IRWIN MAIN BOOK 10th Edition PDF

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This page intentionally left blank irwin_fm_i-xxii-hr.qxd 3-09-2010 15:50 Page i accessible, affordable, active learning :LOH\3/86LVDQLQQRYDWLYHUHVHDUFKEDVHGRQOLQHHQYLURQPHQWIRU HIIHFWLYHWHDFKLQJDQGOHDUQLQJ :LOH\3/86 «PRWLYDWHVVWXGHQWVZLWK «VXSSRUWVLQVWUXFWRUVZLWK FRQÀGHQFHERRVWLQJ UHOLDEOHUHVRXUFHVWKDW IHHGEDFNDQGSURRIRI UHLQIRUFHFRXUVHJRDOV SURJUHVV LQVLGHDQGRXWVLGHRI WKHFODVVURRP ,QFOXGHV ,QWHUDFWLYH 7H[WERRN  5HVRXUFHV :LOH\3/86/HDUQ0RUH ZZZZLOH\SOXVFRP irwin_fm_i-xxii-hr.qxd 3-09-2010 15:50 Page ii $//7+(+(/35(6285&(6$1'3(5621$/6833257 s1. Since s2>s1 , the exponential associated with s2 will decay to zero faster than that asso- ciated with s1. This causes iL(t) to rise—the larger the value of s2 , the faster the rise. After 5A1兾s2 B seconds have elapsed, the exponential associated with s2 is approximately zero and iL(t) decreases exponentially with a time constant of ␶=1/s1. Thus, to slow the fall of iL(t) iL(t) 3.0 A 3.0 A 2.0 A 2.0 A 1.0 A 1.0 A 0.0 A 0.0 A –0.0 s 0.5 s 1.0 s 1.5 s 2.0 s 2.5 s 2.9 s –0.0 s 0.5 s 1.0 s 1.5 s 2.0 s 2.5 s 3.0 s 3.5 s 4.0 s Time Time (a) (b) Figure 7.51 Ignition current as a function of time. irwin07_296-368hr.qxd 28-07-2010 11:34 Page 349 SECTION 7.5 DESIGN EXAMPLES 349 iL(t), we should reduce s1. Hence, let us choose s1=1. Since s1+s2 must equal 20, s2=19. Under these conditions 1 1 C = = = 263 mF Ls1 s2 (0.2)(1)(19) and 60 60 K1 = = = 3.33 s2 - s1 18 Thus, the current is iL(t) = 3.33 C e-t - e-19t D A which is shown in Fig. 7.51b. At 100 ms the current is 2.52 A. Also, at t=1.1 s, the cur- rent is 1.11 A—above the 1-A requirement. Therefore, the choice of C=263 mF meets all starter specifications. A defibrillator is a device that is used to stop heart fibrillations—erratic uncoordinated DESIGN quivering of the heart muscle fibers—by delivering an electric shock to the heart. The Lown defibrillator was developed by Dr. Bernard Lown in 1962. Its key feature, shown in Fig. 7.52a, EXAMPLE 7.21 is its voltage waveform. A simplified circuit diagram that is capable of producing the Lown waveform is shown in Fig. 7.52b. Let us find the necessary values for the inductor and capacitor. vo(t) (V) 3000 L i(t) t=0 + VS R=50  vo(t) 6000 V patient C - 0 5 10 Time (milliseconds) (a) (b) Figure 7.52 Lown defibrillator waveform and simplified circuit. Reprinted with permission from John Wiley & Sons, Inc., Introduction to Biomedical Equipment Technology. Since the Lown waveform is oscillatory in nature, we know that the circuit is underdamped SOLUTION (␨ 6 1) and the voltage applied to the patient is of the form vo(t) = K1 e-␨␻o t sin[␻t] where R ␨␻o = 2L ␻ = ␻o 21 - ␨2 and 1 ␻o = 1LC irwin07_296-368hr.qxd 28-07-2010 11:34 Page 350 350 CHAPTER 7 F I R S T- A N D S E C O N D - O R D E R T R A N S I E N T C I R C U I T S for the series RLC circuit. From Fig. 7.52a, we see that the period of the sine function is T = 10 ms Thus, we have one expression involving ␻o and ␨ 2␲ ␻ = = ␻o 21 - ␨2 = 200␲ rad兾s T A second expression can be obtained by solving for the ratio of vo(t) at t = T兾4 to that at t = 3T兾4. At these two instants of time, the sine function is equal to +1 and -1, respectively. Using values from Fig. 7.52a, we can write vo(t兾4) K1 e-␨␻o(T兾4) 3000 = = e␨␻o(T兾2) L = 12 -vo(3T兾4) K1 e-␨␻o(3T兾4) 250 or ␨␻o = 497.0 Given R = 50 ⍀, the necessary inductor value is L = 50.3 mH Using our expression for ␻, ␻2 = (200␲)2 = ␻2o - (␨␻o)2 or 1 = (200␲)2 = - (497.0)2 LC Solving for the capacitor value, we find C = 31.0 ␮F Let us verify our design using the circuit shown in Fig. 7.53a. The output voltage plot shown in Fig. 7.53b matches the Lown waveform in Fig. 7.52a; thus, we can consider the design to be a success. While this solution is viable, it is not the only one. Like many design problems, there are often various ways to satisfy the design specifications. Figure 7.53 vo(t) Circuit and output 4.0 kV plot for the Lown L=50.3 mH defibrillator. 3.0 kV i(t) 2.0 kV + + vC(t)= R=50  vo(t) 1.0 kV C=31 F 5000 V - 0 kV - –1.0 kV 0s 2 ms 4 ms 6 ms 8 ms 10 ms Time (a) (b) irwin07_296-368hr.qxd 28-07-2010 11:34 Page 351 PROBLEMS 351 SUMMARY First-Order Circuits The response of a first-order transient circuit to an input pulse can be obtained by treating the pulse as a combina- An RC or RL transient circuit is said to be first order if it tion of two step-function inputs. contains only a single capacitor or single inductor. The voltage or current anywhere in the network can be obtained Second-Order Circuits by solving a first-order differential equation. The voltage or current in an RLC transient circuit can be The form of a first-order differential equation with a con- described by a constant coefficient differential equation stant forcing function is of the form d2x(t) dx(t) dx(t) x(t) + 2␨␻0 + ␻20 x(t) = f(t) + = A dt 2 dt dt ␶ where f(t) is the network forcing function. and the solution is The characteristic equation for a second-order circuit is x(t) = A␶ + K2 e-t兾␶ s2 + 2␨␻0 s + ␻20 = 0, where ␨ is the damping ratio and where A␶ is referred to as the steady-state solution and ␶ is ␻0 is the undamped natural frequency. called the time constant. If the two roots of the characteristic equation are The function e-t兾␶ decays to a value that is less than 1% of its initial value after a period of 5␶. Therefore, the time real and unequal, then ␨>1 and the network response is constant, ␶, determines the time required for the circuit to overdamped reach steady state. real and equal, then ␨=1 and the network response is critically damped The time constant for an RC circuit is RTh C and for an RL complex conjugates, then ␨0 in the network in Fig. P7.17. 5 k t=0 2 k 1 k t=0 i(t) + 4 k 4 k 5 mA 1 k 2 k 1 k 4 k 6 mA 4 k vo(t) 1 mH 100 F - Figure P7.13 Figure P7.17 irwin07_296-368hr.qxd 28-07-2010 11:34 Page 354 354 CHAPTER 7 F I R S T- A N D S E C O N D - O R D E R T R A N S I E N T C I R C U I T S 7.18 Find vo(t) for t>0 in the network in Fig. P7.18. 7.23 Find vo(t) for t>0 in the circuit in Fig. P7.23. t=0 3 k 6 k 6 k 2 k 8 k 3 k + 2 k 6 k 6 k 50 F 4 k 1 k vo(t) 12 mA t=0 + 100 F vo(t) 12 V - - Figure P7.18 Figure P7.23 7.19 Use the step-by-step method to find io(t) for t 7 0 in the 7.24 Find vo(t) for t>0 in the network in Fig. P7.24. circuit in Fig. P7.19. t=0 6V 6 + - 8 k + 6 k 12 k 4 k vo(t) 6 2H ± 24 V 50 F – io(t) - t=0 Figure P7.24 Figure P7.19 7.25 Use the step-by-step technique to find io(t) for t 7 0 in the network in Fig. P7.25. 7.20 Use the step-by-step method to find io(t) for t 7 0 in 12 V the circuit in Fig. P7.20. ±– 4 k 4 k t=0 4 k 4 k 200 F 200 F t=0 io(t) 12 k 6 k 12 mA 12 k Figure P7.25 io(t) Figure P7.20 7.26 Use the step-by-step method to find vo(t) for t 7 0 in the network in Fig. P7.26. 7.21 Find vo(t) for t>0 in the circuit in Fig. P7.21. t=0 6 k 80 F + + 12 mA 6 k 6 k vo(t) 100 F - t=0 4 mA 6 k 6 k vo(t) - Figure P7.26 Figure P7.21 7.27 Find io(t) for t>0 in the circuit in Fig. P7.27. 7.22 Find vo(t) for t>0 in the circuit in Fig. P7.22. t=0 io(t) t=0 2 k 1 k 9 k 2 k 4 k + 4 VA + 1 k + 6V 3 k 12 mA 60 F 6 k vo(t) - 150 F - - Figure P7.22 Figure P7.27 irwin07_296-368hr.qxd 28-07-2010 11:34 Page 355 PROBLEMS 355 7.28 Find io(t) for t>0 in the circuit in Fig. P7.28. 7.32 Use the step-by-step method to find vo(t) for t 7 0 in 6V the network in Fig. P7.32. –+ 1 —H 3 t=0 2 k t=0 4 + 2 k 2 k 100 F 2 io(t) ± 12 V 2 vo(t) Figure P7.28 – 7.29 Find io(t) for t>0 in the circuit in Fig. P7.29. – 12 V ± 4 k - Figure P7.32 4 mA 8 k 4 k 120 F 7.33 Find vc(t) for t 7 0 in the network in Fig. P7.33 using t=0 it (t) the step-by-step method. Figure P7.29 2 k 2 k t=0 7.30 Find io(t) for t>0 in the network in Fig. P7.30. 50 F + 2 k vC(t) 100 F - ± 12 V – 6 k 6 k io(t) + Figure P7.33 6V 6 k t=0 - 7.34 Find vo(t) for t 7 0 in the network in Fig. P7.34 using Figure P7.30 the step-by-step method. 7.31 Find io(t) for t>0 in the circuit in Fig. P7.31. 2 k t=0 150 F 50 F ± 24 V 2 k – io(t) + 2 k 4 k 4 k vo(t) 4 k ± 12 V 10 k 6 mA t=0 – - Figure P7.31 Figure P7.34 irwin07_296-368hr.qxd 28-07-2010 11:34 Page 356 356 CHAPTER 7 F I R S T- A N D S E C O N D - O R D E R T R A N S I E N T C I R C U I T S 7.35 Find vo(t) for t 7 0 in the circuit in Fig. P7.35 using the 7.38 Find vo(t) for t 7 0 in the circuit in Fig. P7.38 step-by-step method. using the step-by-step method. t=0 - 4 k 4 k vo(t) 3 6 + + 4 8 k 50 F - 12 V + 6A 6 vo(t) t=0 2H Figure P7.38 - 7.39 The switch in the circuit in Fig. P7.39 is opened at Figure P7.35 t=0. Find i(t) for t>0. 0.5 H t=0 4 7.36 Use the step-by-step method to find io(t) for t 7 0 in the circuit in Fig. P7.36. 4 k 4 k 200 F i(t) 3 6V + 2 24 V ± 4 k 2 k – – t=0 io(t) – + 9V Figure P7.36 Figure P7.39 7.40 The switch in the circuit in Fig. P7.40 is moved at t=0. Find io(t) for t>0. 7.37 Find vo(t) for t 7 0 in the network in Fig. P7.37 using the 18 V step-by-step technique. 1 k 100 F 4 k 4 k 200 F -+ + t=0 24 V ± 4 k 2 k vo(t) 3 k 4 k 4 k – t=0 io(t) 4 k - Figure P7.37 Figure P7.40 7.41 The switch in Fig. P7.41 closes at t=0. Find vo(t) for t>0. t=0 9 k 6 k + 6V + 3 k 120 F 4 k vo(t) - 4 mA - Figure P7.41 irwin07_296-368hr3.qxd 2-09-2010 12:13 Page 357 PROBLEMS 357 7.42 The switch in the circuit in Fig. P7.42 has been closed for 7.46 Find io(t) for t 7 0 in the network in Fig. P7.46 using a long time and is opened at t=0. Find i(t) for t>0. the step-by-step method. 1 k⍀ 1 k⍀ 12 V -+ 2 k⍀ 200 ␮F t=0 0.3 H + t=0 30 V i(t) 2 k⍀ - + 10 V 2 k⍀ 2 k⍀ 2 k⍀ 2 k⍀ - 4 k⍀ io(t) Figure P7.42 7.43 The switch in the circuit in Fig. P7.43 has been closed for Figure P7.46 a long time and is moved at t=0. Find io(t) for t>0. t=0 7.47 Use the step-by-step method to find io(t) for t 7 0 in the 5 k⍀ io(t) network in Fig. P7.47. - 10 V + 6 k⍀ 6 k⍀ 5 mA 5 k⍀ 2 k⍀ 2/3 H 6 k⍀ + 12 V 5 k⍀ io(t) - t=0 50 ␮F Figure P7.43 7.44 The switch in the circuit in Fig. P7.44 has been in posi- tion A for a long time and is moved to position B Figure P7.47 at t=0. Calculate vo(t) for t>0. t=0 7.48 Find vo(t) for t>0 in the network in Fig. P7.48. A 5 k⍀ 3 k⍀ B t=0 + 6V 8 k⍀ - + 60 ␮F 6 k⍀ + vo(t) 10 ␮F + 24 V - 1 k⍀ - 12 V + - 80 ␮F 4 k⍀ vo(t) - Figure P7.44 Figure P7.48 7.45 The switch in the circuit in Fig. P7.45 has been closed for a long time and is opened at t=0. Determine vC(t) for t>0. 7.49 Find vo(t) for t>0 in the network in Fig. P7.49. 5 k⍀ 5 k⍀ 3 k⍀ 9 k⍀ t=0 t=0 2 k⍀ - -+ 30 V + 10 k⍀ 8 k⍀ + + 4V + 18 V vC (t) 6 k⍀ 200 ␮F 4 k⍀ vo(t) - 20 ␮F - - Figure P7.45 Figure P7.49 irwin07_296-368hr.qxd 28-07-2010 11:34 Page 358 358 CHAPTER 7 F I R S T- A N D S E C O N D - O R D E R T R A N S I E N T C I R C U I T S 7.50 Find vo(t) for t>0 in the network in fig. P7.50. 7.54 Find vo(t) for t>0 in the network in Fig. P7.54. 12 V -+ 6 k t=0 12 k 4 k 2 k 3 k 3 k 6 k 3 k 200 F + + 12 V + 160 F 3 k vo(t) - 6V + 2 k 1 k vo(t) - - - Figure P7.50 t=0 7.51 Determine vo(t) for t>0 in the network in Fig. P7.51 Figure P7.54 7.55 Use the step-by-step technique to find vo(t) for t 7 0 in the circuit in Fig. P7.55. t=0 6 k + 6V + - 5 k 3 k 6 k 50 F 6 k vo(t) + 6 k - 12 V + 1 k vo(t) - 200 F t=0 - 6 k Figure P7.51 + 12 V - 7.52 Find vo(t) for t>0 in the circuit in Fig. P7.52. Figure P7.55 4 k 1 k 7.56 Use the step-by-step technique to find io(t) for t 7 0 in + the network in Fig. P7.56. 2 k 6V - 4 mA 3 k 100 F 4 k 100 F 2 k + 2 k 12 V + t=0 4 k vo(t) - ± 2 k 12 k 2 k 4V t=0 – - io(t) Figure P7.52 Figure P7.56 7.53 Find vo(t) for t>0 in the network in Fig. P7.53. 7.57 Find vo(t) for t 7 0 in the network in Fig. P7.57 using the step-by-step technique. + + 6 k 150 F 3 k vo(t) 2 2 2H - 3 vo(t) t=0 ± 6 12 V – 6 mA 12 k 4 k t=0 - Figure P7.53 Figure P7.57 irwin07_296-368hr2.qxd 2-08-2010 16:43 Page 359 PROBLEMS 359 7.58 Find io(t) for t>0 in the network in Fig. P7.58. 7.62 Find vo(t) for t>0 in the circuit in Fig. P7.62. t=0 1 k⍀ 4 k⍀ 150 ␮F 6V + 1 k⍀ 4 mA 3 k⍀ t=0 - 1 k⍀ + 6V 2 k⍀ - 1 k⍀ + io(t) 1 k⍀ 200 ␮F 1 k⍀ vo(t) Figure P7.58 - 7.59 Find vc(t) for t 7 0 in the circuit in Fig. P7.59 using Figure P7.62 the step-by-step method. 5⍀ 7.63 Find io(t) for t>0 in the network in Fig. P7.63. io(t) t=0 5⍀ 20 ⍀ 6⍀ 3⍀ t=0 0.5 F 12 V + 12 ⍀ - 6A – 0.2 H + 6V - vC(t) + 10 ⍀ 10 ⍀ Figure P7.63 5⍀ Figure P7.59 7.64 Find vo(t) for t>0 in the circuit in Fig. P7.64. 7.60 Find i(t) for t 7 0 in the circuit in Fig. P7.60 using 6⍀ the step-by-step method. 6⍀ 12 V 5⍀ t=0 t=0 2⍀ -+ 0.5 H + t=0 ± i(t) ± 6⍀ 0.1 H 6⍀ vo(t) –30 V 2⍀ 2⍀ 30 V – – - Figure P7.60 Figure P7.64 7.61 Find i(t) for t 7 0 in the circuit in Fig. P7.61 using the 7.65 Find io(t) for t>0 in the circuit in Fig. P7.65. step-by-step method. 2H 12 ⍀ 10 ⍀ 15 ⍀ t=0 6V 6⍀ 1H 105 V ± -+ – i(t) t=0 6⍀ 6⍀ 4⍀ 2⍀ io(t) Figure P7.61 Figure P7.65 irwin07_296-368hr.qxd 28-07-2010 11:34 Page 360 360 CHAPTER 7 F I R S T- A N D S E C O N D - O R D E R T R A N S I E N T C I R C U I T S 7.66 Find io(t) for t>0 in the network in Fig. P7.66. 3 2 1H 12 V + 6 t=0 4A 4 – io(t) Figure P7.66 7.67 Find io(t) for t>0 in the circuit in Fig. P7.67. 2H 6 t=0 6V + 12  4 2 2A – io(t) Figure P7.67 7.68 Find vo(t) for t>0 in the network in Fig. P7.68. 1H 2 + t=0 3A 6 vo(t) 3 2 + 24 A – - Figure P7.68 7.69 Find io(t) for t>0 in the circuit in Fig.P7.69. 7.71 Use the step-by-step method to find vo(t) for t 7 0 in io(t) the network in Fig. P7.71. 2 6 k 2 k + t=0 6 F 6 6 6 4 k vo(t) 12 V ± 6 F 6 F – 12 V + 1H 3 t=0 - – Figure P7.71 Figure P7.69 7.70 Use the step-by-step method to find io(t) for t 7 0 in the network in Fig. P7.70. 7.72 Find iL(t) for t 7 0 in the circuit in Fig. P7.72 using the step-by-step method. 4 t=0 iL(t) t=0 12  6 12  + 4 io(t) 10 V ± 4A v1(t) 2 2H 12 V ± – – 3H - ±– 3A 2H 5v1(t) Figure P7.70 Figure P7.72 irwin07_296-368hr.qxd 28-07-2010 11:34 Page 361 PROBLEMS 361 7.73 Find io(t) for t 7 0 in the network in Fig. P7.73 using 7.76 The current source in the network in Fig. P7.76a is the step-by-step method. defined in Fig. P7.76b. The initial voltage across the capacitor must be zero. (Why?) Determine the current t=0 io(t) for t 7 0. 2 k 2 k iA io(t) 2 2F 10 mA 2 k 2 mH 2 k 2 iA i(t) 2 2 io(t) Figure P7.73 (a) i(t) (A) 7.74 Use the step-by-step technique to find vo(t) for t 7 0 in 6 the network in Fig. P7.74. 24 V 200 F 3 k 3 k 2 k 0 4.5 ±– t (s) + (b) – ± ± Figure P7.76 6V t=0 vo(t) 6 k 2000 iA – - iA 7.77 The voltage v(t) shown in Fig. P7.77a is given by the graph shown in Fig. P7.77b. If iL(0) = 0, answer the fol- Figure P7.74 lowing questions: (a) How much energy is stored in the inductor at t = 3 s? (b) How much power is supplied by the source at t = 4 s? (c) What is i(t = 6 s)? (d) How much power is absorbed 7.75 Determine the equation for the voltage vo(t) for t 7 0 in by the inductor at t = 3 s? Fig. P7.75a when subjected to the input pulse shown in Fig. P7.75b. i(t) 3 k 2 k + v(t) ± – 2H 2 v(t) ± 6 k 200 F vo(t) – iL(t) - (a) (a) v(t) v(t) (V) 10 V 12 5 2 t(s) 0 1 –10 V t (s) (b) (b) Figure P7.75 Figure P7.77 irwin07_296-368hr.qxd 28-07-2010 11:34 Page 362 362 CHAPTER 7 F I R S T- A N D S E C O N D - O R D E R T R A N S I E N T C I R C U I T S 7.78 Find the output voltage vo(t) in the network 7.82 The switch in the circuit in Fig. P7.82 is closed at t = 0. If in Fig. P7.78 if the input voltage is i1(0-) = 2 A, determine i2(0+), vR(0+), and i1(t = q). vi(t) = 5(u(t) - u(t - 0.05))V. t=0 i1(t) + i2(t) 1 F + 2H vR(t) 4 3H vi(t) ± 100 k vo(t) – - - Figure P7.82 Figure P7.78 7.83 The switch in the circuit in Fig. P7.83 has been closed for a long time and is opened at t = 0. If vC(t) = 20 - 8e-0.05t V, find R1, R2, and C. 7.79 Given that vC1(0-) = -10 V and vC2(0-) = 20 V in the circuit in Fig. P7.79, find i(0+). + t=0 R2 4 + i(t) + vC(t) 2A R1 C vC1(t) 2F 4F vC2(t) t=0 - - - Figure P7.79 Figure P7.83 7.84 Given that i(t) = 2.5 + 1.5e-4t A for t 7 0 in the 7.80 In the network in Fig. P7.80, find i(t) for t 7 0. If circuit in Fig. P7.84, find R1, R2, and L. vC1(0-) = -10 V, calculate vC2(0-). t=0 vC1(t) i(t) i(t) + - R1 R2 + 0.3 F ± 20 V L – ± 10e–5tu(t) V 0.6 F vC2(t) – - Figure P7.84 Figure P7.80 7.85 Given that i(t) = 13.33e-t - 8.33e-0.5tA for t 7 0 in the network in Fig. P7.85, find the following: (a) vc(0), 7.81 In the circuit in Fig. P7.81, vR(t) = 100e-400t V for (b) vc(t = 1 s), t 6 0. Find vR(t) for t 7 0. (c) the capacitance C. t=0 v (t) i(t) + C - + 4 C 25  100  vR(t) L ± 40e–tu(t) V 2 – - Figure P7.81 Figure P7.85 irwin07_296-368hr.qxd 28-07-2010 11:34 Page 363 PROBLEMS 363 7.86 The differential equation that describes the current io(t) 7.92 For the underdamped circuit shown in Fig. P7.92, in a network is determine the voltage v(t) if the initial conditions on the 2 d io(t) dio(t) storage elements are iL(0) = 1 A and vC(0) = 10 V. 2 + 6 + 4io(t) = 0 dt dt iL(0) + + Find (a) the characteristic equation of the network, 2H v(t) 5 vC (0) 1 —F (b) the network’s natural frequencies 40 - - (c) the expression for io(t). Figure P7.92 7.87 The voltage v1(t) in a network is defined by the equation d 2v1(t) dv1(t) + 4 + 5v1(t) = 0 dt 2 dt 7.93 Find vC(t) for t 7 0 in the circuit in Fig. P7.93. Find t=0 (a) the characteristic equation of the network + iL(t) (b) the circuit’s natural frequencies (c) the expression for v1(t). 1A 8 0.04 F vC(t) 1H - 7.88 The voltage v1(t) in a network is defined by the equation Figure P7.93 d2v1(t) dv1(t) 2 + 2c d + 5v1(t) = 0 dt dt 7.94 Find vC(t) for t 7 0 in the circuit in Fig. P7.94 if Find vC(0) = 0. (a) the characteristic equation of the network. (b) the circuit’s natural frequencies. t=0 1 k 100 mH (c) the expression for v1(t). + ± 12 V 1 F vC(t) – 7.89 A parallel RLC circuit contains a resistor R = 1 ⍀ and an inductor L = 2H. Select the value of the capacitor so - that the circuit is critically damped. Figure P7.94 7.90 A series RLC circuit contains a resistor of R = 2 ⍀ and a capacitor C = 1兾2 F. Select the value of the inductor 7.95 Find vo(t) for t 7 0 in the circuit in Fig. P7.95 and plot so that the circuit is critically damped. the response, including the time interval just prior to closing the switch. 2 t=0 7.91 In the critically damped circuit shown in Fig. P7.91, the initial conditions on the storage elements are iL(0) = 2 A and vC(0) = 5 V. Determine the voltage v(t). + + + iL(0) 12 V ± 1H 10  vo(t) – 1 —F 400 vC (0) 0.01 F v(t) 10  4H - - - Figure P7.91 Figure P7.95 irwin07_296-368hr.qxd 28-07-2010 11:34 Page 364 364 CHAPTER 7 F I R S T- A N D S E C O N D - O R D E R T R A N S I E N T C I R C U I T S 7.96 The switch in the circuit in Fig. P7.96 has been closed for 7.97 The switch in the circuit in Fig P7.97 has been closed for a a long time and is opened at t = 0. Find i(t) for t 7 0. long time and is opened at t = 0. Find i(t) for t 7 0. t=0 6 1.5  5 5 t=0 5 i(t) 12 V ± 0.1 F ± 20 V 0.04 F – – 1H – ± 6V i(t) – ± 10 V 1.25 H Figure P7.96 Figure P7.97 7.98 Find vo(t) for t 7 0 in the circuit in Fig. P7.98 and plot the response, including the time interval just prior to closing the switch. 1 1 —H t=0 —H 8 8 8 1 —F + 25 1.5 A 4 6 1 vo(t) - Figure P7.98 7.99 In the circuit shown in Fig. P7.99, find v(t) 7 0. 7.101 Find vo(t) for t 7 0 in the circuit in Fig. P7.101 and plot the response, including the time interval just prior 12 to moving the switch. —5 H t=0 t=0 2.5 mH 8 k + ± 8V 2 —1 F v(t) – 12 1 k 2 — nF 15 - 6 k 4 k Figure P7.99 + 3 mA – ± 6 k vo(t) 12 V - 7.100 Find vo(t) for t 7 0 in the circuit in Fig. P7.100 and Figure P7.101 plot the response, including the time interval just prior to moving the switch. 7.102 Find vo(t) for t 7 0 in the network in Fig. P7.102 and plot the response, including the time interval just prior t=0 1 mH to moving the switch. 10 k 5 k t=0 2.5 mA 4 k 1 k 0.05 nF 100 V ± 10 k + 2 k – vo(t) 10 F - Figure P7.100 100 mH Figure P7.102 irwin07_296-368hr.qxd 28-07-2010 11:34 Page 365 PROBLEMS 365 7.103 Find vo(t) for t 7 0 in the circuit in Fig. P7.103 and 7.104 Design a parallel RLC circuit with R ⭌ 1 k⍀ that has plot the response, including the time interval just prior the characteristic equation to moving the switch. s 2 + 4 * 107s + 3 * 1014 = 0 2 k 1 mH t=0 ± 12 V ± 1 k vo(t) – 6.25 nF 7.105 Design a parallel RLC circuit with R ⭌ 1 k⍀ that has – the characteristic equation s2 + 4 * 107s + 4 * 1014 = 0 Figure P7.103 Solve the remaining problems using computational methods. 7.106 Given the network in Fig. P7.106 to plot vo(t) over a 10-s interval starting at t = 0, using a 100-ms step size. 2 4 t=0 2H 12 V ± 6 2A 1F – + 2 vo(t) - Figure P7.106 7.107 Given the network in Fig. P7.107, plot vo(t) over a 7.108 Given the network in Fig. P7.108a and the input volt- 10-s interval starting at t = 0, using a 100-ms step size. age shown in Fig. P7.108b, plot the voltage vo(t) over the interval 0 ⱕ t ⱕ 4 s, using a 20-ms step –± size. t=0 12 V 20 k 10 k 100 k 1M + vin(t) ± 2 mH 0.1 F vo(t) – 6V ± 50 F – + - 10 mH 5 k vo(t) (a) - vin(t) V Figure P7.107 10 0 1 t(s) (b) Figure P7.108 irwin07_296-368hr.qxd 28-07-2010 11:34 Page 366 366 CHAPTER 7 F I R S T- A N D S E C O N D - O R D E R T R A N S I E N T C I R C U I T S 7.109 Given the network in Fig. P7.109a, and the input voltage shown in Fig. P7.109b, plot vo(t) over the interval 0 ⱕ t ⱕ 10 s, using a 100-ms step size. –± vin(t) 2 2 2 + 1F 2A 1H 1 4 vo(t) - (a) vin(t) V 10 0 1 t (s) (b) Figure P7.109 7.110 Given the network in Fig. P7.110a and the input in Fig. P 7.110b, plot vo(t) in the inter- val 0 ⱕ t ⱕ 4 s, using a 20-ms step size. 10 k 5 k 100 F + vs(t) ± 10 mH 10 k vo(t) – - (a) vs(t) V 10 0 1 2 3 4 t (s) (b) Figure P7.110 irwin07_296-368hr.qxd 28-07-2010 11:34 Page 367 TYPICAL PROBLEMS FOUND ON THE FE EXAM 367 TYPICAL PROBLEMS FOUND ON THE FE EXAM 7PFE-1 In the circuit in Fig. 7PFE-1, the switch, which has been closed for a long time, opens at t = 0. Find the value of the capacitor voltage vc(t) at t = 2s. a. 0.936 V b. 0.756 V c. 0.264 V d. 0.462 V t=0 8 k 6 k + 12 V ± 6 k vC(t) 100 F 6 k – - Figure 7PFE-1 7PFE-2 In the network in Fig. 7PFE-2, the switch closes at t = 0. Find v0(t) at t = 1s. a. 5.62 V b. 1.57 V c. 4.25 V d. 3.79 V 12 k 4 k t=0 + ± 12 V 12 k 100 F vo(t) – - Figure 7PFE-2 7PFE-3 Assume that the switch in the network Fig. 7PFE-3 has been closed for some time. At t = 0 the switch opens. Determine the time required for the capacitor voltage to decay to one-half of its initially charged value. a. 0.416 s b. 0.625 s c. 0.235 s d. 0.143 s 12 k + t=0 12 V ± vC(t) 100 F 6 k – - Figure 7PFE-3 irwin07_296-368hr.qxd 28-07-2010 11:34 Page 368 368 CHAPTER 7 F I R S T- A N D S E C O N D - O R D E R T R A N S I E N T C I R C U I T S 7PFE-4 Find the inductor current iL(t) for t 7 0 in the circuit in Fig. 7PFE-4. a. iL(t) = 3 - 2e-t兾6 A, t 7 0 b. iL(t) = 1 + 2e-2t兾3 A, t 7 0 c. iL(t) = 6 - e-t兾6 A, t 7 0 d. iL(t) = 3 - e-2t兾3 A, t 7 0 t=0 2 iL(t) ± 10 V 2 4H 2 1A – Figure 7PFE-4 7PFE-5 Find the inductor current iL(t) for t 7 0 in the circuit in Fig. 7PFE-5. a. iL(t) = 1.4 + 0.4e-4t兾3 A, t 7 0 b. iL(t) = 1.2 + 0.4e-5t兾3 A, t 7 0 c. iL(t) = 0.4 + 0.2e-4t兾3 A, t 7 0 d. iL(t) = 2.4 + 0.6e-5t兾3 A, t 7 0 t=0 iL(t) 3H 3 2 4 4 12 V ± – Figure 7PFE-5 irwin08_369-434hr.qxd 28-07-2010 12:03 Page 369 C H A PT E R AC STEADY-STATE 8 ANALYSIS THE LEARNING GOALS FOR THIS CHAPTER ARE: Understand the basic characteristics of sinusoidal functions Be able to perform phasor and inverse phasor transformations and draw phasor diagrams Know how to calculate impedance and admittance for our basic circuit elements: R, L, C Be able to combine impedances and admittances in series and parallel Be able to draw the frequency-domain circuit for a given circuit with a sinusoidal source Know how to apply our circuit analysis techniques to frequency-domain circuits TebNad/iStockphoto S Smart Power Grids What makes a power grid smart? It takes a tricity continues to increase, a smarter grid that provides electricity two-way communication between the electric power generator and more efficiently also helps in the fight against global warming. the consumer. For providers, a smart grid gives rapid information This chapter introduces ac signals—alternating voltages and about power quality and alerts them when blackouts occur. For currents—which is the form of electricity that the power grid deliv- consumers, it promotes the efficient use of electricity by letting ers to the end user. Your background in differential equations, them know the time of day when power is more available and less needed in the last chapter, will again become highly important in expensive. The aim of the smart power grid is to interconnect the finding steady-state voltages and currents for linear circuits having nation within this decade in an efficient power distribution system power sources described by cosine waveforms. A compact nota- that includes conventional and large-scale renewable power tion called phasors will be introduced, and you will perform calcu- sources. Real-time pricing is expected to become commonplace. lations involving complex numbers. Remarkably, dc circuit analysis Smart power grids allocate available power to meet higher pri- techniques that you studied earlier can now be applied directly orities. Wireless sensors, software, and microprocessors are key using phasors for ac circuit analysis. Ac circuits can be analyzed components in this intelligent digital network. As demand for elec- using every basic method—nodal, mesh, superposition, and others. 369 irwin08_369-434hr.qxd 28-07-2010 12:03 Page 370 370 CHAPTER 8 A C S T E A D Y- S TAT E A N A LY S I S 8.1 Let us begin our discussion of sinusoidal functions by considering the sine wave x(t) = XM sin t 8.1 Sinusoids where x(t) could represent either v(t) or i(t). XM is the amplitude, maximum value, or peak value;  is the radian or angular frequency; and t is the argument of the sine function. A plot of the function in Eq. (8.1) as a function of its argument is shown in Fig. 8.1a. Obviously, the function repeats itself every 2 radians. This condition is described mathematically as x(t + 2) = x(t), or in general for period T, as xC (t + T)D = x(t) 8.2 meaning that the function has the same value at time t+T as it does at time t. Figure 8.1 x(t) x(t) Plots of a sine wave as a XM XM function of both t and t. 3