the capacitanceof charge stored per plate for a unit of the capacitor. The unit of capacitance is the farad (F), in honor ofvoltage difference in a capacitor. the English physicist Michael Faraday (1791–1867). From Eq. (6.1), we may derive the following definition. Capacitance is the ratio of the charge on one plate of a capacitor to the voltage difference between the two plates, measured in farads (F). Note from Eq. (6.1) that 1 farad ϭ 1 coulomb/volt.

6.2 Capacitors 217 HistoricalMichael Faraday (1791–1867), an English chemist and physicist,was probably the greatest experimentalist who ever lived. Born near London, Faraday realized his boyhood dream by work-ing with the great chemist Sir Humphry Davy at the Royal Institu-tion, where he worked for 54 years. He made several contributionsin all areas of physical science and coined such words as electroly-sis, anode, and cathode. His discovery of electromagnetic inductionin 1831 was a major breakthrough in engineering because it provideda way of generating electricity. The electric motor and generator oper-ate on this principle. The unit of capacitance, the farad, was namedin his honor. The Burndy Library Collection at The Huntington Library, San Marino, California. Although the capacitance C of a capacitor is the ratio of the charge Capacitor voltage rating and capaci-q per plate to the applied voltage v, it does not depend on q or v. It tance are typically inversely rated duedepends on the physical dimensions of the capacitor. For example, for to the relationships in Eqs. (6.1) andthe parallel-plate capacitor shown in Fig. 6.1, the capacitance is given by (6.2). Arcing occurs if d is small and V is high.C ϭ ⑀A (6.2) dwhere A is the surface area of each plate, d is the distance betweenthe plates, and ⑀ is the permittivity of the dielectric material betweenthe plates. Although Eq. (6.2) applies to only parallel-plate capacitors,we may infer from it that, in general, three factors determine the valueof the capacitance:1. The surface area of the plates—the larger the area, the greater the capacitance.2. The spacing between the plates—the smaller the spacing, the greater the capacitance.3. The permittivity of the material—the higher the permittivity, the greater the capacitance. Capacitors are commercially available in different values and types. iC iCTypically, capacitors have values in the picofarad (pF) to microfarad (mF)range. They are described by the dielectric material they are made of and +v − +v −by whether they are of fixed or variable type. Figure 6.3 shows the cir-cuit symbols for fixed and variable capacitors. Note that according to the (a) (b)passive sign convention, if v 7 0 and i 7 0 or if v 6 0 and i 6 0, thecapacitor is being charged, and if v ؒ i 6 0, the capacitor is discharging. Figure 6.3 Circuit symbols for capacitors: (a) fixed Figure 6.4 shows common types of fixed-value capacitors. Poly- capacitor, (b) variable capacitor.ester capacitors are light in weight, stable, and their change with tem-perature is predictable. Instead of polyester, other dielectric materialssuch as mica and polystyrene may be used. Film capacitors are rolledand housed in metal or plastic films. Electrolytic capacitors producevery high capacitance. Figure 6.5 shows the most common types ofvariable capacitors. The capacitance of a trimmer (or padder) capacitor

218 Chapter 6 Capacitors and Inductors(a) (b) (c)Figure 6.4Fixed capacitors: (a) polyester capacitor, (b) ceramic capacitor, (c) electrolytic capacitor.Courtesy of Tech America. is often placed in parallel with another capacitor so that the equivalent capacitance can be varied slightly. The capacitance of the variable air capacitor (meshed plates) is varied by turning the shaft. Variable capac- itors are used in radio receivers allowing one to tune to various sta- tions. In addition, capacitors are used to block dc, pass ac, shift phase, store energy, start motors, and suppress noise.(a) To obtain the current-voltage relationship of the capacitor, we take the derivative of both sides of Eq. (6.1). Since dq (6.3) iϭ dt differentiating both sides of Eq. (6.1) gives (b) i ϭ C dv (6.4) dtFigure 6.5Variable capacitors: (a) trimmer capacitor, This is the current-voltage relationship for a capacitor, assuming the(b) filmtrim capacitor. passive sign convention. The relationship is illustrated in Fig. 6.6 for a capacitor whose capacitance is independent of voltage. CapacitorsCourtesy of Johanson. that satisfy Eq. (6.4) are said to be linear. For a nonlinear capacitor, the plot of the current-voltage relationship is not a straight line. According to Eq. (6.4), for a capacitor Although some capacitors are nonlinear, most are linear. We will to carry current, its voltage must vary assume linear capacitors in this book. with time. Hence, for constant voltage, i ϭ 0. The voltage-current relation of the capacitor can be obtained by integrating both sides of Eq. (6.4). We get i v(t) ϭ CΎ1 t i (t)dt (6.5) Ϫϱ or Slope = C v(t) ϭ Ύ1 t i (t)dt ϩ v(t0) (6.6) C 0 dv/dt t0Figure 6.6 where v(t0) ϭ q(t0)͞C is the voltage across the capacitor at time t0.Current-voltage relationship of a capacitor. Equation (6.6) shows that capacitor voltage depends on the past history

6.2 Capacitors 219of the capacitor current. Hence, the capacitor has memory—a propertythat is often exploited. The instantaneous power delivered to the capacitor is dv (6.7) p ϭ vi ϭ Cv dtThe energy stored in the capacitor is therefore Ύ Ύ Ύt t dv v(t) v dv ϭ 1 Cv2 ` v(t) (6.8) 2 v(Ϫϱ)w ϭ p(t)dt ϭ C v dt ϭ CϪϱ dt v(Ϫϱ) ϪϱWe note that v(Ϫϱ) ϭ 0, because the capacitor was uncharged att ϭ Ϫϱ. Thus, w ϭ 1 Cv2 (6.9) 2Using Eq. (6.1), we may rewrite Eq. (6.9) as (6.10) q2 wϭ 2CEquation (6.9) or (6.10) represents the energy stored in the electric field vvthat exists between the plates of the capacitor. This energy can beretrieved, since an ideal capacitor cannot dissipate energy. In fact, the ttword capacitor is derived from this element’s capacity to store energy (a) (b)in an electric field. Figure 6.7 We should note the following important properties of a capacitor: Voltage across a capacitor: (a) allowed, (b) not allowable; an abrupt change is not1. Note from Eq. (6.4) that when the voltage across a capacitor is not possible. changing with time (i.e., dc voltage), the current through the capac- itor is zero. Thus, An alternative way of looking at this is using Eq. (6.9), which indicates thatA capacitor is an open circuit to dc. energy is proportional to voltage squared. Since injecting or extracting However, if a battery (dc voltage) is connected across a capacitor, energy can only be done over some the capacitor charges. finite time, voltage cannot change2. The voltage on the capacitor must be continuous. instantaneously across a capacitor.The voltage on a capacitor cannot change abruptly. Leakage resistance The capacitor resists an abrupt change in the voltage across it. Capacitance According to Eq. (6.4), a discontinuous change in voltage requires an infinite current, which is physically impossible. For example, Figure 6.8 the voltage across a capacitor may take the form shown in Circuit model of a nonideal capacitor. Fig. 6.7(a), whereas it is not physically possible for the capacitor voltage to take the form shown in Fig. 6.7(b) because of the abrupt changes. Conversely, the current through a capacitor can change instantaneously.3. The ideal capacitor does not dissipate energy. It takes power from the circuit when storing energy in its field and returns previously stored energy when delivering power to the circuit.4. A real, nonideal capacitor has a parallel-model leakage resistance, as shown in Fig. 6.8. The leakage resistance may be as high as

220 Chapter 6 Capacitors and Inductors 100 M⍀ and can be neglected for most practical applications. For this reason, we will assume ideal capacitors in this book.Example 6.1 (a) Calculate the charge stored on a 3-pF capacitor with 20 V across it. (b) Find the energy stored in the capacitor. Solution: (a) Since q ϭ Cv, q ϭ 3 ϫ 10Ϫ12 ϫ 20 ϭ 60 pC (b) The energy stored is w ϭ 1 Cv2 ϭ 1 ϫ 3 ϫ 10Ϫ12 ϫ 400 ϭ 600 pJ 22Practice Problem 6.1 What is the voltage across a 4.5-mF capacitor if the charge on one plate is 0.12 mC? How much energy is stored? Answer: 26.67 A, 1.6 mJ.Example 6.2 The voltage across a 5-mF capacitor is v(t) ϭ 10 cos 6000t V Calculate the current through it. Solution: By definition, the current is i(t) ϭ C dv ϭ 5 ϫ 10Ϫ6 d (10 cos 6000t) dt dt ϭ Ϫ5 ϫ 10Ϫ6 ϫ 6000 ϫ 10 sin 6000t ϭ Ϫ0.3 sin 6000t APractice Problem 6.2 If a 10-mF capacitor is connected to a voltage source with v(t) ϭ 75 sin 2000t V determine the current through the capacitor. Answer: 1.5 cos 2000t A.Example 6.3 Determine the voltage across a 2-mF capacitor if the current through it is i(t) ϭ 6eϪ3000t mA Assume that the initial capacitor voltage is zero.

6.2 Capacitors 221Solution: Practice Problem 6.3Since v ϭ CΎ1 t i dt ϩ v(0) and v(0) ϭ 0, 0 1 t ϫ 10Ϫ6 Ύv ϭ 2 6eϪ3000t dt ؒ 10Ϫ3 0 ϭ 3 ϫ 103 eϪ3000t ` t ϭ (1 Ϫ eϪ3000t) V Ϫ3000 0The current through a 100-mF capacitor is i(t) ϭ 50 sin 120pt mA.Calculate the voltage across it at t ϭ 1 ms and t ϭ 5 ms. Take v(0) ϭ 0.Answer: 93.14 mV, 1.736 V.Determine the current through a 200-mF capacitor whose voltage is Example 6.4shown in Fig. 6.9. v (t)Solution: 50The voltage waveform can be described mathematically as 50t V 06t61 0 16t63 1 2 3 4t v(t) ϭ dϪ120000 Ϫ 50t V 36t64 ϩ 50t V otherwise −50 0 Figure 6.9 For Example 6.4.Since i ϭ C dv͞dt and C ϭ 200 mF, we take the derivative of v to obtain i (mA) 50 06t61 10 i(t) ϭ 200 ϫ 10Ϫ6 ϫ dϪ50 16t63 36t64 50 otherwise 0 10 mA 06t61 0 4t ϭ dϪ10 mA 16t63 1 23 36t64 10 mA otherwise −10 0 Figure 6.10Thus the current waveform is as shown in Fig. 6.10. For Example 6.4.An initially uncharged 1-mF capacitor has the current shown in Practice Problem 6.4Fig. 6.11 across it. Calculate the voltage across it at t ϭ 2 ms andt ϭ 5 ms. i (mA)Answer: 100 mV, 400 mV. 100 0 6 t (ms) 24 Figure 6.11 For Practice Prob. 6.4.

222 Chapter 6 Capacitors and InductorsExample 6.5 Obtain the energy stored in each capacitor in Fig. 6.12(a) under dc conditions. 2 mF + v1 − 2 kΩ i 2 kΩ 5 kΩ 5 kΩ 6 mA 3 kΩ + 4 kΩ 3 kΩ v26 mA 4 kΩ − 4 mF (a) (b)Figure 6.12For Example 6.5. Solution: Under dc conditions, we replace each capacitor with an open circuit, as shown in Fig. 6.12(b). The current through the series combination of the 2-k⍀ and 4-k⍀ resistors is obtained by current division as i ϭ 3 ϩ 3 ϩ (6 mA) ϭ 2 mA 2 4 Hence, the voltages v1 and v2 across the capacitors are v1 ϭ 2000i ϭ 4 V v2 ϭ 4000i ϭ 8 V and the energies stored in them are w1 ϭ 1 C1v12 ϭ 1 ϫ 10Ϫ3)(4)2 ϭ 16 mJ 2 (2 2 w2 ϭ 1 C2v22 ϭ 1 ϫ 10Ϫ3)(8)2 ϭ 128 mJ 2 (4 2Practice Problem 6.5 Under dc conditions, find the energy stored in the capacitors in Fig. 6.13. 3 kΩ Answer: 20.25 mJ, 3.375 mJ.1 kΩ 6.3 Series and Parallel Capacitors 30 ␮F We know from resistive circuits that the series-parallel combination is a powerful tool for reducing circuits. This technique can be extended to50 V +− 20 ␮F 6 kΩ series-parallel connections of capacitors, which are sometimes encoun- tered. We desire to replace these capacitors by a single equivalentFigure 6.13 capacitor Ceq.For Practice Prob. 6.5. In order to obtain the equivalent capacitor Ceq of N capacitors in parallel, consider the circuit in Fig. 6.14(a). The equivalent circuit is

6.3 Series and Parallel Capacitors 223in Fig. 6.14(b). Note that the capacitors have the same voltage v across i i1 i2 i3 iN +them. Applying KCL to Fig. 6.14(a), C1 C2 C3 CN v − i ϭ i1 ϩ i2 ϩ i3 ϩ p ϩ iN (6.11)But ik ϭ Ck dv͞dt. Hence, (a) i ϭ dv ϩ dv ϩ dv ϩ p ϩ dv + C1 dt C2 dt C3 dt CN dt Ceq v (6.12) i − ϭ N dv dv a a Ckb ϭ Ceq dt kϭ1 dt (b)where Figure 6.14 (a) Parallel-connected N capacitors, Ceq ϭ C1 ϩ C2 ϩ C3 ϩ p ϩ CN (b) equivalent circuit for the parallel capacitors. (6.13)The equivalent capacitance of N parallel-connected capacitors is thesum of the individual capacitances.We observe that capacitors in parallel combine in the same manner as i C1 C2 C3 CNresistors in series. + v1 − + v2 − + v3 − + vN − We now obtain Ceq of N capacitors connected in series by com- v +−paring the circuit in Fig. 6.15(a) with the equivalent circuit inFig. 6.15(b). Note that the same current i flows (and consequentlythe same charge) through the capacitors. Applying KVL to the loopin Fig. 6.15(a), v ϭ v1 ϩ v2 ϩ v3 ϩ p ϩ vN (6.14) (a) iBut vk ϭ CkΎ1 t i(t) dt ϩ vk(t0). Therefore, t0 + Ceq v 1t v +− Ύ Ύ1 t − vϭ t0 i(t) dt ϩ v1(t0) ϩ C2 i(t) dt ϩ v2 (t0) C1 t0 (b) ϩpϩ Ύ1 t i(t) dt ϩ vN (t0) CN Figure 6.15 t0 (a) Series-connected N capacitors, (6.15) (b) equivalent circuit for the series Ύ1 1 1 t capacitor. b ϭa i(t) dt ϩ v1(t0) ϩ v2(t0) ϩ ϩpϩ CN C1 C2 t0 ϩ p ϩ vN(t0) ϭ Ύ1 t i(t) dt ϩ v(t0) Ceq t0where 1 ϭ 1 ϩ 1 ϩ 1 ϩpϩ 1 (6.16) Ceq C1 C2 C3 CN

224 Chapter 6 Capacitors and Inductors Example 6.6 The initial voltage v(t0) across Ceq is required by KVL to be the sum of the capacitor voltages at t0. Or according to Eq. (6.15), v(t0) ϭ v1(t0) ϩ v2(t0) ϩ p ϩ vN (t0) Thus, according to Eq. (6.16), The equivalent capacitance of series-connected capacitors is the reciprocal of the sum of the reciprocals of the individual capacitances. Note that capacitors in series combine in the same manner as resistors in parallel. For N ϭ 2 (i.e., two capacitors in series), Eq. (6.16) becomes 1 11 ϭϩ Ceq C1 C2 or Ceq ϭ C1C2 (6.17) C1 ϩ C2 Find the equivalent capacitance seen between terminals a and b of the circuit in Fig. 6.16. 5 ␮F 60 ␮F 20 ␮F 6 ␮F 20 ␮F a Ceq b Figure 6.16 For Example 6.6. Solution: The 20-mF and 5-mF capacitors are in series; their equivalent capaci- tance is 20 ϫ 5 ϭ 4 mF 20 ϩ 5 This 4-mF capacitor is in parallel with the 6-mF and 20-mF capacitors; their combined capacitance is 4 ϩ 6 ϩ 20 ϭ 30 mF This 30-mF capacitor is in series with the 60-mF capacitor. Hence, the equivalent capacitance for the entire circuit is 30 ϫ 60 Ceq ϭ 30 ϩ 60 ϭ 20 mF

6.3 Series and Parallel Capacitors 225Find the equivalent capacitance seen at the terminals of the circuit in Practice Problem 6.6Fig. 6.17. 50 ␮FAnswer: 40 mF. 60 ␮F Ceq 70 ␮F 20 ␮F 120 ␮F Figure 6.17 For Practice Prob. 6.6.For the circuit in Fig. 6.18, find the voltage across each capacitor. Example 6.7Solution: 20 mF 30 mFWe first find the equivalent capacitance Ceq, shown in Fig. 6.19. The two + v1 − + v2 − +parallel capacitors in Fig. 6.18 can be combined to get 40 ϩ 20 ϭ 60 mF. 30 V +− 40 mF v3This 60-mF capacitor is in series with the 20-mF and 30-mF capacitors. − 20 mFThus, Ceq ϭ 1 mF ϭ 10 mF 1 ϩ 1 ϩ 1 Figure 6.18 60 30 20 For Example 6.7.The total charge is q ϭ Ceqv ϭ 10 ϫ 10Ϫ3 ϫ 30 ϭ 0.3 CThis is the charge on the 20-mF and 30-mF capacitors, because they are 30 V +− Ceqin series with the 30-V source. (A crude way to see this is to imaginethat charge acts like current, since i ϭ dq͞dt.) Therefore, q 0.3 q 0.3 Figure 6.19v1 ϭ C1 ϭ 20 ϫ 10Ϫ3 ϭ 15 V v2 ϭ C2 ϭ 30 ϫ 10Ϫ3 ϭ 10 V Equivalent circuit for Fig. 6.18.Having determined v1 and v2, we now use KVL to determine v3 by v3 ϭ 30 Ϫ v1 Ϫ v2 ϭ 5 V Alternatively, since the 40-mF and 20-mF capacitors are in parallel,they have the same voltage v3 and their combined capacitance is 40 ϩ20 ϭ 60 mF. This combined capacitance is in series with the 20-mF and30-mF capacitors and consequently has the same charge on it. Hence, v3 ϭ q ϭ 60 0.3 ϭ 5 V 60 mF ϫ 10Ϫ3Find the voltage across each of the capacitors in Fig. 6.20. Practice Problem 6.7Answer: v1 ϭ 45 V, v2 ϭ 45 V, v3 ϭ 15 V, v4 ϭ 30 V. 40 ␮F 60 ␮F + v1 − + v3 − 90 V +− + + 30 ␮F 20 ␮F v4 v2 − − Figure 6.20 For Practice Prob. 6.7.

226 Chapter 6 Capacitors and InductorsLength, ᐍ 6.4 Inductors Cross-sectional area, A An inductor is a passive element designed to store energy in its mag- netic field. Inductors find numerous applications in electronic and Core material power systems. They are used in power supplies, transformers, radios, TVs, radars, and electric motors. Number of turns, N Any conductor of electric current has inductive properties and mayFigure 6.21 be regarded as an inductor. But in order to enhance the inductive effect,Typical form of an inductor. a practical inductor is usually formed into a cylindrical coil with many turns of conducting wire, as shown in Fig. 6.21. An inductor consists of a coil of conducting wire. If current is allowed to pass through an inductor, it is found that the voltage across the inductor is directly proportional to the time rate of change of the current. Using the passive sign convention,In view of Eq. (6.18), for an inductor v ϭ L di (6.18)to have voltage across its terminals, its dtcurrent must vary with time. Hence,v ϭ 0 for constant current through where L is the constant of proportionality called the inductance of thethe inductor. inductor. The unit of inductance is the henry (H), named in honor of the American inventor Joseph Henry (1797–1878). It is clear from Eq. (6.18) that 1 henry equals 1 volt-second per ampere. Inductance is the property whereby an inductor exhibits opposition to the change of current flowing through it, measured in henrys (H). (a) The inductance of an inductor depends on its physical dimension and construction. Formulas for calculating the inductance of inductors (b) of different shapes are derived from electromagnetic theory and can be found in standard electrical engineering handbooks. For example, for (c) the inductor, (solenoid) shown in Fig. 6.21,Figure 6.22 N 2m A (6.19)Various types of inductors: (a) solenoidal Lϭwound inductor, (b) toroidal inductor,(c) chip inductor. /Courtesy of Tech America. where N is the number of turns, / is the length, A is the cross-sectional area, and m is the permeability of the core. We can see from Eq. (6.19) that inductance can be increased by increasing the number of turns of coil, using material with higher permeability as the core, increasing the cross-sectional area, or reducing the length of the coil. Like capacitors, commercially available inductors come in differ- ent values and types. Typical practical inductors have inductance values ranging from a few microhenrys (mH), as in communication systems, to tens of henrys (H) as in power systems. Inductors may be fixed or variable. The core may be made of iron, steel, plastic, or air. The terms coil and choke are also used for inductors. Common inductors are shown in Fig. 6.22. The circuit symbols for inductors are shown in Fig. 6.23, following the passive sign convention. Equation (6.18) is the voltage-current relationship for an inductor. Figure 6.24 shows this relationship graphically for an inductor whose

6.4 Inductors 227 Historical NOAA’s People CollectionJoseph Henry (1797–1878), an American physicist, discovered induc-tance and constructed an electric motor. Born in Albany, New York, Henry graduated from Albany Acad-emy and taught philosophy at Princeton University from 1832 to 1846.He was the first secretary of the Smithsonian Institution. He conductedseveral experiments on electromagnetism and developed powerful elec-tromagnets that could lift objects weighing thousands of pounds. Inter-estingly, Joseph Henry discovered electromagnetic induction beforeFaraday but failed to publish his findings. The unit of inductance, thehenry, was named after him.inductance is independent of current. Such an inductor is known as a i i ilinear inductor. For a nonlinear inductor, the plot of Eq. (6.18) willnot be a straight line because its inductance varies with current. We + + +will assume linear inductors in this textbook unless stated otherwise. vL vL vL − − − The current-voltage relationship is obtained from Eq. (6.18) as di ϭ 1 v dt (a) (b) (c) L Figure 6.23Integrating gives Ύ1 t (6.20) Circuit symbols for inductors: (a) air-core,or (b) iron-core, (c) variable iron-core. i ϭ v(t) dt L Ϫϱ v Ύi ϭ 1 t (6.21) L v(t) dt ϩ i(t0) t0 Slope = Lwhere i(t0) is the total current for Ϫϱ 6 t 6 t0 and i(Ϫϱ) ϭ 0. The 0 di/dtidea of making i(Ϫϱ) ϭ 0 is practical and reasonable, because theremust be a time in the past when there was no current in the inductor. Figure 6.24 Voltage-current relationship of an inductor. The inductor is designed to store energy in its magnetic field. Theenergy stored can be obtained from Eq. (6.18). The power delivered tothe inductor is p ϭ vi ϭ aL di bi (6.22) dtThe energy stored is Ύ Ύt t diw ϭ p(t) dt ϭ L dt idt Ϫϱ Ϫϱ (6.23) Ύϭ L t i di ϭ 1 Li2(t) Ϫ 1 Li2(Ϫϱ) 22 Ϫϱ

228 Chapter 6 Capacitors and Inductors Since i(Ϫϱ) ϭ 0, w ϭ 1 Li2 (6.24) 2 ii We should note the following important properties of an inductor. tt 1. Note from Eq. (6.18) that the voltage across an inductor is zero (a) (b) when the current is constant. Thus,Figure 6.25 An inductor acts like a short circuit to dc.Current through an inductor: (a) allowed,(b) not allowable; an abrupt change is not 2. An important property of the inductor is its opposition to thepossible. change in current flowing through it. Since an inductor is often made of a The current through an inductor cannot change instantaneously. highly conducting wire, it has a very small resistance. According to Eq. (6.18), a discontinuous change in the current through an inductor requires an infinite voltage, which is not phys- L Rw ically possible. Thus, an inductor opposes an abrupt change in the current through it. For example, the current through an inductor Cw may take the form shown in Fig. 6.25(a), whereas the inductor cur- rent cannot take the form shown in Fig. 6.25(b) in real-life situa-Figure 6.26 tions due to the discontinuities. However, the voltage across anCircuit model for a practical inductor. inductor can change abruptly. 3. Like the ideal capacitor, the ideal inductor does not dissipate energy. The energy stored in it can be retrieved at a later time. The inductor takes power from the circuit when storing energy and delivers power to the circuit when returning previously stored energy. 4. A practical, nonideal inductor has a significant resistive component, as shown in Fig. 6.26. This is due to the fact that the inductor is made of a conducting material such as copper, which has some resistance. This resistance is called the winding resistance Rw, and it appears in series with the inductance of the inductor. The pres- ence of Rw makes it both an energy storage device and an energy dissipation device. Since Rw is usually very small, it is ignored in most cases. The nonideal inductor also has a winding capacitance Cw due to the capacitive coupling between the conducting coils. Cw is very small and can be ignored in most cases, except at high fre- quencies. We will assume ideal inductors in this book.Example 6.8 The current through a 0.1-H inductor is i(t) ϭ 10teϪ5t A. Find the volt- age across the inductor and the energy stored in it. Solution: Since v ϭ L di͞dt and L ϭ 0.1 H, v ϭ 0.1 d (10teϪ5t) ϭ eϪ5t ϩ t(Ϫ5)eϪ5t ϭ eϪ5t(1 Ϫ 5t) V dt

6.4 Inductors 229The energy stored is w ϭ 1 Li2 ϭ 1 (0.1)100t 2eϪ10t ϭ 5t 2eϪ10t J 22If the current through a 1-mH inductor is i(t) ϭ 60 cos 100t mA, find Practice Problem 6.8the terminal voltage and the energy stored.Answer: Ϫ6 sin 100t mV, 1.8 cos2 (100t) mJ.Find the current through a 5-H inductor if the voltage across it is Example 6.9 30t 2, t70 v(t) ϭ b t60 0,Also, find the energy stored at t ϭ 5 s. Assume i(v) 7 0.Solution: Ύ1 tSince i ϭ v(t) dt ϩ i (t0) and L ϭ 5 H, L t0 Ύ1 t 30t 2 dt ϩ 0 ϭ 6 ϫ t 3 ϭ 2t 3 A iϭ 53 0The power p ϭ vi ϭ 60t 5, and the energy stored is then Ύ Ύw ϭ 5 t6 5 p dt ϭ 60t 5 dt ϭ 60 6 2 ϭ 156.25 kJ 00Alternatively, we can obtain the energy stored using Eq. (6.24), bywritingw05 ϭ 1 Li2(5) Ϫ 1 ϭ 1 ϫ 53)2 Ϫ 0 ϭ 156.25 kJ02 Li(0) (5)(2 22as obtained before.The terminal voltage of a 2-H inductor is v ϭ 10(1 Ϫ t) V. Find the Practice Problem 6.9current flowing through it at t ϭ 4 s and the energy stored in it at t ϭ 4 s.Assume i(0) ϭ 2 A.Answer: Ϫ18 A, 320 J.

230 Chapter 6 Capacitors and InductorsExample 6.10 Consider the circuit in Fig. 6.27(a). Under dc conditions, find: (a) i, vC, and iL, (b) the energy stored in the capacitor and inductor. i 1Ω 5Ω iL Solution: iL12 V +− 4Ω (a) Under dc conditions, we replace the capacitor with an open circuit and the inductor with a short circuit, as in Fig. 6.27(b). It is evident + 2H from Fig. 6.27(b) that vC 1 F − (a) i ϭ iL ϭ 1 12 5 ϭ 2 A ϩ i 1Ω 5Ω The voltage vC is the same as the voltage across the 5-⍀ resistor. Hence, vC ϭ 5i ϭ 10 V12 V +− 4Ω (b) The energy in the capacitor is + vC wC ϭ 1 CvC2 ϭ 1 (1)(102) ϭ 50 J − 2 2 (b) and that in the inductor isFigure 6.27 wL ϭ 1 LiL2 ϭ 1 (2)(22) ϭ 4 JFor Example 6.10. 2 2Practice Problem 6.10 Determine vC, iL, and the energy stored in the capacitor and inductor in the circuit of Fig. 6.28 under dc conditions. iL 6 H + 6Ω 2Ω vC 4 F Answer: 15 V, 7.5 A, 450 J, 168.75 J. −10 AFigure 6.28 6.5 Series and Parallel InductorsFor Practice Prob. 6.10. i L1 L2 L3 LN Now that the inductor has been added to our list of passive elements, it is necessary to extend the powerful tool of series-parallel combination. We+ + v1 − + v2 − + v3 − . . . + vN − need to know how to find the equivalent inductance of a series-connectedv or parallel-connected set of inductors found in practical circuits.− Consider a series connection of N inductors, as shown in Fig. 6.29(a), with the equivalent circuit shown in Fig. 6.29(b). The inductors have the same current through them. Applying KVL to the loop, (a) v ϭ v1 ϩ v2 ϩ v3 ϩ p ϩ vN (6.25) Substituting vk ϭ Lk di͞dt results in i + di di di di v Leq v ϭ L1 dt ϩ L2 dt ϩ L3 dt ϩ p ϩ LN dt − ϭ (L1 ϩ L2 ϩ L3 ϩ p ϩ di (6.26) LN) dt (b) N di diFigure 6.29 ϭ a Lk b ϭ Leq dt(a) A series connection of N inductors, a dt(b) equivalent circuit for the seriesinductors. kϭ1 where Leq ϭ L1 ϩ L2 ϩ L3 ϩ p ϩ LN (6.27)

6.5 Series and Parallel Inductors 231Thus, iThe equivalent inductance of series-connected inductors is the sum + i1 i2 i3 L3 iNof the individual inductances. v L1 L2 LN − (a)Inductors in series are combined in exactly the same way as resistorsin series. i We now consider a parallel connection of N inductors, as shown +in Fig. 6.30(a), with the equivalent circuit in Fig. 6.30(b). The induc-tors have the same voltage across them. Using KCL, v Leq i ϭ i1 ϩ i2 ϩ i3 ϩ p ϩ iN (6.28) −But ik ϭ LkΎ1 t v dt ϩ ik(t0); hence, (b) t0 Figure 6.30 (a) A parallel connection of N inductors, 1t (b) equivalent circuit for the parallel Ύ Ύ1 t inductors. iϭ t0 v dt ϩ i1(t0) ϩ L2 v dt ϩ i2(t0) L1 t0 Ύϩ p ϩ 1 t LN v dt ϩ iN (t0) t0 1t Ύ1 1 ϭa ϩ ϩpϩ b v dt ϩ i1(t0) ϩ i2(t0) L1 L2 LN t0 ϩ p ϩ iN (t0) t 1t Ύ ΎN 1 N ϭ a a b v dt ϩ a ik(t0) ϭ Leq v dt ϩ i(t0) (6.29) Lk kϭ1 t0 kϭ1 t0where 1 ϭ 1 ϩ 1 ϩ 1 ϩpϩ 1 (6.30) Leq L1 L2 L3 LNThe initial current i(t0) through Leq at t ϭ t0 is expected by KCL to bethe sum of the inductor currents at t0. Thus, according to Eq. (6.29), i(t0) ϭ i1(t0) ϩ i2(t0) ϩ p ϩ iN(t0)According to Eq. (6.30),The equivalent inductance of parallel inductors is the reciprocal of thesum of the reciprocals of the individual inductances.Note that the inductors in parallel are combined in the same way asresistors in parallel. For two inductors in parallel (N ϭ 2), Eq. (6.30) becomes 1 11 or Leq ϭ L1L2 (6.31) ϭϩ L1 ϩ L2 Leq L1 L2As long as all the elements are of the same type, the ¢-Y transforma-tions for resistors discussed in Section 2.7 can be extended to capacitorsand inductors.

232 Chapter 6 Capacitors and Inductors TABLE 6.1 Important characteristics of the basic elements.† Relation Resistor (R) Capacitor (C) Inductor (L) v-i: v ϭ iR Ύv ϭ 1 t di C vϭL i-v: i(t)dt ϩ v(t0) dt p or w: t0 Series: Parallel: i ϭ v͞R dv Ύi ϭ 1 t At dc: iϭC L v(t)dt ϩ i(t0) dt t0 p ϭ i2R ϭ v2 w ϭ 1 Cv2 w ϭ 1 Li2 R 2 2 Req ϭ R1 ϩ R2 Ceq ϭ C1C2 Leq ϭ L1 ϩ L2 C1 ϩ C2 Req ϭ R1R2 Ceq ϭ C1 ϩ C2 Leq ϭ L1L2 R1 ϩ R2 L1 ϩ L2 Same Open circuit Short circuit Circuit variable i that cannot change abruptly: Not applicable v † Passive sign convention is assumed. It is appropriate at this point to summarize the most important characteristics of the three basic circuit elements we have studied. The summary is given in Table 6.1. The wye-delta transformation discussed in Section 2.7 for resistors can be extended to capacitors and inductors. Example 6.11 Find the equivalent inductance of the circuit shown in Fig. 6.31. 4 H 20 H 12 H Solution: The 10-H, 12-H, and 20-H inductors are in series; thus, combiningL eq them gives a 42-H inductance. This 42-H inductor is in parallel with 7H the 7-H inductor so that they are combined, to give 8H 10 H 7 ϫ 42 ϭ 6 H 7 ϩ 42Figure 6.31For Example 6.11. This 6-H inductor is in series with the 4-H and 8-H inductors. Hence, Leq ϭ 4 ϩ 6 ϩ 8 ϭ 18 HPractice Problem 6.11 Calculate the equivalent inductance for the inductive ladder network in Fig. 6.32. 20 mH 100 mH 40 mH L eq 20 mH 50 mH 40 mH 30 mH Figure 6.32 For Practice Prob. 6.11. Answer: 25 mH.

6.6 Applications 233For the circuit in Fig. 6.33, i(t) ϭ 4(2 Ϫ eϪ10t) mA. If i2(0) ϭ Ϫ1 mA, Example 6.12find: (a) i1(0); (b) v(t), v1(t), and v2(t); (c) i1(t) and i2(t). i 2HSolution: + + v1 − i1 i2(a) From i(t) ϭ 4(2 Ϫ eϪ10t) mA, i(0) ϭ 4(2 Ϫ 1) ϭ 4 mA. Since i ϭ + 12 Hi1 ϩ i2, v 4 H v2 i1(0) ϭ i(0) Ϫ i2(0) ϭ 4 Ϫ (Ϫ1) ϭ 5 mA −(b) The equivalent inductance is − Figure 6.33 For Example 6.12. Leq ϭ 2 ϩ 4 ʈ 12 ϭ 2 ϩ 3 ϭ 5 HThus, v(t) ϭ Leq di ϭ 5(4)(Ϫ1)(Ϫ10)eϪ10t mV ϭ 200eϪ10t mV dtand v1(t) ϭ di ϭ 2(Ϫ4)(Ϫ10)eϪ10t mV ϭ 80eϪ10t mV 2 dtSince v ϭ v1 ϩ v2, v2(t) ϭ v(t) Ϫ v1(t) ϭ 120eϪ10t mV(c) The current i1 is obtained as 1 t 120 tΎ Ύi1(t) 4 4 ϭ v2 dt ϩ i1(0) ϭ eϪ10t dt ϩ 5 mA 00 ϭ Ϫ3eϪ10t 0 t ϩ 5 mA ϭ Ϫ3eϪ10t ϩ 3 ϩ 5 ϭ 8 Ϫ 3eϪ10t mA 0Similarly, Ύ Ύi2(t)ϭ 1 t ϩ ϭ 120 t 12 dt i2(0) 12 v2 eϪ10t dt Ϫ 1 mA 00 ϭ ϪeϪ10t 0 t Ϫ 1 mA ϭ ϪeϪ10t ϩ 1 Ϫ 1 ϭ ϪeϪ10t mA 0Note that i1(t) ϩ i2(t) ϭ i(t).In the circuit of Fig. 6.34, i1(t) ϭ 0.6eϪ2t A. If i(0) ϭ 1.4 A, find: Practice Problem 6.12(a) i2(0); (b) i2(t) and i(t); (c) v1(t), v2(t), and v(t).Answer: (a) 0.8 A, (b) (Ϫ0.4 ϩ 1.2eϪ2t) A, (Ϫ0.4 ϩ 1.8eϪ2t) A, i2 3 H +(c) Ϫ36eϪ2t V, Ϫ7.2eϪ2t V, Ϫ28.8eϪ2t V. i v2 8 H −6.6 Applications + v1 − +Circuit elements such as resistors and capacitors are commerciallyavailable in either discrete form or integrated-circuit (IC) form. Unlike v i1 6 Hcapacitors and resistors, inductors with appreciable inductance are dif-ficult to produce on IC substrates. Therefore, inductors (coils) usually − Figure 6.34 For Practice Prob. 6.12.

234 Chapter 6 Capacitors and Inductors i2 Rf come in discrete form and tend to be more bulky and expensive. For this reason, inductors are not as versatile as capacitors and resistors, i1 R1 v1 0 A − and they are more limited in applications. However, there are several+ 1− applications in which inductors have no practical substitute. They are routinely used in relays, delays, sensing devices, pick-up heads, tele- phone circuits, radio and TV receivers, power supplies, electric motors, microphones, and loudspeakers, to mention a few. Capacitors and inductors possess the following three special prop- erties that make them very useful in electric circuits: 1. The capacity to store energy makes them useful as temporary volt- age or current sources. Thus, they can be used for generating a large amount of current or voltage for a short period of time. 2. Capacitors oppose any abrupt change in voltage, while inductors oppose any abrupt change in current. This property makes induc- tors useful for spark or arc suppression and for converting pulsat- ing dc voltage into relatively smooth dc voltage. 3. Capacitors and inductors are frequency sensitive. This property makes them useful for frequency discrimination. The first two properties are put to use in dc circuits, while the third one is taken advantage of in ac circuits. We will see how useful these properties are in later chapters. For now, consider three applications involving capacitors and op amps: integrator, differentiator, and analog computer. 6.6.1 Integrator Important op amp circuits that use energy-storage elements include integrators and differentiators. These op amp circuits often involve resistors and capacitors; inductors (coils) tend to be more bulky and expensive. The op amp integrator is used in numerous applications, especially in analog computers, to be discussed in Section 6.6.3. 0Vvi v2 + + + An integrator is an op amp circuit whose output is proportional to the vo integral of the input signal. −− (a) If the feedback resistor Rf in the familiar inverting amplifier of iC C Fig. 6.35(a) is replaced by a capacitor, we obtain an ideal integrator, as shown in Fig. 6.35(b). It is interesting that we can obtain a mathe- − matical representation of integration this way. At node a in Fig. 6.35(b), a iR R iR ϭ iC (6.32)+ +vi But− + iR ϭ vi, iC ϭ ϪC dvo vo R dt − Substituting these in Eq. (6.32), we obtain (b) vi ϭ ϪC dvo (6.33a) R dtFigure 6.35Replacing the feedback resistor in the dvo ϭ Ϫ1 vi dt (6.33b)inverting amplifier in (a) produces an RCintegrator in (b).

6.6 Applications 235Integrating both sides givesvo(t) Ϫ vo(0) ϭ Ϫ Ύ1 t vi(t) dt (6.34) RC 0To ensure that vo(0) ϭ 0, it is always necessary to discharge the integra-tor’s capacitor prior to the application of a signal. Assuming vo(0) ϭ 0, 1 t Ϫ vi(t) dt RC Ύvo ϭ (6.35) 0which shows that the circuit in Fig. 6.35(b) provides an output voltageproportional to the integral of the input. In practice, the op amp inte-grator requires a feedback resistor to reduce dc gain and prevent satu-ration. Care must be taken that the op amp operates within the linearrange so that it does not saturate.If v1 ϭ 10 cos 2t mV and v2 ϭ 0.5t mV, find vo in the op amp circuit Example 6.13in Fig. 6.36. Assume that the voltage across the capacitor is initially zero. 3 MΩ 2 ␮F v1Solution:This is a summing integrator, and Ύ Ύ1 1 − vo +vo ϭ Ϫ R1C v1 dt Ϫ R2C v2 dt v2 100 kΩΎ1 tϭ Ϫ 3 ϫ 106 ϫ 2 ϫ 10Ϫ6 10 cos (2t)dt Figure 6.36 For Example 6.13. 0 Ύ1 tϪ 100 ϫ 103 ϫ 2 ϫ 10Ϫ6 0.5tdt 0ϭ Ϫ 1 10 sin 2t Ϫ 1 0.5t 2 ϭ Ϫ0.833 sin 2t Ϫ 1.25t 2 mV 6 2 0.2 2The integrator in Fig. 6.35(b) has R ϭ 100 k⍀, C ϭ 20 mF. Determine Practice Problem 6.13the output voltage when a dc voltage of 2.5 mV is applied at t ϭ 0.Assume that the op amp is initially nulled.Answer: Ϫ1.25t mV.6.6.2 DifferentiatorA differentiator is an op amp circuit whose output is proportional tothe rate of change of the input signal. In Fig. 6.35(a), if the input resistor is replaced by a capacitor, theresulting circuit is a differentiator, shown in Fig. 6.37. Applying KCLat node a, iR ϭ iC (6.36)

236 Chapter 6 Capacitors and Inductors But iR ϭ Ϫvo, iC ϭ C dvi R dt Substituting these in Eq. (6.36) yields iR R iC C − vo ϭ ϪRC dvi (6.37) + dt+avi +− v−o showing that the output is the derivative of the input. Differentiator cir- cuits are electronically unstable because any electrical noise within theFigure 6.37 circuit is exaggerated by the differentiator. For this reason, the differ-An op amp differentiator. entiator circuit in Fig. 6.37 is not as useful and popular as the inte- grator. It is seldom used in practice.Example 6.14 Sketch the output voltage for the circuit in Fig. 6.38(a), given the input voltage in Fig. 6.38(b). Take vo ϭ 0 at t ϭ 0. 0.2 ␮F 5 kΩ vi +− Solution: − This is a differentiator with + + RC ϭ 5 ϫ 103 ϫ 0.2 ϫ 10Ϫ6 ϭ 10Ϫ3 s vo − For 0 6 t 6 4 ms, we can express the input voltage in Fig. 6.38(b) as ϭ 2000t 0 6 t 6 2 ms vi e (a) 8 Ϫ 2000t 2 6 t 6 4 msvo(V) This is repeated for 4 6 t 6 8 ms. Using Eq. (6.37), the output is 4 obtained as vo ϭ ϪRC dvi ϭ e Ϫ2 V 0 6 t 6 2 ms dt 2V 2 6 t 6 4 ms 024 6 8 t (ms) Thus, the output is as sketched in Fig. 6.39. (b) vo (V)Figure 6.38 2For Example 6.14. 0 2 4 6 8 t (ms) −2 Figure 6.39 Output of the circuit in Fig. 6.38(a).Practice Problem 6.14 The differentiator in Fig. 6.37 has R ϭ 100 k⍀ and C ϭ 0.1 mF. Given that vi ϭ 1.25t V, determine the output vo. Answer: Ϫ12.5 mV.

6.6 Applications 2376.6.3 Analog Computer Example 6.15Op amps were initially developed for electronic analog computers.Analog computers can be programmed to solve mathematical models ofmechanical or electrical systems. These models are usually expressed interms of differential equations. To solve simple differential equations using the analog computerrequires cascading three types of op amp circuits: integrator circuits,summing amplifiers, and inverting/noninverting amplifiers for negative/positive scaling. The best way to illustrate how an analog computer solvesa differential equation is with an example. Suppose we desire the solution x(t) of the equationd 2x dx ϩ ϭa dt 2 ϩ b cx f (t), t70 (6.38) dtwhere a, b, and c are constants, and f (t) is an arbitrary forcing func-tion. The solution is obtained by first solving the highest-order deriv-ative term. Solving for d 2x͞dt 2 yields d 2x f (t) b dx c (6.39) dt 2 ϭ a Ϫ a dt Ϫ a xTo obtain dx͞dt, the d 2x͞dt 2 term is integrated and inverted. Finally,to obtain x, the dx͞dt term is integrated and inverted. The forcing func-tion is injected at the proper point. Thus, the analog computer for solv-ing Eq. (6.38) is implemented by connecting the necessary summers,inverters, and integrators. A plotter or oscilloscope may be used to viewthe output x, or dx͞dt, or d 2x͞dt 2, depending on where it is connectedin the system. Although the above example is on a second-order differential equa-tion, any differential equation can be simulated by an analog computercomprising integrators, inverters, and inverting summers. But care mustbe exercised in selecting the values of the resistors and capacitors, toensure that the op amps do not saturate during the solution time interval. The analog computers with vacuum tubes were built in the 1950s and1960s. Recently their use has declined. They have been superseded bymodern digital computers. However, we still study analog computers fortwo reasons. First, the availability of integrated op amps has made it pos-sible to build analog computers easily and cheaply. Second, understand-ing analog computers helps with the appreciation of the digital computers.Design an analog computer circuit to solve the differential equation:d 2vo ϩ 2 dvo ϩ vo ϭ 10 sin 4t, t70dt 2 dtsubject to vo(0) ϭ Ϫ4, v¿o(0) ϭ 1, where the prime refers to the timederivative.Solution: 1. Define. We have a clearly defined problem and expected solution. I might remind the student that many times the problem is not so well defined and this portion of the problem-solving process could

238 Chapter 6 Capacitors and Inductors require much more effort. If this is so, then you should always keep in mind that time spent here will result in much less effort later and most likely save you a lot of frustration in the process.2. Present. Clearly, using the devices developed in Section 6.6.3 will allow us to create the desired analog computer circuit. We will need the integrator circuits (possibly combined with a summing capability) and one or more inverter circuits.3. Alternative. The approach for solving this problem is straight- forward. We will need to pick the correct values of resistances and capacitors to allow us to realize the equation we are repre- senting. The final output of the circuit will give the desired result.4. Attempt. There are an infinite number of possibilities for picking the resistors and capacitors, many of which will result in correct solutions. Extreme values of resistors and capacitors will result in incorrect outputs. For example, low values of resistors will overload the electronics. Picking values of resistors that are too large will cause the op amps to stop functioning as ideal devices. The limits can be determined from the characteristics of the real op amp. We first solve for the second derivative as d2vo ϭ 10 sin 4t Ϫ 2 dvo Ϫ vo (6.15.1) dt2 dtSolving this requires some mathematical operations, includingsumming, scaling, and integration. Integrating both sides ofEq. (6.15.1) givesΎdvo ϭ Ϫ t 2 dvo(t) dtdt aϪ10 sin(4t) ϩ ϩ vo(t)b dt ϩ v¿o(0)0 (6.15.2)where v¿o(0) ϭ 1. We implement Eq. (6.15.2) using the summingintegrator shown in Fig. 6.40(a). The values of the resistors andcapacitors have been chosen so that RC ϭ 1 for the term Ϫ Ύ1 t vo(t)dt RC 0Other terms in the summing integrator of Eq. (6.15.2) areimplemented accordingly. The initial condition dvo(0)͞dt ϭ 1 isimplemented by connecting a 1-V battery with a switch across thecapacitor as shown in Fig. 6.40(a). The next step is to obtain vo by integrating dvo͞dt andinverting the result, Ύvo ϭ Ϫ t aϪdvo(t) b dt ϩ v(0) (6.15.3) dt 0This is implemented with the circuit in Fig. 6.40(b) with thebattery giving the initial condition of Ϫ4 V. We now combine thetwo circuits in Fig. 6.40(a) and (b) to obtain the complete circuitshown in Fig. 6.40(c). When the input signal 10 sin 4t is applied,we open the switches at t ϭ 0 to obtain the output waveform vo,which may be viewed on an oscilloscope.

6.6 Applications 239 − 1V + t=0 1 MΩ 1 ␮F − 4V + t=0−10 sin (4t) 1 ␮F − 1 MΩ + dvo dvo 1 MΩ 1 MΩ 1 MΩ vo dt dt −vo − − dvo 0.5 MΩ + + vo dt (a) − 1V + t=0 − 4V + (b) 1 MΩ t=0 1 MΩ 1 ␮F10 sin (4t) +− 1 V 1 ␮F 1 MΩ vo dvo − 1 MΩ − dt + 1 MΩ 0.5 MΩ + vo − + (c)Figure 6.40For Example 6.15.5. Evaluate. The answer looks correct, but is it? If an actual solution for vo is desired, then a good check would be to first find the solution by realizing the circuit in PSpice. This result could then be compared with a solution using the differential solution capability of MATLAB. Since all we need to do is check the circuit and confirm that it represents the equation, we have an easier technique to use. We just go through the circuit and see if it generates the desired equation. However, we still have choices to make. We could go through the circuit from left to right but that would involve differentiating the result to obtain the original equation. An easier approach would be to go from right to left. This is the approach we will use to check the answer. Starting with the output, vo, we see that the right-hand op amp is nothing more than an inverter with a gain of one. This means that the output of the middle circuit is Ϫvo. The following represents the action of the middle circuit.ΎϪvo t dvo vo(0)b ϭ 2 t vo(0)b ϭ Ϫa 0 dt dt ϩ Ϫavo 0 ϩ ϭ Ϫ(vo(t) Ϫ vo(0) ϩ vo(0))where vo(0) ϭ Ϫ4 V is the initial voltage across the capacitor. We check the circuit on the left the same way.Ύdvo ϭ Ϫa t Ϫ d 2vo dt Ϫ v¿o(0)b ϭ ϪaϪdvo ϩ v¿o(0) Ϫ v¿o(0)b 0 dt 2 dtdt

240 Chapter 6 Capacitors and Inductors Now all we need to verify is that the input to the first op amp is Ϫd 2vo͞dt 2. Looking at the input we see that it is equal to Ϫ10 sin(4t) ϩ vo ϩ 1͞10Ϫ6 dvo ϭ Ϫ10 sin(4t) ϩ vo ϩ 2 dvo 0.5 M⍀ dt dt which does produce Ϫd2vo͞dt2 from the original equation. 6. Satisfactory? The solution we have obtained is satisfactory. We can now present this work as a solution to the problem.Practice Problem 6.15 Design an analog computer circuit to solve the differential equation: d 2vo ϩ 3 dvo ϩ 2vo ϭ 4 cos 10t, t70 dt 2 dt subject to vo(0) ϭ 2, v¿o(0) ϭ 0. Answer: See Fig. 6.41, where RC ϭ 1 s. R C 2V t=0 R −R d2vo + C R − dt2 2 + d2vo R − + vo dt2 R R cos (10t) +− R 3 − R + 4 R − + Figure 6.41 For Practice Prob. 6.15. 6.7 Summary 1. The current through a capacitor is directly proportional to the time rate of change of the voltage across it. i ϭ C dv dt The current through a capacitor is zero unless the voltage is chang- ing. Thus, a capacitor acts like an open circuit to a dc source.

Review Questions 2412. The voltage across a capacitor is directly proportional to the time integral of the current through it. t 1 t i dt ϭ Ύ Ύ1 C i dt ϩ v(t0) vϭ C Ϫϱ t0 The voltage across a capacitor cannot change instantly.3. Capacitors in series and in parallel are combined in the same way as conductances.4. The voltage across an inductor is directly proportional to the time rate of change of the current through it. di vϭL dt The voltage across the inductor is zero unless the current is chang- ing. Thus, an inductor acts like a short circuit to a dc source.5. The current through an inductor is directly proportional to the time integral of the voltage across it. 1t Ύ Ύ1 t iϭ v dt ϭ v dt ϩ i(t0) L L Ϫϱ t0 The current through an inductor cannot change instantly.6. Inductors in series and in parallel are combined in the same way resistors in series and in parallel are combined.7. At any given time t, the energy stored in a capacitor is 1 Cv2, while 2 1 Li2. the energy stored in an inductor is 28. Three application circuits, the integrator, the differentiator, and the analog computer, can be realized using resistors, capacitors, and op amps. Review Questions6.1 What charge is on a 5-F capacitor when it is v (t) connected across a 120-V source? 10 (a) 600 C (b) 300 C 0 (c) 24 C (d) 12 C 16.2 Capacitance is measured in: −10 2 t (a) coulombs (b) joules Figure 6.42 (c) henrys (d) farads For Review Question 6.4.6.3 When the total charge in a capacitor is doubled, the energy stored: (a) remains the same (b) is halved 6.5 The total capacitance of two 40-mF series-connected (c) is doubled (d) is quadrupled capacitors in parallel with a 4-mF capacitor is:6.4 Can the voltage waveform in Fig. 6.42 be associated (a) 3.8 mF (b) 5 mF (c) 24 mF with a real capacitor? (d) 44 mF (e) 84 mF (a) Yes (b) No

242 Chapter 6 Capacitors and Inductors6.6 In Fig. 6.43, if i ϭ cos 4t and v ϭ sin 4t, the 6.9 Inductors in parallel can be combined just like element is: resistors in parallel. (a) a resistor (b) a capacitor (c) an inductor (a) True (b) False i 6.10 For the circuit in Fig. 6.44, the voltage divider Element formula is:v +− (a) v1 ϭ L1 ϩ L2 vs (b) v1 ϭ L1 ϩ L2 vs L1 L2Figure 6.43 (c) v1 ϭ L1 L2 L2 vs (d) v1 ϭ L1 L1 L2 vsFor Review Question 6.6. ϩ ϩ L16.7 A 5-H inductor changes its current by 3 A in 0.2 s. The + v1 − + voltage produced at the terminals of the inductor is:(a) 75 V (b) 8.888 V vs +− v2 L2(c) 3 V (d) 1.2 V −6.8 If the current through a 10-mH inductor increases Figure 6.44 from zero to 2 A, how much energy is stored in the For Review Question 6.10. inductor? Answers: 6.1a, 6.2d, 6.3d, 6.4b, 6.5c, 6.6b, 6.7a, 6.8b,(a) 40 mJ (b) 20 mJ 6.9a, 6.10d.(c) 10 mJ (d) 5 mJProblemsSection 6.2 Capacitors 6.6 The voltage waveform in Fig. 6.46 is applied across a 55-mF capacitor. Draw the current waveform 6.1 If the voltage across a 7.5-F capacitor is 2teϪ3t V, through it. find the current and the power. v (t) V 6.2 A 50-mF capacitor has energy w(t) ϭ 10 cos2 377t J. Determine the current through the capacitor. 10 6.3 Design a problem to help other students better 0 2 4 6 8 10 12 t (ms) understand how capacitors work. −10 6.4 A current of 4 sin 4t A flows through a 5-F capacitor. Find the voltage v(t) across the capacitor given that Figure 6.46 v(0) ϭ 1 V. For Prob. 6.6. 6.5 The voltage across a 4-mF capacitor is shown in Fig. 6.45. Find the current waveform. v(t) V 6.7 At t ϭ 0, the voltage across a 25-mF capacitor is 10 V. 10 Calculate the voltage across the capacitor for t 7 0 when current 5t mA flows through it. 0 2 4 6 8 t (ms) 6.8 A 4-mF capacitor has the terminal voltage −10 50 V, tՅ0 v ϭ b AeϪ100t ϩ BeϪ600t V, tՆ0Figure 6.45For Prob. 6.5. If the capacitor has an initial current of 2 A, find: (a) the constants A and B, (b) the energy stored in the capacitor at t ϭ 0, (c) the capacitor current for t 7 0.

Problems 2436.9 The current through a 0.5-F capacitor is 6(1 Ϫ eϪt) A. 6.15 Two capacitors (25 mF and 75 mF) are connected Determine the voltage and power at t ϭ 2 s. Assume to a 100-V source. Find the energy stored in each v(0) ϭ 0. capacitor if they are connected in:6.10 The voltage across a 5-mF capacitor is shown in (a) parallel (b) series Fig. 6.47. Determine the current through the capacitor. 6.16 The equivalent capacitance at terminals a-b in the v (t) (V) circuit of Fig. 6.50 is 30 mF. Calculate the value of C. 16 a C 0 1 23 4 t (␮s) 14 ␮FFigure 6.47 80 ␮FFor Prob. 6.10.6.11 A 4-mF capacitor has the current waveform shown in b Fig. 6.48. Assuming that v(0) ϭ 10 V, sketch the voltage waveform v(t). Figure 6.50 For Prob. 6.16.i(t) (mA) 15 6.17 Determine the equivalent capacitance for each of the 10 circuits of Fig. 6.51. 5 0 4 F 12 F −5 −10 3F 6F 24 6 8 t (s) 4F (a) 6FFigure 6.48 5F 4F 2FFor Prob. 6.11. (b) 6.12 A voltage of 30eϪ2000t V appears across a parallel combination of a 100-mF capacitor and a 12-⍀ 3F 6F resistor. Calculate the power absorbed by the parallel 2F combination. 4F 3F 6.13 Find the voltage across the capacitors in the circuit of Fig. 6.49 under dc conditions. 10 Ω 50 Ω (c) + 20 Ω + Figure 6.51 +− 60 V v2 For Prob. 6.17. C1 v140 Ω − − C2 6.18 Find Ceq in the circuit of Fig. 6.52 if all capacitors are 4 mF.Figure 6.49For Prob. 6.13.Section 6.3 Series and Parallel Capacitors Ceq 6.14 Series-connected 20-pF and 60-pF capacitors are Figure 6.52 placed in parallel with series-connected 30-pF and For Prob. 6.18. 70-pF capacitors. Determine the equivalent capacitance.

244 Chapter 6 Capacitors and Inductors6.19 Find the equivalent capacitance between terminals 40 ␮F a and b in the circuit of Fig. 6.53. All capacitances are in mF. 10 ␮F 10 ␮F 80 35 ␮F 5 ␮F 12 40 20 ␮Fa 15 ␮F 15 ␮F 50 20 12 30 10 a b b Figure 6.56 For Prob. 6.22.Figure 6.53For Prob. 6.19. 60 6.23 Using Fig. 6.57, design a problem that will help other students better understand how capacitors work together when connected in series and in parallel.6.20 Find the equivalent capacitance at terminals a-b of C1 the circuit in Fig. 6.54. C3 a V +− C2 C4 1 ␮F 1 ␮F Figure 6.57 For Prob. 6.23. 2 ␮F 2 ␮F 2 ␮F 6.24 For the circuit in Figure 6.58, determine (a) the voltage across each capacitor and (b) the energy stored in each capacitor. 60 ␮F 20 ␮F3 ␮F 3 ␮F 3 ␮F 3 ␮F 90 V +− 30 ␮F 14 ␮F 80 ␮F b Figure 6.58 For Prob. 6.24.Figure 6.54 6.25 (a) Show that the voltage-division rule for twoFor Prob. 6.20. capacitors in series as in Fig. 6.59(a) is6.21 Determine the equivalent capacitance at terminals v1 ϭ C2 C2 vs, v2 ϭ C1 C1 C2 vs a-b of the circuit in Fig. 6.55. C1 ϩ ϩ assuming that the initial conditions are zero. 5 ␮F 6 ␮F 4 ␮F C1 a 2 ␮F 3 ␮F 12 ␮F + v1 − i1 i2 b + C1 C2 vs +− C2 isFigure 6.55 v2For Prob. 6.21. − (a) (b)6.22 Obtain the equivalent capacitance of the circuit in Figure 6.59 Fig. 6.56. For Prob. 6.25.

Problems 245(b) For two capacitors in parallel as in Fig. 6.59(b), 6.30 Assuming that the capacitors are initially uncharged, show that the current-division rule is find vo(t) in the circuit of Fig. 6.62.i1 ϭ C1 C1 C2 is, i2 ϭ C1 C2 C2 is ϩ ϩassuming that the initial conditions are zero. is (mA) 906.26 Three capacitors, C1 ϭ 5 mF, C2 ϭ 10 mF, and 6 ␮F C3 ϭ 20 mF, are connected in parallel across a is 3 ␮F + 150-V source. Determine: vo ( t )(a) the total capacitance, 0 2 t (s) − 1(b) the charge on each capacitor, Figure 6.62(c) the total energy stored in the parallel For Prob. 6.30. combination.6.27 Given that four 4-mF capacitors can be connected in 6.31 If v(0) ϭ 0, find v(t), i1(t), and i2(t) in the circuit of series and in parallel, find the minimum and Fig. 6.63. maximum values that can be obtained by such series/parallel combinations.*6.28 Obtain the equivalent capacitance of the network shown in Fig. 6.60.40 ␮F 30 ␮F 50 ␮F is (mA)10 ␮F 20 ␮F 30 1 2 3 4 5t 0 −30Figure 6.60 is 6 ␮F i1 i2For Prob. 6.28. 4 ␮F Figure 6.63 + 6.29 Determine Ceq for each circuit in Fig. 6.61. For Prob. 6.31. v − CCeq C C C C (a) C 6.32 In the circuit of Fig. 6.64, let is ϭ 50eϪ2t mA and C C v1(0) ϭ 50 V, v2(0) ϭ 20 V. Determine: (a) v1(t) and v2(t), (b) the energy in each capacitor at C t ϭ 0.5 s. (b) Ceq 12 ␮F + +– v2 – 40 ␮FFigure 6.61For Prob. 6.29. v1 is 20 ␮F* An asterisk indicates a challenging problem. Figure 6.64 For Prob. 6.32.

246 Chapter 6 Capacitors and Inductors6.33 Obtain the Thevenin equivalent at the terminals, a-b, 6.41 The voltage across a 2-H inductor is 20(1 Ϫ eϪ2t) V. of the circuit shown in Fig. 6.65. Please note that If the initial current through the inductor is 0.3 A, Thevenin equivalent circuits do not generally exist find the current and the energy stored in the inductor for circuits involving capacitors and resistors. This is at t ϭ 1 s. a special case where the Thevenin equivalent circuit does exist. 6.42 If the voltage waveform in Fig. 6.67 is applied across the terminals of a 5-H inductor, calculate the current through the inductor. Assume i(0) ϭ Ϫ1 A. 45 V −+ 5F a v (t) (V) 3F 2F b 10Figure 6.65For Prob. 6.33. 0 1 2 3 4 5t Figure 6.67 For Prob. 6.42.Section 6.4 Inductors 6.43 The current in an 80-mH inductor increases from 0 to 60 mA. How much energy is stored in the6.34 The current through a 10-mH inductor is 10eϪt͞2 A. inductor? Find the voltage and the power at t ϭ 3 s. *6.44 A 100-mH inductor is connected in parallel with a6.35 An inductor has a linear change in current from 2-k⍀ resistor. The current through the inductor is 50 mA to 100 mA in 2 ms and induces a voltage of i(t) ϭ 50eϪ400t mA. (a) Find the voltage vL across 160 mV. Calculate the value of the inductor. the inductor. (b) Find the voltage vR across the resistor. (c) Does vR(t) ϩ vL(t) ϭ 0? (d) Calculate6.36 Design a problem to help other students better the energy in the inductor at t ϭ 0. understand how inductors work. 6.45 If the voltage waveform in Fig. 6.68 is applied to a6.37 The current through a 12-mH inductor is 4 sin 100t A. 10-mH inductor, find the inductor current i(t). Assume i(0) ϭ 0.Find the voltage, across the inductor for 0 6 t 6 v (t)p͞200 s, and the energy stored at t ϭ p s. 200 56.38 The current through a 40-mH inductor is 0 1 2t 0, t60 i(t) ϭ b teϪ2t A, t70 –5Find the voltage v(t). Figure 6.68 For Prob. 6.45.6.39 The voltage across a 200-mH inductor is given by 6.46 Find vC, iL, and the energy stored in the capacitorv(t) ϭ 3t2 ϩ 2t ϩ 4 V for t 7 0. and inductor in the circuit of Fig. 6.69 under dc conditions. Determine the current i(t) through the inductor. Assume that i(0) ϭ 1 A.6.40 The current through a 5-mH inductor is shown in Fig. 6.66. Determine the voltage across the inductor at t ϭ 1, 3, and 5 ms. i(A) 3A 2Ω 2F iL 10 5Ω 0.5 H Figure 6.69 + 0 For Prob. 6.46. vC 2 4 6 t (ms) − 4ΩFigure 6.66For Prob. 6.40.

Problems 2476.47 For the circuit in Fig. 6.70, calculate the value of R that 6.52 Using Fig. 6.74, design a problem to help other will make the energy stored in the capacitor the same students better understand how inductors behave as that stored in the inductor under dc conditions. when connected in series and when connected in parallel. R 160 ␮F L2 L4 5 A 2 Ω 4 mH L3 Leq L1Figure 6.70 L5For Prob. 6.47. Figure 6.74 L6 For Prob. 6.52. 6.48 Under steady-state dc conditions, find i and v in the circuit in Fig. 6.71.5 mA i 2 mH 6 ␮F 20 kΩ 6.53 Find Leq at the terminals of the circuit in Fig. 6.75. + 30 kΩ v −Figure 6.71 6 mH 8 mHFor Prob. 6.48. a 5 mHSection 6.5 Series and Parallel Inductors 8 mH 6 mH6.49 Find the equivalent inductance of the circuit in b 8 mH 12 mH Fig. 6.72. Assume all inductors are 10 mH. 10 mH 4 mH Figure 6.75 For Prob. 6.53.Figure 6.72 6.54 Find the equivalent inductance looking into theFor Prob. 6.49. terminals of the circuit in Fig. 6.76. 6.50 An energy-storage network consists of series- 9H connected 16-mH and 14-mH inductors in parallel 10 H with series-connected 24-mH and 36-mH inductors. Calculate the equivalent inductance. 6.51 Determine Leq at terminals a-b of the circuit in Fig. 6.73. 10 mH 60 mH 12 H 3H 25 mH 20 mH 4H 6Ha b 30 mH a bFigure 6.73 Figure 6.76For Prob. 6.51. For Prob. 6.54.

248 Chapter 6 Capacitors and Inductors6.55 Find Leq in each of the circuits in Fig. 6.77. 6.58 The current waveform in Fig. 6.80 flows through a L 3-H inductor. Sketch the voltage across the inductor over the interval 0 6 t 6 6 s. Leq L L i(t) L 2 L L L L L 0 123456 tFigure 6.77 (a) Figure 6.80For Prob. 6.55. For Prob. 6.58. L Leq 6.59 (a) For two inductors in series as in Fig. 6.81(a), show that the voltage division principle is (b) v1 ϭ L1 L1 L2 vs, v2 ϭ L1 L2 L2 vs ϩ ϩ assuming that the initial conditions are zero. (b) For two inductors in parallel as in Fig. 6.81(b), show that the current-division principle is6.56 Find Leq in the circuit of Fig. 6.78. i1 ϭ L2 L2 is, i2 ϭ L1 L2 is ϩ ϩ L1 L1 assuming that the initial conditions are zero. L L L L L1 L LFigure 6.78 + v1 −For Prob. 6.56. LL vs +− + L2 is i1 i2 v2 L1 L2 − (b) L eq (a) Figure 6.81 For Prob. 6.59.*6.57 Determine Leq that may be used to represent the 6.60 In the circuit of Fig. 6.82, io(0) ϭ 2 A. Determine inductive network of Fig. 6.79 at the terminals. io(t) and vo(t) for t 7 0. i 2 dia dtL eq 4H io (t) +− + 3H 5H 4e–2t A 3 H 5 H vo Figure 6.82 − For Prob. 6.60. bFigure 6.79For Prob. 6.57.

Problems 2496.61 Consider the circuit in Fig. 6.83. Find: (a) Leq, i1(t), 6.64 The switch in Fig. 6.86 has been in position A for a and i2(t) if is ϭ 3eϪt mA, (b) vo(t), (c) energy stored long time. At t ϭ 0, the switch moves from position in the 20-mH inductor at t ϭ 1 s. A to B. The switch is a make-before-break type so that there is no interruption in the inductor current. Find: i1 i2 (a) i(t) for t 7 0, + 4 mH (b) v just after the switch has been moved to position B, vo 20 mH (c) v(t) long after the switch is in position B. – 6 mH is 4Ω B L eq t=0AFigure 6.83For Prob. 6.61. 12 V + i 5Ω 6A – + 0.5 H v –6.62 Consider the circuit in Fig. 6.84. Given that Figure 6.86 v(t) ϭ 12eϪ3t mV for t 7 0 and i1(0) ϭ Ϫ10 mA, For Prob. 6.64. find: (a) i2(0), (b) i1(t) and i2(t). 25 mH i1(t) i2(t) 6.65 The inductors in Fig. 6.87 are initially charged and are + 20 mH 60 mH connected to the black box at t ϭ 0. If i1(0) ϭ 4 A, i2(0) ϭ Ϫ2 A, and v(t) ϭ 50eϪ200t mV, t Ն 0, find: v (t) (a) the energy initially stored in each inductor, – (b) the total energy delivered to the black box fromFigure 6.84 t ϭ 0 to t ϭ ϱ,For Prob. 6.62. (c) i1(t) and i2(t), t Ն 0, (d) i(t), t Ն 0. i(t) + t=0 i1 i2 Black box v 5H 20 H6.63 In the circuit of Fig. 6.85, sketch vo. − Figure 6.87 For Prob. 6.65. + i2(t) 6.66 The current i(t) through a 20-mH inductor is equal, i1(t) vo 2 H 24 in magnitude, to the voltage across it for all values of time. If i(0) ϭ 2 A, find i(t). – Section 6.6 Applications i1(t) (A) i2(t) (A) 3 4 6.67 An op amp integrator has R ϭ 50 k⍀ and C ϭ 0.04 mF. If the input voltage is vi ϭ 10 sin 50t mV, 03 6 t (s) 0 6 t (s) obtain the output voltage.Figure 6.85For Prob. 6.63.

250 Chapter 6 Capacitors and Inductors 6.68 A 10-V dc voltage is applied to an integrator with 6.73 Show that the circuit in Fig. 6.90 is a noninverting R ϭ 50 k⍀, C ϭ 100 mF at t ϭ 0. How long will it integrator. take for the op amp to saturate if the saturation voltages are ϩ12 V and Ϫ12 V? Assume that the R R + initial capacitor voltage was zero. R vo − − 6.69 An op amp integrator with R ϭ 4 M⍀ and vi +− + C ϭ 1 mF has the input waveform shown in Fig. 6.88. Plot the output waveform. R vi (mV) C 20 Figure 6.90 10 For Prob. 6.73. 0 6.74 The triangular waveform in Fig. 6.91(a) is applied to 1 2 3 4 5 6 t (ms) the input of the op amp differentiator in Fig. 6.91(b). Plot the output. –10 v (t) –20 10Figure 6.88For Prob. 6.69.6.70 Using a single op amp, a capacitor, and resistors of 0 100 k⍀ or less, design a circuit to implement 1 2 3 4 t (ms) t −10 (a) Ύvo ϭ Ϫ50 vi(t) dt 0 0.01 ␮F 20 kΩ + vi +− vo Assume vo ϭ 0 at t ϭ 0. − −6.71 Show how you would use a single op amp to generate + t (b) Ύvo ϭ Ϫ (v1 ϩ 4v2 ϩ 10v3) dt Figure 6.91 0 For Prob. 6.74. If the integrating capacitor is C ϭ 2 mF, obtain the other component values.6.72 At t ϭ 1.5 ms, calculate vo due to the cascaded integrators in Fig. 6.89. Assume that the integrators are reset to 0 V at t ϭ 0. 2 ␮F 0.5 ␮F 10 kΩ 20 kΩ 6.75 An op amp differentiator has R ϭ 250 k⍀ and C ϭ1 V +− − − 10 mF. The input voltage is a ramp r(t) ϭ 12t mV. + + + Find the output voltage. vo 6.76 A voltage waveform has the following characteristics: − a positive slope of 20 V/s for 5 ms followed by a negative slope of 10 V/s for 10 ms. If the waveformFigure 6.89 is applied to a differentiator with R ϭ 50 k⍀,For Prob. 6.72. C ϭ 10 mF, sketch the output voltage waveform.