Analisis de Circuitos en Ingenieria Solucionario, 7° ED. - William H. Hayt Jr., Jack E. Kemmerly,...

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Transcript of Analisis de Circuitos en Ingenieria Solucionario, 7° ED. - William H. Hayt Jr., Jack E. Kemmerly,...

CHAPTER 2 ENGINEERING CIRCUIT ANALYSIS 1. (a) 12 s (b) 750 mJ (c) 1.13 k (d) 3.5 Gbits (e) 6.5 nm (f) 13.56 MHz (g) 39 pA (h) 49 k (i) 11.73 pA

SELECTED ANSWERS

3. 5. 7. 9. 11. 13. 15. 17. 19. 21. 23. 25. 27. 29. 31. 33. 35. 37. 39. 41. 43. 45.

300 kW; 3.7 m; 25 mm; 71 kJ; 290 fs 131 kW; 1.4 GJ; 1 battery 13 GW; 100 mW 290 kJ; 1.5 kJ 6.2 A; 3.5 A; The current is never negative; 34 C 12 MV; 0; -18.7 MV; -6.2 MV -6.4 mW; -120 W; 60 W; 12 W 73 W; -36 W; 28 W 5 mW, 0, -2 mW; 36 J; 22 J 64 W, 256 W, -640 W, 800 W, -480 W -1 mV 58 W; 4.8 A 5.6 mA, 4.5 mA; 23 mW, 28 mW 43.5 mW; 231 mW; 253 mW Since we know that the total power supplied is equal to the total power absorbed, we may write: Vs I = I2R1 + I2R2. Now invoke Ohms law. 500 A, 2.5 mW; -500 A, 2.5 mW; -500 A, 2.5 mW; 500 A, 2.5 mW -2 V (at t = 0.324 s) 2 km. Hmmmm. 1.7 .cm 560 m, 1.3 W 266 m; 514 mA Design. Many possible solutions. Hint: Start with finding resistivity, then choose geometry.

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CHAPTER 3 ENGINEERING CIRCUIT ANALYSIS 1. 3. 5. 7. 9. 11. Circuit diagram not shown. (a) 4 nodes; (b) 5 branches; (c) yes, path; no, loop. (a) 4; (b) 5; (c) yes,no,yes,no,no (a) 3 A; (b) -3 A; (c) 0 ix = 1 A; iy = 5 A.

SELECTED ANSWERS

If the DMM appears as a short, then all 5 A flows through the DMM, and none through the resistors, resulting in a (false) reading of 0 V for the circuit undergoing testing. (a) 12 V; (b) -2.2 V R = 34 ; G = 90 mS Circuit I: i = 0; Circuit II: i = 1.1 A -23.5 V (a) v1 = 60 V v2 = 60 V v3 = 15 V v4 = 45 V v5 = 45 V i1 = 27 A i2 = 3 A i3 = 24 A i4 = 15 A i5 = 9 A (b) = -1.62 kW = 180 W = 360 W = 675 W = 405 W

13. 15. 17. 19. 21.

23. 25. 27. 29. 31. 33.

(a) 8 V, -4 V, -12 V; (b) 14 V, 2 V, -6 V; (c) 2 V, -10 V, -18 V (a) 25 W; (b) 24 W; (c) 16 W; (d) 18.4 W; (e) -600 W None of the conditions specified in (a) to (d) can be met by this circuit. 5.0 A; 10.4 V (a) 2.4 k; (b) R = 0 -250 cos 5t mV

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CHAPTER 3 ENGINEERING CIRCUIT ANALYSIS 35. (a) P5A P100 P25 Pdep (b) P5A P100 P25 Pdep 37. P8A P6 P8A P12 P4 39. 41. 43. 45. 47. 49. 51. 53. 55. = 8 vx = (vx)2 / 6 = 7 vx = (vx) / 12 = (vx) / 42 2

SELECTED ANSWERS

= 5 vx = (vx) / 100 = (vx) / 25 = vx(0.8 ix) = 0.8 (vx)2 / 252 2

= 1.389 kW = = 771.7 W 3.087 kW

= 2.470 kW

= 5 vx = (vx) / 100 = (vx) / 25 = vx(0.8 iy)2 2

= 776.0 W = = 240.9 W 963.5 W

= 428.1 W

= 240 W = = = 150 W 75 W 225 W = 210 W

(a) 50 mA; (b) Can set vS = 50 V. 638 mW 1.45E-3 miles (a) 1 A; (b) 9 A (a) 10 mA; (b) 3.8 A (a) 570 mA; (b) 0; (c) 71 mA -515 V Req = 1 k (a) 10 k || 10 k; (b) 47 k + 10 k + 1 k || 1k || 1k; (c) 47 k || 47 k + 10 k || 10 k + 1 k 5.5 k 60 ; 213 ; 52

57. 59.

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CHAPTER 3 ENGINEERING CIRCUIT ANALYSIS 61. 63. 65. 67. 69. 71. 73. 75. 77. 250 W; 188 W; 338 W; 180 W; 45 W (a) 850 mS; (b) 136 mS Proof 607 mV 22 A One possible solution: 11 mA, 1 k, 1 k 139 A; 868 W 18 WR 2 (R 3 + R 4 ) ; R 1 (R 2 + R 3 + R 4 ) + R 2 (R 3 + R 4 ) R 1 (R 2 + R 3 + R 4 ) (b) VS ; R 1 (R 2 + R 3 + R 4 ) + R 2 (R 3 + R 4 ) R2 (c) VS . R 1 (R 2 + R 3 + R 4 ) + R 2 (R 3 + R 4 )

SELECTED ANSWERS

(a) VS

79. 81.

(a) 42 A; (b) 11.9 V; (c) 0.238

R3 R5 VS R (R + R + R ) + R (R + R ) 4 5 3 4 5 2 3vout = -56.02 sin 10t V

83.

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CHAPTER 4 ENGINEERING CIRCUIT ANALYSIS 1. 3. (a) -8.4 V; (b) 32 (a) v1 = 264 V, v2 = 184 V and v3 = 397 V; (b) >> e1 = '4 = v1/100 + (v1 - v2)/20 + (v1 - vx)/50'; >> e2 = '10 - 4 - (-2) = (vx - v1)/50 + (vx - v2)/40'; >> e3 = '-2 = v2/25 + (v2 - vx)/40 + (v2 - v1)/20'; >> a = solve(e1,e2,e3,'v1','v2','vx'); >> a.v1 -1.74 V 172 V (a) 58.5 V, 64.4 V; (b) 543 W -28 V -8.1 V v1 = 3.4 V v2 = 7.1 V v3 = 7.5 V v4 = 4.9 V 17. 19. 21. 23. 25. 27. 29. 31. 33. 35. (a) 26 V, (b) 83 mW -3.25 -91 V 45 W v1 = -8.6 V, v2 = -3.9 V and v3 = 6.1 V (a) 143 mA; (b) 16 W (a) 3.1 A; (b) 370 W 2.79 A -380 W v5 = 1.7 V v6 = 3.8 V v7 = 3.5 V v8 = 2.4 V

SELECTED ANSWERS

5. 7. 9. 11. 13. 15.

i1 = 239 A, i2 = 1.08 mA, i3 = -1.20 mA and i4 = -480 A

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CHAPTER 4 ENGINEERING CIRCUIT ANALYSIS 37. 39. 41. (a) -5700 ; (b) this value is unique. (a) 330 A; (b) 330 A; (c) units of resistance. P2mA P4V P6V PdepV PdepI = 5000(i1 i2)(i1) = 4 (-i2) = 6 (-i3) = 1000 i3 (i3 i2) = 10,000(i3 i4)(0.5 i2) = 5 mW = -6 mW = 9 mW = 4.5 mW = -5.6 mW

SELECTED ANSWERS

43. 45. 47. 49. 51. 53. 55. 57. 59. 61. 63. 65. 67.

-3.65 W -1.03 V 5 (a) 0; (b) 96 V; (c) -38 V 3.55 A; 1.69 A 121 mA; 4.70 A Hint: i3 = 1.24 A and i4 = 1.42 A by mesh analysis. 350 mA i1 = 2.65 A, i2 = 3.20 A, i3 = -3.80 A, i4 = -1 mA -4 mA -16 V 3.14 V, 1.71 V, 714 mA, -143 mA, -2.14 A, 857 mA One possible solution:

where R = 5/3 = 1 + 2/3 = 1 + 1 || 1 || 1 + 1 || 1 || 1. 69. One possible solution: 9 V in series with 5 1- resistors (R1) and 5 1- resistors (R2 R5). Take V1 across R2-R5, V2 across R3-R5, and V3 across R4-R5.

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CHAPTER 5 ENGINEERING CIRCUIT ANALYSIS 1. 3. 5. 7. 9. 11. 13. 15. 17. 19. 21. 23. 25. 27. 29. 31.

SELECTED ANSWERS

Define percent error as 100 [ex (1 + x)]/ ex. If we choose x < 0.1, we ensure that the error is less than 1%. 4.7 V, 2.0 A 4 V 40 V and 10 V 100 V. 10.8 V (a) 1.3 A; (b) 60 W, 18 W, -130 W, 32 W, 20 W (a) 200 V; (b) -143 V 957 W Impossible; 76 mW (a) 18 V 2.46 V; 0.546 V, 1.91 V. (a) 42 V voltage source in series with 6 and in series with 10 ; (b) 26 V; (c) Cannot remove the resistor across which v appears or v may become lost. 10 mW 33 W (a) 12.8 mV 764 nA Current source is 7.25 A, resistor is 2 ohms.

33. 35. 37.

1.57 V, 811 m The final circuit is an 8.5 V voltage source in series with a 2.0 M resistor. (a) An 8/5 A current source in parallel with 5 , in parallel with RL.

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CHAPTER 5 ENGINEERING CIRCUIT ANALYSIS 39. 41. 43. -2 V

SELECTED ANSWERS

(a) The Thvenin equivalent is a 9.3 V source in series with a 17 resistor, which is in series with the 5 resistor of interest; (b) 928 mW. (a) 25 ; (b) 303 ; (c) Increased current leads to increased filament temperature, which results in a higher resistance (as measured). This means the Thvenin equivalent must apply to the specific current of a particular circuit one model is not suitable for all operating conditions. (a) 6.7 , -300 mA, arrow upwards; (b) 6.7 , -150 mA, arrow upwards. (a) 38.9 V, 178 ; (b) 1.96 W. VTH = 0, RTH = 192 . 15 , 15 VTH = 0; The Norton equivalent is 0 A in parallel with 1.3 . VTH (and hence IN) = 0; RTH = RN = 198 m. 2 M VTH =vin Ri ( Ro AR f ) R1 Ro + Ri Ro + R1 R f + Ri R f + R1 Ri + AR1 Ri

45. 47. 49. 51. 53. 55. 57. 59.

; RTH =

Ro (Ri Rf + R1 Rf + R1 Ri) -------------------------------------------------------------Ri Ro + R1 Ro + Ri Rf + R1 Rf + R1 Ri + A R1 Ri. 61. 63. 65. 67. 16 , 6.3 W 65 V, 15 , 70 W (a) 200 V; (b) 125 W; (c) 80 There is no conflict with our derivation concerning maximum power. While a dead short across the battery terminals will indeed result in maximum current draw from the battery, and power is indeed proportional to i2, the power delivered to the load is i2RLOAD = i2(0) = 0 watts. This is the minimum, not the maximum, power that the battery can deliver to a load. Select R1 = RTH = 8 k

69.

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CHAPTER 5 ENGINEERING CIRCUIT ANALYSIS 71. 73. 75. 77. 79. 81. 83. 85. 1.2 , 0.54 , 4.9 9.9 5.5 V, 1.0 -13 V, 27

SELECTED ANSWERS

Although the network may be simplified, it is not possible to replace it with a three-resistor equivalent. IS(max) = 224 mA 1.4 One possible solution of many:

87.

One possible current-limiting scheme is to connect a 9-V battery in series with a resistor Rlimiting and in series with the LED; Rlimiting = 220 .

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CHAPTER 6 ENGINEERING CIRCUIT ANALYSIS 1. 3. 5. 7. (a) -30 V; (b) -2.5 V; (c) 1.4 V

SELECTED ANSWERS

(a) vout = 10vin = 20 sin 5t ; (b) vout = 10vin = 10 5 sin 5t One possible design is to use a simple inverting op amp circuit with Rf = 9.1 k and Rin = 5.1 k. To get a positive output that is smaller than the input, the easiest way is to use inverting amplifier with an inverted voltage supply to give a negative voltage, where Rf = 1.5 k and Rin = 5.1 k (a) 1.7 V; (b) 3 V; (c) -2.4 V (a) vout = 2vin = 8 sin 10t ; (b) vout = 2vin = 2 + 0.5 sin 10t -2.2 V One possible solution of many: a non-inverting op amp circuit with the microphone connected to the non-inverting input terminal, the switch connected between the op amp output pin and ground, a feedback resistor Rf = 133 , and a resistor R1 = 1 . V1 = 21 V

9. 11. 13. 15.

17. 19. 21. 23. 25. 27. 29. 31.

vout

= -4 (1 + sin 3t ) V ; -5.6 V

Rf = 236 k and R1 = 1 k. (a) B must be the non-inverting input; (b) Choose R2 = RB = 1 ; (c) A is the inverting input. vout(0.25 s) = 0.93 V 4.2 V

- Rf

Ri =1

N

vii

Pick R1 = 10 k. Then vS = -0.21 V.

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CHAPTER 6 ENGINEERING CIRCUIT ANALYSIS 33. One possible solution of many:

SELECTED ANSWERS

35.

Set R = 10 k:

Then connect several into:

after setting Rf2 = Rf1 = Rin = R =10 k. 37. 39. 41. 43. 1 kV -179 kV 1.7 V Rf = 0, Rin = 100 k, R2 = 51 .

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CHAPTER 6 ENGINEERING CIRCUIT ANALYSIS 45. Rf = 120 k and Rin = 200 k, R = 560 .

SELECTED ANSWERS

47.

R = 400 , R1 = 82 .

I

Is

49.

R = 91 , R1 = 560 , 467 > RL > 67 .

51. 53. 55.

(a) 3.7 mV; (b) 28 mV;vout - 100A ; A = 9999. = vin 101 + A

(c) 3.7 V.

vout = -16 mV

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CHAPTER 6 ENGINEERING CIRCUIT ANALYSIS 57. (a)

SELECTED ANSWERS

(b) vout = 105(-0.00004v2 - 9.9998010-6v1)+5v2 = 1.00008v2 - 0.99998v1 = 0.0005 1.99996 sin t (c) vout = 105 vd = 105 (v2 / 2 va ) =0.99998v2-0.99998v1 = 1.99996 sin t 59. 61. 63. 65. (a) V3 = 27 V; Positive voltage supply, negative voltage supply, inverting input, ground, output pin. This is a non-inverting op amp circuit, so we expect a gain of 214. For vx = -10 mV, PSpice predicts vd = 6 V, where the hand calculations based on the detailed model predict 50 V, which is about one order of magnitude larger. For the same input voltage, PSpice predicts an input current of -1 A, whereas the hand calculations predict 99.5vx mA = -995 nA (which is reasonably close). (a) Negative saturation begins at Vin = 4.72 V, and positive saturation begins at Vin = +4.67 V. (b) 40.6 mA.

67. 69.

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CHAPTER 6 ENGINEERING CIRCUIT ANALYSIS 71. (a)15

SELECTED ANSWERS

12 V10 5 V out (V) 0 -2 -5 -10 -15 V active (V) -1 0 1 2

-12 V

73.

75.

R2 R3 ; (b) Vout = 0; (c) R = 4.3 k and R (a) Vout = V1 V2 = Vref R +R R +R 1 2 3 Gauge = 4.7 k, gain of 5.39 for R = 4.7 k, so R = 11.5 k.

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CHAPTER 7 ENGINEERING CIRCUIT ANALYSIS 1. 3. 5. 7. (a) 0; (b) -613sin120t mA; (c) -40e-t nA

SELECTED ANSWERS

(a) 30 (1 t ) e t mA ; (b) 4e5t (100 cos100t 5sin100t ) mA (a) 6.95 pF; (b) 17 kV; (c) 72 Design problem: more than one solution. Hint:

9. 11. 13. 15. 17. 19.

(a) 33.4 mV; (b) 33.4 mV; (c) 50.1 mV (a) 120 sin 400t A ; (b) 6.4 J; (c) 400(1 e100t )V ; (d) vc = 500 400e 100t V (a) 2 k; (b) 20 mJ (a) 0; (b) -613sin120t V; (c) 240e 6t pV (a) 150 (1 t ) et fV ; (b) 100e5t ( 20 cos100t sin100t ) pV (a)

(b) 40 ms; (c) t = 20, 40 ms; (d) 2.5 J 21. 23. 25. 27. (a) 4t 2 + 4t V ; (b) 4t 2 + 4t + 5 A (a) 2 A; (b) 5.6 J; (c) 1 A (a) 2.33 V; (b) 480 mA: (c) 1.1 A (a) 6.4 J; (b) 100 mJ; (c) Left to right (magnitudes): 100, 0, 100, 116, 16, 16, 0 (V); (d) Left to right (magnitudes): 0, 0, 2, 2, 0.4, 1.6, 0 (A) (a) 0, 400 mW 4.3 F

29. 31.

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CHAPTER 7 ENGINEERING CIRCUIT ANALYSIS 33. (a)

SELECTED ANSWERS

(b) 3.6 V

35. 37. 39. 41. 43. 45.

Cequiv = 85 nF 140 nF (a) 3 H; (b) N 292 pH (a) 11.4 ; (b) 11.4 H; (c) 8.8 F (a)

(b)

(c)

47. 49.

(a) -6.4e-80t mA; (b) 80e 80t 60V ; (c) 20e 80 t + 60V 9.2 V, 2.4sin103t V

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CHAPTER 7 ENGINEERING CIRCUIT ANALYSIS (a) vs =1 1+ A idt + vi ; (b) vo + RC vo + Avs = 0 c

SELECTED ANSWERS

51. 53. 55.

(a) vs = 10.0sin10t + 0.0005 0.0005cos10t ; (b) 10sin10t V L 0 inductor values. (a) Vout = Rft

vs dt ' ; (b) Capacitor values are more readily available than

57.

One possible solution of many (with C = 1 mF, R = 600 k):

59.

One possible solution of many (with C = 1 F, R = 1 M):

61.

(a)

(b)20v20 + 1 5 106 1 5 106

(vo

t

20

vc )dt + 12 = is

(vo

t

c

v20 )dt 12 + 10vc + 8 103 vc = 0

(c)

iL is i i + 5 106 iL + L c = 0 20 10 t ic iL 1 + i dt + 2 = 0 3 o c 10 8 10

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CHAPTER 7 ENGINEERING CIRCUIT ANALYSIS 63.

SELECTED ANSWERS

65.

iout = iS

G in (V2 - V1 ) + G f V2 1 t V1dt + G in (V1 - V2 ) L1 0

67. 69. 71. 73. 75.

32 J 2.6 mJ 221 J R = 1 and L = 1 H 558 pJ

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CHAPTER 8 ENGINEERING CIRCUIT ANALYSIS 1. 3. 5. 7. 9. 11. 13. (a) 1.25 mA; (b) 740 mA; (c) -6.6 V; -6.6 V 50 mH 3.5 (a) 4 A, 0 V; (b) 4 A, -48 V (a) 2e 400t A, t > 0 ; (b) 37 mA: (c) 1.7 ms (a) 6.9; (b) 2

SELECTED ANSWERS

15. 17. 19. 21. 23. 25. 27. 29. 31. 33. 35. 37.

6.3 k, measuring to 5 (a) 100 s; (b) 366 nA (a) 4.999 V; (b) 4.998 mA; (c) 49.9 mJ (a) 9.95 (a) 69 s; (b) 35 s 20 V, 100 mA; 4.5 V, 0 A; 1 V, 0 A. (a) 2.7 A; (b) 1.9 A (a) 85 V; (b) 29 V; (c) 35 s (a) iL (t ) = 0.4e 750 t A, t > 0 5 A; 2.3 A; 1.9 A (a) 30 A; (b) 1.7 ms; (c) i (t ) = 30e 600t A ; (d) 1440e 600t V ; (e) 6e 600 t + 14A (a) 290 mA; (b) 200 mA; (c) 50 mA; (d) 277 mA; (e) 34 mA

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CHAPTER 8 ENGINEERING CIRCUIT ANALYSIS 39. 41. 43. 45. (a) -6 mA; (b) 12e-100t mA (a) 20e 250,000t V (t > 0) ; (b) 9.4 V (a) 99.8 V; (b) 88e-2539t V

SELECTED ANSWERS

(a) 100 V, 0; (b) 100 V, 100 V; (c) 80 ms; (d) 100e-12.5t V; (e) 5e-12.5t mA; (f) 20e 12.5t + 80V , 80e 12.5t + 80V ; (g) 16 mJ, 100 mJ, 20 mJ (a) 20 mA; (b) 20e 10000t 2e 5000t mA, t < 0v (t ) = 6u (t ) 6u (t 2) + 3u (t 4) V

47. 49. 51. 53. 55. 57. 59. 61. 63.

(a) 9; (b) 9; (c) 9; (d) 3; (e) 3 1 A; 600 mA; 600 mA (a) 1; (b) 12; (c) 1.47 2.5 A; 3 A; 2.5 A; 2 A; -2 A (a) 9.8 V; (b) 2 (a) iL (t ) = (2 2e 200000t ) u (t )m mA; (b) 6e 200000t u (t )V (a) iL (t ) = 4(1 e 1000t )u (t )A ; (b) v1 (t ) = (100 80e 1000t )u (t )V10 t 9 (a) i (t ) = 0.9e 9 A; (b) 1.04 A 56

65. 67. 69. 71.

2.5 V (a) 2 A; (b) iL (t ) = 5 3e 40t A, t > 0 (a) 80 mA; (b) 0.08(1 e 25t )A, t > 0 ; (c) 0.16 0.08e 25t A, t > 0 ; (d) 0.016 cos 50t + 0.032sin 50t 0.016e 25t A, t > 0iL (t ) = 0.1 + (0.1 0.1e 9000 t ) u (t )A

73. 75. 77.

(a) 3 A; (b) 2.4 A; (c) 2.6 A (a) 20(1 e 40t ) u (t )A ; (b) 10 8e 40t u (t )A

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CHAPTER 8 ENGINEERING CIRCUIT ANALYSISv(t ) = 0.94 10e 0.1t + 10 cos 4t + 400sin 4t 1601

SELECTED ANSWERS

79. 81. 83. 85. 87. 89. 91. 93. 95. 97. 99. 101.

4.5 1 e 10t

(

)8 3 5

iA = 10 + 7.5e 10 t /10 = 10 + 7.5e 10 t mA, t > 0, iA = 2.5mA t < 0 (a) vc (t ) = 8u (t ) + (16 24e 500 t ) u (t ) ; (b) 0.4u (t ) + (0.8 + 2.4e 500t ) u (t )mA 6.32 V; 15.7 V (a) 80 V; (b) 80 + 160e 100000t V, t> 0 ; (c) 80 V; (d) 80 32e 20000t V, t > 0 693 ns (a) 242 mV: (b) 3.11 mW; (c) 15 J 1.0e-t/10 u(t) V vo(t) = -0.2[1 + e-2010 t]u(t) V 2.5 F (a)3

103.

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CHAPTER 9 ENGINEERING CIRCUIT ANALYSIS 1. 3. 5. 7. (a) 175 103 s 1 ; (b) 22.4 krad/s; (c) overdamped.

SELECTED ANSWERS

(a) 5108 s-1; (b) 32 Trad/s; (c) 0.5 j 32 Grad/s ; (d) underdamped 1.44 H; 14 mF; 4.9 (a) 100 aF; (b) 1 M; (c) 5 Gs-1; (d) 5 109 + j 70.71 1012 s 1 , 5 109 j 70.711012 s -1 ; (e) 7.1 10 5 (a) 800 rad/s (b) 954103 s-1; (c) 95300% (a) 158 m; (b) i (t ) = 4.169e 158.5t 0.169e6.31104

9. 11. 13. 15. 17. 19. 21. 23. 25. 27. 29. 31. 33. 35.

t

A

(a) 20e 10t + 60e 40t V, t > 0 ; (b) 160 e 10 t 120 e 40 t A (a) v (t ) = 18 e 0.069t e 0.181t V ; (b) 8.6 s, -6.1 V 2.025e 50t 0.025e 450t A, t > 0 v(t ) = 170e 8t 42e2t , t > 0 (a) 50 V; (b) 2 A; (c) vc (t ) = 25e 2000t + 75e 6000t , t > 0 ; (e) 270 s; (f) 2 ms R < 160 ohms (a) 1.6 m; (b) iL ( t ) = e 3.210 t 3.2 106 t + 105

(

)

(a) 8 mH: (b) 930 mA; (c) 24 ms 160 ohms8.11 1013 cm

e 4000t (2 cos 2000t + 4sin 2000t )A, t > 0 (a) e 5000t (200 cos104 t + 100sin104 t ) V, t > 0 ; (b) 10 e 5000t (10 cos104 t 7.5sin104 t ) mA, t > 0 0.6e 100t sin1000t mA, t > 0 R = 10.4 ohms; 2.15 s

37. 39.

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CHAPTER 9 ENGINEERING CIRCUIT ANALYSIS 41. 43. 45. 47. 49. 51. 53. 55. 57. 59. 61. 63. 65. 67. 69. 71. (a,b) e-t (4 cos 5t + 0.8 sin 5t) A; (c) 4.7 s e 4t (10 cos 2t + 20sin 2t )A, t > 0 (2.25e 200t 0.25e 6000t ) u (t ) + 2u (t ) V (a) 0.5e 10t A, t > 0 ; (b) 100e 10t V t > 0 4.7 kV 1.5 ohms; 23 J

SELECTED ANSWERS

vC (t ) = e 2500t 100 cos 1.6 105 t + 1.6sin 1.6 105 t V

(

)

(

)

vC (t ) = e 0.21t [13cos18t + 0.14sin18t ] V for t > 0 and 13 V, t < 0iL(t) = 10 - e-4t (20 sin 2t + 10 cos 2t) A, t > 0 e 4000t (2 cos 2000t 4sin 2000t ) A, t > 0 12 e t (t + 2) V, t > 0 (a) 2.5e 500t 22.5e 1500t mA, t > 0 ; (b) 25e 500t + 22.5e 1500t mA, t > 0 (a) 30 V; (b) 51 V; (c) 44 V; (d) 44 V (a) 0; (b) 0; (c) 920 mA; (d) -1.03 A 1003 ohms one possible solution:

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CHAPTER 9 ENGINEERING CIRCUIT ANALYSISdv 1 = v dt 3.3 (b) one possible solution:

SELECTED ANSWERS

73.

(a)

75.

(a)

diL = - 4iL ; dt

(b) one possible solution:

Copyright 2007 The McGraw-Hill Companies, Inc. All Rights Reserved.

CHAPTER 10 ENGINEERING CIRCUIT ANALYSIS 1.

SELECTED ANSWERS

(a) f (t ) = 8.5sin (290.9t + 325.0) ; (b) 8.5 cos (290.9t 125) ; (c) 4.875+ cos 290.9t + 6.963sin 290.9t (a) 58, 57; (b) 134o 85 Mrad/s, 39 V, pi (a) -6 cos (260t + 9o) lags 6 cos (260t 9o) by 360 9 189 = 162o; (b) -cos (t - 100o) lags cos (t - 100o) by 180o; (c) -sin t lags sin t by 180o; (d) 7000 cos (t ) lags 9 cos (t 3.14o) by 180 3.14 = 176.9o. (a) 800 mV; (b) 771 mV; (c) 814 mV; (d) 805 mV 13.3 cos (5t 89.6o) V 743 cos (500t 22o) mA (a) 26 s; (b) 10 or 26 s; (c) 16 or 26 s 12.5cos(500t 0.11o) mA1.4 cos (400t 45) + 1.3cos (200t 27) A

3. 5. 7.

9. 11. 13. 15. 17. 19.

21.

(a) Vm sin t = Ri +

1 i ; (b) C

1 cos t + tan 1 CR 1 + 2 C2 R 2

CVm

23. 25. 27. 29. 31. 33. 35.

(a) 16.8 j 5.9; (b) j 204; (b) 0.31 + j 1.7 (a) 18.7 -16o; (b) 3.2 46o (a) 39 76 ; (b) 4 70 ; (c) 2.4 + j8.9 ; (d) 0.67 + j 0.21 65e j (10t +126 ) A (a) 1220A ; (b) 7.6113 A ; (c) 3.9 108 A ; (d) -65 V; (d) 54 V 35 mV (a) 18.3 cos (5000t 41o) V; (b) 76 cos (5000t + 79) V ; (c) 58 cos (5000t + 118) V9.9 cos (400t + 79) V

37.

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CHAPTER 10 ENGINEERING CIRCUIT ANALYSIS 39. 41. 43. 45. 47. 49. 51. 53. 55. 57. 59. 61. 63. 65. 67. 69. (a) j292 ; (b) j2.92 ; (c) j292 m; (d) j292 n (a) 478 + j176 ; (b) 588 + j120 212 cos (800t 46o) mA (a) 196 11 ; (b) 72 F; (c) 11.3 and 444 rad/s R 2 = 4.3 , R1 = 3.2 (a) 10.6 j1.9 ; (b) 10 + j0.25

SELECTED ANSWERS

(a) 1 + 4 H; (b) 5 + 2 H + 500 mF; (c) 1.2 + 69 mH; (d) 5 (a) j88 mS; (b) j8.8 S; (c) j880 S; (d) j8.8 GS 2 , 2 H (a) 105 rad/s; (b) 105 rad/s; (c) 102 krad/s; (d) 52 krad/s, 134 krad/s (a) 250 F; (b) 100 F (a) 1 S + 250 mH; (b) 5 || 1 F || 1 H; (c) 820 m || 69 mF; (d) 5 3423 V 70 cos(1000t 45) V 1.2 cos (100t 76) A

(a)

j C1R f A j C1R f A Vo V ; (b) o = = Vs 1 + A + j C1R f Vs (1 + A) (1 + j C f R f ) + j C1R f

71. 73.

16 mW 4.9 F g C + C + C + C m jC RL RS 1 1 (a) ang(Vout) = tan g R 2 - tan - 1 2 2 m S + 2C + C C RS R L

75.

(

)

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CHAPTER 10 ENGINEERING CIRCUIT ANALYSIS

SELECTED ANSWERS

77.

, 2 , 2 H + j (2 2 1)2.5 , 1.25 H, 0.89 63 A158108 V , j150

79. 81. 83. 85.

(a) 88 cos (t 107o) mV

87.

(a)

(b)

Vout VS

=

0.802 1 + 6.4 1024 2

89.

v1(t) = 3.210-3 cos (2104t 87o) + 31010-12 cos (2105t + 177o) V and v2(t) = 3110-9 cos(2104t 177o) + 11610-12 cos(2105t 93o) V

91. 93.

57 77 , 26 140 , 51 50 , 14313 , 51 140 , 51 140-40.5o, 28o

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CHAPTER 11 ENGINEERING CIRCUIT ANALYSIS

SELECTED ANSWERS

1. 117 W; 137 W; -19.7 W 3. -8 W; -0.554 W; 0.422 W 5. -23.5 W; 4.31 W; 32.1 W; -12.9 W 7. 54 kW; 7.31 kW; 134 W 9. 226 mW; 294 K, representing temp increase of 111 mK 11. 297 W; 0; 186 W; 0 13. 10.9 W; 20.8 W 15. 26 W 17. 8 + j14 ; 180 W 19. 96 W 21. 52 W; 15 W, 31 W 23. 289 W, 145 W; 90.3 W, 181 W 25. 54 W, 1.6 W, 0, 0 27. 1.4, 1.4, 1.4, 1.4 29. 4.04 A 31. 12.6 V; 12; 10 33. 8.5;12.4 35. 42.7 W; 25 W; 7.32 W; 55.2 W; 80.2 W 37. 30 V, 30 V; 34.6 V, 34.2 V 39. 9.88 41. 655 W; 320 W; 335 W; 800 VA; 320 VA; 568 VA; 0.6 lagging 43. 1230 VA, 774 VA, 86.5 VA, 865 VA, 3020 VA 45. 4.79 Arms; 0.91 lagging 47. 7.5 F; 40 F 49. 211 + j442 VA, 289 + j0 VA, 0 + j192 VA, 562 + j0 VA, 640 j390 VA, 142-90o VA 51. 1600 + j1800 VA; 0.66 lagging; 0.95 lagging 53. 70 kW; 81.4 kVA; 0.86 lagging P ( tan old - tan new ) 55. C = 2 Vrms 57.5.1 Arms; 1200 W; -1200 W; 1200 VA; 1200 VA; j1200 VA 59. 5203o kVA, 38 kVA, -j49.6 kVA, j77 kVA, 480 + j0 kVA; 5203o kVA; NO!; 520 kW; 28 kVAR

PROPRIETARY MATERIAL. 2007 The McGraw-Hill Companies, Inc.

CHAPTER 12 ENGINEERING CIRCUIT ANALYSIS 1. -9.3 V; -0.7 V; 9.3 V 3. Van = |Vp| 0o Vbn = |Vp| -60o Vcn = |Vp| -120o Van = |Vp| 0o Vbn = |Vp| 60o Vcn = |Vp| 120o 5. 56.7 -11.5 V; 190 35.0 V 7.o o

SELECTED ANSWERS

Vdn = |Vp| -180o Ven = |Vp| -240o Vfn = |Vp| -300o Vdn = |Vp| 180o Ven = |Vp| 240o Vfn = |Vp| 300o

The phase sequence is negative, since sequence is acbacb. A positive sequence would be abcabc

9. The temptation is to extend the procedure for voltages, but without the specific circuit topology, we do not have sufficient information to determine I31. 11. 22.8 -18.5 A; 34.4 -12.1 A, 7.60 -109 A, 36 180 A 13. 91.5 F ; 6.68 kVA 15. 1040 W; 81.3 144o V 17. 1.18 ; 282 20.8 V ; 450 173 V ; 15.8 j6.00 kVA 19. 6.80 96.1 A rms 21. 2.9717 A ; 52.8 W; 1990 W; 0.96 leadingo o o o

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CHAPTER 12 ENGINEERING CIRCUIT ANALYSIS

SELECTED ANSWERS

23. 0.894; 22 F; 541 VAR 25. 346 30 V ; 48 j 24 ; 11.286.6 A rms 27. 5.48 A rms; 3.16 A rms; 240 V rms 29. 40.245 A rms; 60.5 170 A rms; 36 30 A rms; 4320 + j 4320 VA 31. 243 30 V ; 24 1 A rms ; 41.7 31 A rms 33. 33.9 45 ; 25.2 7.6 A ; 53 157 A rms ; 6100 + j 3300 VA 35. 1.54 kW; 2.16 kW; 615 W. 37. 186 W. 39. 862 W 41. We assume that the wire resistance cannot be separated from the load, so we measure from the source connection.

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CHAPTER 13 ENGINEERING CIRCUIT ANALYSIS 1. M 21 = 663 H 3. 1 and 3, 2 and 4; 1 and 4, 2 and 3; 3 and 1, 2 and 4 5. 60.8 sin 800t pV; 36 sin 800t pV

SELECTED ANSWERS

7. 2300e t + 3400e 3t A/s ; 1700e t + 4600e 3t A/s ; i2 = 1700e t + 4600e 3t A 9. 10.4 W; 4.8 W; 0 each; 0 11. 106 + j 76 ; 25 W 13. ic (t ) =30t A, t >0 (t + 0.01) 22

15. (6 + j5) I j2 I 6 I = 100, -j2 I + (4 + j5) I j4 I = 0, -6 I - j4 I +1 2 3

(11 + j6) I = 0 ; I = 4.32 -54 A3 3

o

1

2

3

1

2

0.22 0.022 + j 0.1 17.+ ; 2.8 + j1.2 25 + 0.252 25 + 0.252 19. 27.3cos(10 t + 69 ) V; 23.6cos(10 t + 66 ) V; 9.6 W, 5.76 W 21. 1.3 60 A 23. 2.16K 2 W K 4 1.82K 2 + 1.1881

25. 0.84 W; 0.26 W; 1.1 W 27. 1.7 42 ; 0.39 -80o; 2.2 0.05 29. V2 = j1.7 k L1L 2 + 1

31. 4.56 j 4 n ; 10 + j 63 33. M = 5 H, L 1 = 9 H, L 2 = 11 H 35. 600 mH; 880 mH; 750 mH

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CHAPTER 13 ENGINEERING CIRCUIT ANALYSIS 37. OC T Z oc A = j4 M SC

SELECTED ANSWERS

Z Z

T B oc TA in

= j4 M = j4 j10 + j8 M

T T ; Z SS A = Z SS B = j4 M + j8 j10 M ;

T Z in B = j26 j12 j8 M

39.

j 4.9 1 + j 0.5

41. 25 + j 0.62 ; j 24 ; j 25 43. 20 + j 31 ; 20 + j 28 ; 20 + j 25 ; 21 + j 24 45. 192 W, 73 W, 61 W, 550 W 47. 8 W; 2.1 W; 5 kW 49. 0.89, 5 51. 9.2 V 53. 4.8 A 1 120 57. IQ = 28.8 10 3 + 576 Age 576 Age ; half a century 1000 2

59. You need to purchase (and wire in) a three-phase transformer rated at 3 (208 )(10 ) = 3.6 kVA.

( )

Copyright 2007 The McGraw-Hill Companies, Inc. All Rights Reserved.

CHAPTER 14 ENGINEERING CIRCUIT ANALYSIS

SELECTED ANSWERS

1. s = 0; s = j9 s-1; s = -8 s-1; s = -1000 j1000 s-1; s = 0, s = 2 s-1 3. 8et; 19; 1138o ; 10 ; 10 ; 10 ; 889 5. 6.6 C ; 9 C ; No. 7. 8.1e 3t cos(15t 60) ; 8.1e 3t cos(15t 60) ; 4.1 ; 4.1 9. 30230 V ; 36e 2t cos (50t 56) V ; -19 V; s = 2 + j 50 s 1 ; 2 j 50 s 1 11. impedance R; an impedance Z L = sL = ( + j ) L , an impedance 1 1 Zc= = ; Z R = 100 . Z L = ( 2 + j10)(0.002) = 20101 ; Yes. ( + j )C sC 13. 0.35 105 A ; 350e 2t cos(10t 105) mA 15. 185 48 V ; 185e 3t cos(4t 48) V 17.K s5 3 K ; ;0; s s s+8

19.

21.

5 (1 e2s ) ; 5 (1 e2s ) s s 8 8 8 6 3s 6 3 s 1 e 63s ; s 2 1 e e ; s + 2 1 e 2+s

23.

25. e 3t u(t); (t ) ; t u(t) ; 275 (t ) ; u(t) 27. 0.047 + j 0.11 ; 0.18 + j 0.20 ; (0.47 + j 6.5) 103 29.1 2 s 5 s 4 4 2 s 6 ; 4e 2 s ; 2.9 ( e e ) ; e 2 s ; F (s ) = e s+3 s s 2 2 s 3 4 s 8 ; 3e 5 s ; 4e s e e 3s ; e 4 s ; e s s s+2

31.

(

)

33.

1 1 90 (t ) 4.5u (t ) ; 11 (t ) + 2u (t ) ; te t u (t ) ; e t e 2t + e 3t u (t ) 2 2

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CHAPTER 14 ENGINEERING CIRCUIT ANALYSIS 35. 2.5 mA

SELECTED ANSWERS

37. (t ) + u (t ) + 2e t u (t ) ; (t 2) + 2 (t 1) + (t ) ; 2e 1 (t 1) ; (t 1) + (t 5) 39. 5e t u (t ) ; (5e t 2e 4t )u (t ) ; 6(e t e 4t )u (t ) ; 6(4e 4t e t )u (t ) ; 18 (t ) + 6(e t 16e 4t )u (t ) 41. 2 u (t ) 3 e t u (t ) ; 2 (t ) + 4e 3t u (t ) ; 3 (t 0.8) ; 3(e 2t e 6t )u (t ) ; (3te 2t 0.75e 2t + 0.75e 6t )u (t )2 2 43. f (t ) = (t ) + u (t ) e 3t u (t ) ; f (t ) = 0.5tu (t ) + 0.25u (t ) + 0.35 cos(2t + 135) 3 3

45. h(t ) = (t ) e 2t u (t ) ; h(t ) = 2e t e 2t u (t ) ; d 1 9 81 h(t ) = 2 (t ) + 6 (t ) te t u (t ) + e t u (t ) e 3t u (t ) dt 2 4 4 47. f (t ) = (1.9 5.59e 4t / 3 )u (t ) 20 130 3t 49. 50 V; 0.1v c' + 0.2v c + 0.1(v c 20) = 0 ; vc (t ) = + e u (t ) V 3 3

[

]

51. (4 2e 0.15t )u (t ) 53. y (t ) = (2 + 6e t )u (t ) ; (6e t 1)u (t ) 55. -600 mA; 40 = 100ic + 50 57. R = 250 m, C = 1 F, L =

0

ic dt + 100 ; 0.6e 0.5t u (t )

1 H ; v(t ) = (75e 3t 12.5e t 62.5e 5t )u (t ) V 3

59. ic (t ) + 0.4 (t ) 1.6e 2t u (t ) A 61. STABLE; UNSTABLE. 63. STABLE; STABLE. 65. 7 V ; 7 V 67. 2, 3, 0, 0, 69.n s+b ; b a, b a s+a

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CHAPTER 15 ENGINEERING CIRCUIT ANALYSIS 1.

SELECTED ANSWERS

= 0.032s 0.032s = 384 V

12 mA s

3.

20s 2 + 11, 000s + 200, 000 s + 200 s + 500 20s 2 + 11, 000s + 200, 000 ; 11 ; 8.154 ; ; ; s 2 + 700s + 100, 000 s 2 + 700s + 100, 000 20s 100016s 2 + 50s + 4000 ; 0.16 j 4.7 ; 6.85 114 ; 910 m ; 1 s 2 + 80sr R B (1 + Z L C s ) Z L r R BC C s 2 + (g m Z L r R BC + r R BC + r R BC + Z L r C +Z L R BC )s + r + R B

5.

7.

9. 4.5 e-0.28t u(t) A 11. v1(t) = 5.6e6.67t + 3.6 V, t 0 v2(t) = 3.73e6.67t + 4.4 V, t 02 1 t4 e A, t 0 3 6 2 1 t4 + e A, t 0 3 12

13. i1 (t ) = 15.

and

i2 (t ) =

35s - 131 ; ( s + 2 ) + 100 ( s + 6 )2 + 100 2

e-6t [0.092cos 2t - 1.5 sin 2t] - e-2t [0.092 cos10t - 0.34 sin 10t] A 17. [0.63e-0.61t + 0.79 e-0.55t cos(0.34t + 99o)] u(t) 200s(s 2 + 9s + 12) 19. ; [185 e-3t cos (4t - 48o) + 86e-1.25t cos (1.9t + 107o)] u(t) 4 3 2 2s + 17s + 90s + 185s + 250 21. I1 = 271.7-96.5o A and I2 = 272-96.5o A

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CHAPTER 15 ENGINEERING CIRCUIT ANALYSIS 23.

SELECTED ANSWERS

2[1.301 e-142.8t cos (742.3t + 12.54o) + 0.00202 cos (2t 6.538o) 6.60110-5 (t) - 1.564 e-142.8t cos (742.3t 33.56o) - 2.998 cos (2t + 179.9o)]2 W (a)

25.

(b)2500s + 0.5 7.5 106s + 1500 , V 0.001s 2 + 5s + 500 s s 2 + 5000s + 5 105

(

)

(c) 27.

[-3 e

-2.5106t

+ 3 e-0.2t + 310-3 + 21 cos(711t + 89.9o)] u(t)

420s 4 + 133s3 + 21s 2 + 60s + 9 70 70 ; V; A 2 2 4 3 60s + 19s + 3 60s + 19s + 3 420s + 133s + 21s 2 + 60s + 9 29.

30303(0.2239 1013 + 0.1613 1013s + 98700s 2 ) V1 = , s(0.4639 1010s3 + 0.7732 1015s 2 + 0.56911018s + 0.1936 1018 )

V2 =

7609(705000s3 + 0.1175 1012s 2 + 0.6359 1014s + 0.8897 1014 ) ; s(0.4639 1010s3 + 0.7732 1015s 2 + 0.56911018s + 0.1936 1018 )

v1(t) = [3.504 + 0.380510-2 e-165928t 0.8618 e-739t 2.646 e-0.3404t] u(t) V, v2(t) = [3.496 0.136510-2 e-165928t + 0.309 e-739t 2.647 e-0.3404t] u(t) V

31. 4.639 109s3 + 7.732 1014s 2 + 5.6911017s + 1.936 1017 , 98700s3 + 1.645 1010s 2 + 1.21 1013s + 2.059 1012

2.059 1012 (5s - 3) , 201 A s(4.639 109s3 + 7.732 1014s 2 + 5.6911017s + 1.936 1017 )

Copyright 2007 The McGraw-Hill Companies, Inc. All Rights Reserved.

CHAPTER 15 ENGINEERING CIRCUIT ANALYSIS

SELECTED ANSWERS

33.

(

5000s ; [-0.76 e-6385t + 0.77 cos (103t 8.9o)] u(t) A; 2 6 s + 10 ( s + 6385 )

)

0.77 cos (103t 8.9o) A.1 3 35. Poles at j 2, 1 ; zeroes at s = 0, .; Poles at s = 1, j , double at s = 0 , Zeroes at 4 4 1 j 2,

37. (a.) zeros at s = -25 and -12.5 s-1, and poles at s = 0 and s = -1.7 s-1. (b.) zeros at s = -9. 1 and -105 s-1, and poles at s = -1.55105 and s = -3.2 s-1. 39. Z in =10

5(s + 1)(s + 4) ; 1.5, ; 1, 4 s 1 6(s + 1.5)20

41. 0,

(1 cos t ) ,

, (10/ ) (1 + cos t), 0

43. 8t 8 V, 16 V, zero 45.15 15 15 15 30 - 15e-2s , 15 u(t) 15 u(t - 2), 2 e 2 s , 15 t u(t) 15 u (t - 2), 2 e 2 s + 15e 4 s , 15 t s s s s s 15 15s -2s u(t) 30 u (t - 2) + 15 u(t), 2 e , 5 sin 3t u(t) 15 cos [3(t 2)] u(t - 2), s + 9 s2 + 9

47. h(t) = (t); 8e t u (t ) V 49. (a) (b)

(c) 5

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CHAPTER 15 ENGINEERING CIRCUIT ANALYSIS

SELECTED ANSWERS

51.

; 4.7, 10; 15 cm 10s3 + 30s 2 10s 30 ; -2.3, ; 2.2 117 , 2.263, 3.634 , 5 53 , 30 53. s 2 + 6s + 13 55.

;

2 + 4 100( 2 + j ) ; 100 ; graph; 2 rad/s, 69 (5 2 ) + j 2 4 6 2 + 25

57. 5e 6t A (all t); [5e 6t + e 2t (5 cos 4t + 3 sin 4t ]u (t ) A 59. 1.7 and 24 s -1 ; iin(t) = [10 2.1e-24t 0.885 e-1.7t] u(t) A 61.2.5 ; [1 + 0.066 e-6.4t 1.1e-0.39t ] u(t) V s + 6.75s + 2.52

63.

H ( s) =

10s + 10 5 R + 10 5 5s ; ; 5s s + 10 5 s + 10 5

65.

0 F, 400 ; 5 nF, R 1 = 200 k ; 50 nF, 200 k ; R fA = 1 k , C fA = 10 nF, R fB = 100 , R 1B = R 1 A = 10k

67. 69.

One possible design: If we use a 1-F capacitor, then R = 159 . To complete the design, select Rf = 2 k and R1 = 1 k. One possible design: If we use 100-nF capacitors, then R = 3.167 k.

Copyright 2007 The McGraw-Hill Companies, Inc. All Rights Reserved.

CHAPTER 17 ENGINEERING CIRCUIT ANALYSIS

SELECTED ANSWERS

1.

4 8 9 I1 12 (a) 5 0 7 I 2 = 4 ; (b) 651; (c) 21; (d) 600 mA; (e) -141 mA 7 3 1 I 3 0 (a) 390 W; (b) 250 W; (c) 180 W 2s 2 + 15s + 20 2s + 5 15s + 25 s(s + 4) (a) j CR 1R 3 R 4 ; (b) j 0.8 103 (Lin = 0.8 mH) R2

3. 5.

7.

9. 11. 13. 15. 17. 19.

142 mS, -77 mS 40 mS, -40 mS, 40 mS, -30 mS (a) 50 ; (b) 60 200 mS, -300 mS, -400 mS, 150 mS (a) Input is applied between g-s and output taken from d-s; 1 (b) j ( C gs + C gd ) , jC gd , g m jC gd , + j ( C gs + C gd ) ; rd (c) j 4.8 pS , j1.4 pS , 4.7 103 j (1.4) 1012 S , 104 + j (0.4 + 1.4) 1012 S

21. 23. 25. 27.

784 , 1.72 k, 367 9.9 (a) 56; (b) -9.6; (c) 530; (d) 3.4 ; (e) 35 (a) (b)

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CHAPTER 17 ENGINEERING CIRCUIT ANALYSIS SELECTED ANSWERS 29. 31. 33. 35. (a) 1.55 V; (b) -17.5 A 7.5 1.1 4.5 11 ()

(a) -2; (b) 4; (c) 8; (d) 1 ; (e) 1.3 z11 = 133 48 z12 = 94 2.6 z21 = 940087 z22 = 565 3.6

37. 39. 41.

10 2 42 1.7 (a) ; (b) 17 0.17 S 20 0.2 S

(a) 1.2; (b) 9.6 ; (c) 240 mS 0.01 1000 (a) ; (b) 8.6 k 2 104 S 10 (a)

43.

( jC ) (1 + j r C ) + g 1 + j r ( C + C )

m

1 + j r ( C + C )

j r C

+

( g m jC ) r 1 ; (b) rd 1 + j r ( C + C )

(c) rx +

jC r r ; (d) 1 + j r ( C + C ) 1 + j r ( C + C )

45.

3.3 0.61 (a) ; (b) 11 0.053 S 0.81 1.5 3 1.4 2 , 11/ 7 4 ; (b) (a) , 1 S 1 1/ 7 S 1 0.2 S 1 6

47.

4.71 15.9 0.962 S 3.47

49.

0 1 R 1/ a 0 1 0.58 14 (a) , 0 1 , 0 a ; (b) 0.115 S 4.5 1/ R 1

Copyright 2007 The McGraw-Hill Companies, Inc. All Rights Reserved.