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CHAPTER 2 ENGINEERING CIRCUIT ANALYSIS SELECTED ANSWERS 1. (a) 12 μs (d) 3.5 Gbits (g) 39 pA (b) 750 mJ (e) 6.5 nm (h) 49 kΩ (c) 1.13 kΩ (f) 13.56 MHz (i) 11.73 pA 3. 300 kW; 3.7 m; 25 mm; 71 kJ; 290 fs 5. 131 kW; 1.4 GJ; 1 battery 7. 13 GW; 100 mW 9. 290 kJ; 1.5 kJ 11. 6.2 A; 3.5 A; The current is never negative; 34 C 13. 12 MV; 0; -18.7 MV; -6.2 MV 15. -6.4 mW; -120 W; 60 W; 12 W 17. 73 W; -36 W; 28 W 19. 5 mW, 0, -2 mW; 36 J; 22 J 21. 64 W, 256 W, -640 W, 800 W, -480 W 23. -1 mV 25. 58 W; 4.8 A 27. 5.6 mA, 4.5 mA; 23 mW, 28 mW 29. 43.5 mW; 231 mW; 253 mW 31. Since we know that the total power supplied is equal to the total power absorbed, we may write: V s I = I 2 R 1 + I 2 R 2 . Now invoke Ohm’s law. 33. 500 μA, 2.5 mW; -500 μA, 2.5 mW; -500 μA, 2.5 mW; 500 μA, 2.5 mW 35. -2 V (at t = 0.324 s) 37. 2 km. Hmmmm…. 39. 1.7 μΩ . cm 41. 560 mΩ, 1.3 W 43. 266 mΩ; 514 mA 45. Design. Many possible solutions. Hint: Start with finding resistivity, then choose geometry. Copyright ©2007 The McGraw-Hill Companies, Inc. All Rights Reserved.

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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 SELECTED ANSWERS

1. (a) 12 μs (d) 3.5 Gbits (g) 39 pA (b) 750 mJ (e) 6.5 nm (h) 49 kΩ (c) 1.13 kΩ (f) 13.56 MHz (i) 11.73 pA 3. 300 kW; 3.7 m; 25 mm; 71 kJ; 290 fs 5. 131 kW; 1.4 GJ; 1 battery 7. 13 GW; 100 mW 9. 290 kJ; 1.5 kJ 11. 6.2 A; 3.5 A; The current is never negative; 34 C 13. 12 MV; 0; -18.7 MV; -6.2 MV 15. -6.4 mW; -120 W; 60 W; 12 W 17. 73 W; -36 W; 28 W 19. 5 mW, 0, -2 mW; 36 J; 22 J 21. 64 W, 256 W, -640 W, 800 W, -480 W 23. -1 mV 25. 58 W; 4.8 A 27. 5.6 mA, 4.5 mA; 23 mW, 28 mW 29. 43.5 mW; 231 mW; 253 mW 31. Since we know that the total power supplied is equal to the total power absorbed,

we may write: Vs I = I2R1 + I2R2. Now invoke Ohm’s law.

33. 500 μA, 2.5 mW; -500 μA, 2.5 mW; -500 μA, 2.5 mW; 500 μA, 2.5 mW 35. -2 V (at t = 0.324 s) 37. 2 km. Hmmmm….

39. 1.7 μΩ.cm

41. 560 mΩ, 1.3 W

43. 266 mΩ; 514 mA 45. Design. Many possible solutions. Hint: Start with finding resistivity, then choose geometry.

CHAPTER 3 ENGINEERING CIRCUIT ANALYSIS SELECTED ANSWERS

1. Circuit diagram not shown. 3. (a) 4 nodes; (b) 5 branches; (c) yes, path; no, loop. 5. (a) 4; (b) 5; (c) yes,no,yes,no,no 7. (a) 3 A; (b) -3 A; (c) 0 9. ix = 1 A; iy = 5 A. 11. 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. 13. (a) 12 V; (b) -2.2 V 15. R = 34 Ω; G = 90 mS 17. Circuit I: i = 0; Circuit II: i = 1.1 A 19. -23.5 V 21. (a) (b)

v1 = 60 V i1 = 27 A v2 = 60 V i2 = 3 A v3 = 15 V i3 = 24 A v4 = 45 V i4 = 15 A v5 = 45 V i5 = 9 A

= -1.62 kW = 180 W = 360 W = 675 W = 405 W

23. (a) 8 V, -4 V, -12 V; (b) 14 V, 2 V, -6 V; (c) 2 V, -10 V, -18 V 25. (a) 25 W; (b) 24 W; (c) 16 W; (d) 18.4 W; (e) -600 W 27. None of the conditions specified in (a) to (d) can be met by this circuit. 29. 5.0 A; 10.4 V 31. (a) 2.4 kΩ; (b) R = 0 33. -250 cos 5t mV

CHAPTER 3 ENGINEERING CIRCUIT ANALYSIS SELECTED ANSWERS

35. (a) P5A = –5 vx = –1.389 kW

P100Ω = (vx)2 / 100 = 771.7 W

P25Ω = (vx)2 / 25 = 3.087 kW Pdep = –vx(0.8 ix) = –0.8 (vx)2 / 25 = –2.470 kW

(b)

P5A = –5 vx = –776.0 W

P100Ω = (vx)2 / 100 = 240.9 W

P25Ω = (vx)2 / 25 = 963.5 W Pdep = –vx(0.8 iy) = –428.1 W

37.

P8A = –8 vx = –240 W

P6Ω = (vx)2 / 6 = 150 W P8A = –7 vx = –210 W

P12Ω = (vx)2 / 12 = 75 W

P4Ω = (vx)2 / 4 = 225 W 39. (a) – 50 mA; (b) Can set vS = 50 V. 41. 638 mW 43. 1.45E-3 miles 45. (a) 1 A; (b) 9 A 47. (a) 10 mA; (b) 3.8 A 49. (a) 570 mA; (b) 0; (c) – 71 mA 51. -515 V 53. Req = 1 kΩ 55. (a) 10 kΩ || 10 kΩ; (b) 47 kΩ + 10 kΩ + 1 kΩ || 1kΩ || 1kΩ; (c) 47 kΩ || 47 kΩ + 10 kΩ || 10 kΩ + 1 kΩ 57. 5.5 kΩ 59. 60 Ω; 213 Ω; 52 Ω

CHAPTER 3 ENGINEERING CIRCUIT ANALYSIS SELECTED ANSWERS

61. 250 W; 188 W; 338 W; 180 W; 45 W 63. (a) 850 mS; (b) 136 mS 65. Proof 67. 607 mV 69. 22 A 71. One possible solution: 11 mA, 1 kΩ, 1 kΩ 73. 139 μA; 868 μW 75. 18 μW

77. (a) ( ) )R (RR R R RR)R (R R V

4324321

432S ++++

+ ;

(b) ( ) )R (RR R R RR)R R (R R V

4324321

4321S ++++

++ ;

(c) ( ) )R (RR R R RRR V

4324321

2S ++++

.

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

81. ⎟⎟⎠

⎞⎜⎜⎝

⎛++++ )R (RR )R R(R R

R R V54354 32

53S

83. vout = -56.02 sin 10t V

CHAPTER 4 ENGINEERING CIRCUIT ANALYSIS SELECTED ANSWERS

1. (a) -8.4 V; (b) 32 3. (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 5. -1.74 V 7. 172 V 9. (a) 58.5 V, 64.4 V; (b) 543 W 11. -28 V 13. -8.1 V 15.

1

2

3

4

3.4 V7.1 V7.5 V4.9 V

vvvv

====

5

6

7

8

1.7 V3.8 V3.5 V2.4 V

vvvv

====

17. (a) 26 V, (b) 83 mW 19. -3.25 21. -91 V 23. 45 W 25. v1 = -8.6 V, v2 = -3.9 V and v3 = 6.1 V 27. (a) 143 mA; (b) 16 W 29. (a) 3.1 A; (b) 370 W 31. 2.79 A 33. -380 W 35. i1 = 239 μA, i2 = 1.08 mA, i3 = -1.20 mA and i4 = -480 μA

CHAPTER 4 ENGINEERING CIRCUIT ANALYSIS SELECTED ANSWERS

37. (a) -5700 Ω; (b) this value is unique. 39. (a) 330 μA; (b) 330 μA; (c) units of resistance. 41. P2mA = 5000(i1 – i2)(i1) = 5 mW P4V = 4 (-i2) = -6 mW P6V = 6 (-i3) = 9 mW PdepV = 1000 i3 (i3 – i2) = 4.5 mW PdepI = 10,000(i3 – i4)(0.5 i2) = -5.6 mW 43. -3.65 W 45. -1.03 V 47. 5 Ω 49. (a) 0; (b) 96 V; (c) -38 V 51. 3.55 A; 1.69 A 53. 121 mA; 4.70 A 55. Hint: i3 = 1.24 A and i4 = 1.42 A by mesh analysis. 57. 350 mA 59. i1 = 2.65 A, i2 = 3.20 A, i3 = -3.80 A, i4 = -1 mA 61. -4 mA 63. -16 V 65. 3.14 V, 1.71 V, 714 mA, -143 mA, -2.14 A, 857 mA 67. One possible solution:

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

CHAPTER 5 ENGINEERING CIRCUIT ANALYSIS SELECTED ANSWERS

1. Define percent error as 100 [ex – (1 + x)]/ ex. If we choose x < 0.1, we ensure that the error is less than 1%. 3. 4.7 V, 2.0 A 5. 4 V → 40 V and 10 V → 100 V. 7. 10.8 V 9. (a) 1.3 A; (b) 60 W, 18 W, -130 W, 32 W, 20 W 11. (a) 200 V; (b) -143 V 13. 957 μW 15. Impossible; 76 mW 17. (a) 18 V 19. 2.46 V; 0.546 V, 1.91 V. 21. (a) 42 V voltage source in series with 6 Ω and in series with 10 Ω; (b) 26 V; (c) Cannot remove the resistor across which v appears or v may become lost. 23. 10 mW 25. 33 μW 27. (a) 12.8 mV 29. 764 nA 31. Current source is 7.25 A, resistor is 2 ohms.

33. 1.57 V, 811 mΩ 35. The final circuit is an 8.5 V voltage source in series with a 2.0 MΩ resistor. 37. (a) An 8/5 A current source in parallel with 5 Ω, in parallel with RL.

CHAPTER 5 ENGINEERING CIRCUIT ANALYSIS SELECTED ANSWERS

39. -2 V 41. (a) The Thévenin equivalent is a 9.3 V source in series with a 17 Ω resistor, which is in series with the 5 Ω resistor of interest; (b) 928 mW. 43. (a) 25 Ω; (b) 303 Ω; (c) Increased current leads to increased filament temperature, which results in a higher resistance (as measured). This means the Thévenin equivalent must apply to the specific current of a particular circuit – one model is not suitable for all operating conditions. 45. (a) 6.7 Ω, -300 mA, arrow upwards; (b) 6.7 Ω, -150 mA, arrow upwards. 47. (a) 38.9 V, 178 Ω; (b) 1.96 W. 49. VTH = 0, RTH = 192 Ω. 51. 15 Ω, 15 Ω 53. VTH = 0; The Norton equivalent is 0 A in parallel with 1.3 Ω. 55. VTH (and hence IN) = 0; RTH = RN = 198 mΩ. 57. 2 MΩ

59. VTH = 1 1 1

( )in i o f

o i o f i f i

v R R AR

1 iR R R R R R R R R R AR R−

+ + + + +; RTH =

Ro (Ri Rf + R1 Rf + R1 Ri) -------------------------------------------------------------- Ri Ro + R1 Ro + Ri Rf + R1 Rf + R1 Ri + A R1 Ri. 61. 16 Ω, 6.3 W 63. 65 V, 15 Ω, 70 W 65. (a) 200 V; (b) 125 W; (c) 80 Ω 67. 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. 69. Select R1 = RTH = 8 kΩ

CHAPTER 5 ENGINEERING CIRCUIT ANALYSIS SELECTED ANSWERS

71. 1.2 Ω, 0.54 Ω, 4.9 Ω 73. 9.9 Ω 75. 5.5 V, 1.0 Ω 77. -13 V, 27 Ω 79. Although the network may be simplified, it is not possible to replace it with a three-resistor equivalent. 81. IS(max) = 224 mA 83. 1.4 Ω 85. 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 Ω.

CHAPTER 6 ENGINEERING CIRCUIT ANALYSIS SELECTED ANSWERS

1. (a) -30 V; (b) -2.5 V; (c) 1.4 V 3. (a) ; (b) tvv inout 5sin2010 −=−= tvv inout 5sin51010 −−=−= 5. One possible design is to use a simple inverting op amp circuit with Rf = 9.1 kΩ and Rin = 5.1 kΩ. 7. 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 kΩ and Rin = 5.1 kΩ 9. (a) 1.7 V; (b) 3 V; (c) -2.4 V 11. (a) ; (b) tvv inout 10sin82 == tvv inout 10sin5.022 +== 13. -2.2 V 15. 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 Ω, and a resistor R1 = 1 Ω. 17. V1 = 21 V 19. ; -5.6 V (out -4 1 sin 3 Vv t= + ) 21. Rf = 236 kΩ and R1 = 1 kΩ. 23. (a) B must be the non-inverting input; (b) Choose R2 = RB = 1 Ω; (c) A is the inverting input. 25. vout(0.25 s) = 0.93 V 27. 4.2 V

29. ∑=

N

1f R R-

i i

iv

31. Pick R1 = 10 kΩ. Then vS = -0.21 V.

CHAPTER 6 ENGINEERING CIRCUIT ANALYSIS SELECTED ANSWERS

33. One possible solution of many:

35. Set R = 10 kΩ:

Then connect several into:

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

CHAPTER 6 ENGINEERING CIRCUIT ANALYSIS SELECTED ANSWERS

45. Rf = 120 kΩ and Rin = 200 kΩ, R = 560 Ω.

47. R = 400 Ω, R1 = 82 Ω.

I Is

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

51. (a) –3.7 mV; (b) 28 mV; (c) –3.7 V.

53. A 101

100A- in

out

+=

vv ; A = 9999.

55. vout = -16 mV

CHAPTER 6 ENGINEERING CIRCUIT ANALYSIS SELECTED ANSWERS

57. (a)

(b)

vout = 105(-0.00004v2 - 9.99980×10-6v1)+5v2 = 1.00008v2 - 0.99998v1 = 0.0005 – 1.99996 sin t

(c) =0.99998v)2/(1010 2

55adout vvvv −×== 2-0.99998v1 = 1.99996 sin t

59. (a) V3 = 27 V; 61. Positive voltage supply, negative voltage supply, inverting input, ground, output pin. 63. This is a non-inverting op amp circuit, so we expect a gain of 214. 65. For vx = -10 mV, PSpice predicts vd = 6 μV, where the hand calculations based on the detailed model predict 50 μV, which is about one order of magnitude larger. For the same input voltage, PSpice predicts an input current of -1 μA, whereas the hand calculations predict 99.5vx mA = -995 nA (which is reasonably close). 67. (a) Negative saturation begins at Vin = –4.72 V, and positive saturation begins at Vin = +4.67 V. (b) 40.6 mA. 69.

CHAPTER 6 ENGINEERING CIRCUIT ANALYSIS SELECTED ANSWERS

71. (a)

-15

-10

-5

0

5

10

15

-2 -1 0 1 2

Vactive (V)

Vou

t (V

)

12 V

-12 V

73.

75. (a) ⎟⎟⎠

⎞⎜⎜⎝

+−

+=−=

Gaugerefout RR

RRR

RVVVV3

3

21

221 ; (b) Vout = 0; (c) R = 4.3 kΩ and R

= 4.7 kΩ, gain of 5.39 for R = 4.7 kΩ, so R = 11.5 kΩ.

CHAPTER 7 ENGINEERING CIRCUIT ANALYSIS SELECTED ANSWERS

1. (a) 0; (b) -613sin120πt mA; (c) -40e-t nA 3. (a) ; (b) ( )30 1 mAtt e−− ( )54 100cos100 5sin100 mAte t t− − 5. (a) 6.95 pF; (b) 17 kV; (c) 72 7. Design problem: more than one solution. Hint:

9. (a) 33.4 mV; (b) 33.4 mV; (c) 50.1 mV 11. (a) 120sin 400 At μ− ; (b) 6.4 μJ; (c) ; (d) 100400(1 )Vte− 100500 400 Vt

cv e−= − 13. (a) 2 kΩ; (b) 20 mJ 15. (a) 0; (b) -613sin120πt μV; (c) 6240 pVte−− 17. (a) ; (b) ( )150 1 fVtt e−− ( )5100 20cos100 sin100 pVte t t− − 19. (a)

(b) 40 ms; (c) t = 20, 40 ms; (d) 2.5 J 21. (a) ; (b) 24 4 Vt t+ 24 4 + 5 At t+ 23. (a) 2 A; (b) 5.6 J; (c) 1 A 25. (a) 2.33 V; (b) 480 mA: (c) 1.1 A 27. (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) 29. (a) 0, 400 mW 31. 4.3 μF

CHAPTER 7 ENGINEERING CIRCUIT ANALYSIS SELECTED ANSWERS

33. (a) (b) 3.6 V

35. Cequiv = 85 nF 37. 140 nF 39. (a) 3 H; (b) N 41. 292 pH 43. (a) 11.4 Ω; (b) 11.4 H; (c) 8.8 F 45. (a) (b) (c) 47. (a) -6.4e-80t mA; (b) ; (c) 8080 60Vte− − 8020 60Vte− + 49. 9.2 V, 2.4sin103t V

CHAPTER 7 ENGINEERING CIRCUIT ANALYSIS SELECTED ANSWERS

51. (a) 1s iv idt

c= +∫ v ; (b) 1 A A 0

RCo o sv v v+′ ′+ + =

53. (a) ; (b) 10sin10t V 10.0sin10 0.0005 0.0005cos10sv t= + − t

55. (a) 0

'tf

out s

RV

L−

= ∫ v dt ; (b) Capacitor values are more readily available than

inductor values.

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

59. One possible solution of many (with C = 1 μF, R = 1 MΩ):

61. (a)

(b)

20 206

3206

120 ( ) 125 10

1 ( ) 12 10 8 10 05 10

t

c so

t

c co

v v v dt i

v v dt v v

−−

+ − + =×

′− − + + × =×

∫ c

(c)

6

3

5 10 020 10

1 2 010 8 10

L s L cL

tc Lco

i i i ii

i i i dt

− −′+ × + =

−+ +

× ∫ =

CHAPTER 7 ENGINEERING CIRCUIT ANALYSIS SELECTED ANSWERS

63.

65. )V - (V G V

L1

V G )V - (V G 21in0 1

1

2f12in

S

out

+′

+=

∫t

tdii

67. 32 J 69. 2.6 mJ 71. 221 μJ 73. R = 1 Ω and L = 1 H 75. 558 pJ

CHAPTER 8 ENGINEERING CIRCUIT ANALYSIS SELECTED ANSWERS

1. (a) 1.25 mA; (b) 740 mA; (c) -6.6 V; -6.6 V 3. 50 mH 5. 3.5 Ω 7. (a) 4 A, 0 V; (b) 4 A, -48 V 9. (a) ; (b) 37 mA: (c) 1.7 ms 4002 A,te t− >0 11. (a) 6.9; (b) 2 13.

15. 6.3 kΩ, measuring to 5τ 17. (a) 100 s; (b) 366 nA 19. (a) 4.999 V; (b) 4.998 mA; (c) 49.9 mJ 21. (a) 9.95 Ω 23. (a) 69 μs; (b) 35 μs 25. 20 V, 100 mA; 4.5 V, 0 A; 1 V, 0 A. 27. (a) 2.7 A; (b) 1.9 A 29. (a) 85 V; (b) 29 V; (c) 35 μs 31. (a) 750( ) 0.4 A, 0t

Li t e t−= > 33. 5 A; 2.3 A; 1.9 A 35. (a) 30 A; (b) 1.7 ms; (c) ; (d) ; (e) 600( ) 30 Ati t e−= 6001440 Vte−− 6006 14te− + A 37. (a) 290 mA; (b) 200 mA; (c) 50 mA; (d) 277 mA; (e) 34 mA

CHAPTER 8 ENGINEERING CIRCUIT ANALYSIS SELECTED ANSWERS

39. (a) -6 mA; (b) 12e-100t mA 41. (a) ; (b) 9.4 V 250,00020 V ( 0)te t− > 43. (a) 99.8 V; (b) 88e-2539t V 45. (a) 100 V, 0; (b) 100 V, 100 V; (c) 80 ms; (d) 100e-12.5t V; (e) 5e-12.5t mA; (f) , ; (g) 16 mJ, 100 mJ, 20 mJ 12.520 80Vte−− + 12.580 80Vte−− + 47. (a) 20 mA; (b) 10000 500020 2 mA, 0t te e t− −− < 49. ( ) 6 ( ) 6 ( 2) 3 ( 4) Vv t u t u t u t= − − + − 51. (a) 9; (b) 9; (c) 9; (d) 3; (e) 3 53. 1 A; 600 mA; 600 mA 55. (a) 1; (b) 12; (c) 1.47 57. 2.5 A; 3 A; 2.5 A; 2 A; -2 A 59. (a) 9.8 V; (b) 2 Ω 61. (a) 200000( ) (2 2 ) ( )mt

L e u t−= − )Vi t mA; (b) 2000006 (te u t−

63. (a) ; (b) 1000( ) 4(1 ) ( )At

Li t e u t−= − 10001( ) (100 80 ) ( )Vtv t e u t−= −

65. (a) 610

99( ) 0.95

ti t e

−= − A; (b) 1.04 A

67. 2.5 V 69. (a) 2 A; (b) 40( ) 5 3 A, 0t

Li t e t−= − > 71. (a) 80 mA; (b) ; (c) 250.08(1 )A, 0te t−− > 250.16 0.08 A, 0te t−− > ; (d) 250.016cos50 0.032sin 50 0.016 A, 0tt t e−+ − t> 73. 9000( ) 0.1 (0.1 0.1 ) ( )At

Li t e u t−= + − 75. (a) 3 A; (b) 2.4 A; (c) 2.6 A 77. (a) ; (b) 4020(1 ) ( )Ate u t−− 4010 8 ( )Ate u t−−

CHAPTER 8 ENGINEERING CIRCUIT ANALYSIS SELECTED ANSWERS

79. 0.10.94( ) 10 10cos 4 400sin 41601

tv t e t t−⎡ ⎤= − + +⎣ ⎦

81. ( )104.5 1 te−−

83. 8 3 510 /10 1010 7.5 10 7.5 mA, 0, 2.5mA 0t t

A Ai e e t i− −= + = + > = <t

t>

85. (a) ; (b) 500( ) 8 ( ) (16 24 ) ( )tcv t u t e u t−= − − + − 5000.4 ( ) (0.8 2.4 ) ( )mAtu t e u t−− + +

87. 6.32 V; 15.7 V

89. (a) 80 V; (b) ; (c) 80 V; (d) 10000080 160 V, 0te−+ 2000080 32 V, 0te t−− > 91. 693 ns

93. (a) 242 mV: (b) 3.11 mW; (c) 15 μJ 95. 1.0e-t/10 u(t) V

97. vo(t) = -0.2[1 + e-20×103t]u(t) V

99. 2.5 μF 101. (a)

103.

CHAPTER 9 ENGINEERING CIRCUIT ANALYSIS SELECTED ANSWERS

1. (a) ; (b) 22.4 krad/s; (c) overdamped. 3 175 10 s−× 1

3. (a) 5×108 s-1; (b) 32 Trad/s; (c) 0.5 32 Grad/sj− ± ; (d) underdamped 5. 1.44 H; 14 mF; 4.9 Ω 7. (a) 100 aF; (b) 1 MΩ; (c) 5 Gs-1; (d) 9 15 10 70.71 10 sj 2 1−− × + × , − × ; (e) 9 125 10 70.71 10 sj− × 57.1 10-1 −×

>

⎤⎦

t

>

9. (a) 800 rad/s (b) 954×103 s-1; (c) 95300% 11. (a) 158 mΩ; (b)

4158.5 6.31 10( ) 4.169 0.169 At ti t e e− − ×= − 13. (a) ; (b) A 10 4020 60 V, 0t te e t− −− + tt ee 40 10 120 160 −− − 15. (a) ; (b) 8.6 s, -6.1 V 0.069 0.181( ) 18 Vt tv t e e− −⎡= − −⎣ 17. 50 4502.025 0.025 A, 0t te e− −− > 19. 8 2( ) 170 42 , 0t tv t e e t− −= − 21. (a) 50 V; (b) –2 A; (c) 2000 6000( ) 25 75 , 0− −= − +t t

cv t e e t > ; (e) 270 μs; (f) 2 ms 23. R < 160 ohms 25. (a) 1.6 mΩ; (b) ( ) ( )53.2 10 63.2 10 10t

Li t e t− ×= × +

t>

tt>

t

27. (a) 8 mH: (b) 930 mA; (c) 24 ms 29. 160 ohms 31. 138.11 10 cm× 33. 4000 ( 2cos 2000 4sin 2000 )A, 0te t t− − + 35. (a) ; 5000 4 4(200cos10 100sin10 ) V, 0te t t− + > (b) 5000 4 410 (10cos10 7.5sin10 ) mA, 0te t t−− − 37. 1000.6 sin1000 mA, 0te t− > 39. R = 10.4 ohms; 2.15 s

CHAPTER 9 ENGINEERING CIRCUIT ANALYSIS SELECTED ANSWERS

41. (a,b) e-t (4 cos 5t + 0.8 sin 5t) A; (c) 4.7 s 43. 4 (10cos 2 20sin 2 )A, 0te t t− + >t

t− 45. 200 6000(2.25 0.25 ) ( ) 2 ( ) Vt te e u t u− −− + 47. (a) ; (b) 100.5 A, 0te t− > 10100 V 0te t− > 49. 4.7 kV 51. 1.5 ohms; 23 J 53. ( ) ( )2500 5 5( ) 100cos 1.6 10 1.6sin 1.6 10 Vt

Cv t e t t− ⎡ ⎤= × + ×⎣ ⎦

55. [ ]0.21( ) 13cos18 0.14sin18 Vt

Cv t e t t−= + for t > 0 and 13 V, t < 0 57. iL(t) = 10 - e-4t (20 sin 2t + 10 cos 2t) A, t > 0 59. 4000 (2cos 2000 4sin 2000 ) A, 0te t t− − >t

t t

61. 12 ( 2) V, 0te t t−− + > 63. (a) ; (b) 500 15002.5 22.5 mA, 0t te e− −− > 500 150025 22.5 mA, 0t te e− −+ > 65. (a) 30 V; (b) 51 V; (c) 44 V; (d) 44 V 67. (a) 0; (b) 0; (c) 920 mA; (d) -1.03 A 69. 1003 ohms 71. one possible solution:

CHAPTER 9 ENGINEERING CIRCUIT ANALYSIS SELECTED ANSWERS

73. (a) 1 -3.3

dv vdt

=

(b) one possible solution:

75. (a) LL 4- i

dtdi

= ;

(b) one possible solution:

CHAPTER 10 ENGINEERING CIRCUIT ANALYSIS SELECTED ANSWERS

1. (a) ; (b) 8.5( ) 8.5sin (290.9 325.0 )f t t= + ° cos (290.9 125 )t ° ; − (c) 4.875 cos 290.9 6.963sin 290.9t t+− + 3. (a) 58, 57; (b) 134o

5. 85 Mrad/s, 39 V, pi 7. (a) -6 cos (2π60t + 9o) lags 6 cos (2π60t – 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 – π) lags 9 cos (t – 3.14o) by 180 – 3.14 = 176.9o. 9. (a) 800 mV; (b) 771 mV; (c) 814 mV; (d) 805 mV 11. 13.3 cos (5t – 89.6o) μV 13. 743 cos (500t – 22o) mA 15. (a) 26 μs; (b) 10 or 26 μs; (c) 16 or 26 μs 17. 12.5cos(500t – 0.11o) mA 19. 1.4cos (400 45 ) 1.3cos (200 27 ) At t− ° + − °

21. (a) 1V sin Rm t iC

ω ω ′− = i+ ; (b) 1

2 2 2

CV 1cos tanCR1 C R

m tω ωωω

−⎛ ⎞+⎜ ⎟

+ ⎝ ⎠

23. (a) 16.8 – j 5.9; (b) – j 204; (b) 0.31 + j 1.7 25. (a) 18.7 ∠ -16o; (b) 3.2 ∠ 46o

27. (a) 39 ; (b) ; (c) 76∠ − ° 4 70∠ − ° 2.4 8.9j+ ; (d) 0.67 0.21j+ 29. (10 126 )65 Aj te + °

31. (a) 12 ; (b) ; (c) 3.20 A∠ ° 7.6 113 A∠ ° 9 108 A∠ − ° ; (d) -65 V; (d) 54 V 33. 35 mV 35. (a) 18.3 cos (5000t – 41o) V; (b) 76cos (5000 79 ) Vt + ° ; (c) 58 cos (5000 118 ) Vt + ° 37. 9.9cos (400 79 ) Vt + °

CHAPTER 10 ENGINEERING CIRCUIT ANALYSIS SELECTED ANSWERS

39. (a) –j292 Ω; (b) –j2.92 Ω; (c) –j292 mΩ; (d) –j292 nΩ 41. (a) 478 + j176 Ω; (b) 588 + j120 Ω 43. 212 cos (800t – 46o) mA 45. (a) 196 ; (b) 72 μF; (c) 11.3 and 444 rad/s 11∠ − °Ω 47. 2 1R 4.3 , R 3.2 = Ω = Ω 49. (a) 10.6 – j1.9 Ω; (b) 10 + j0.25 Ω 51. (a) 1 Ω + 4 H; (b) 5 Ω + 2 H + 500 mF; (c) 1.2 Ω + 69 mH; (d) 5 Ω 53. (a) j88 mS; (b) j8.8 S; (c) j880 S; (d) j8.8 GS 55. 2 Ω, 2 H 57. (a) 105 rad/s; (b) 105 rad/s; (c) 102 krad/s; (d) 52 krad/s, 134 krad/s 59. (a) 250 μF; (b) 100 μF 61. (a) 1 S + 250 mH; (b) 5 Ω || 1 F || 1 H; (c) 820 mΩ || 69 mF; (d) 5 Ω 63. 34 23 V∠ ° 65. 70cos(1000 45 ) Vt − ° 67. 1.2cos (100 76 ) At − °

69. (a) 1

1

C R AVV 1 A C R

fo

s f

jj

ωω

= −+ +

; (b) 1

1

C R AVV (1 A)(1 C R ) C R

fo

s f f f

jj j

ωω ω

−=

+ + +

71. 16 mW 73. 4.9 F

75. (a) ang(Vout) = ⎟⎟⎠

⎞⎜⎜⎝

⎛ −−2

1tanSmRgCj μω

- ( )

⎟⎟⎟⎟⎟⎟

⎜⎜⎜⎜⎜⎜

++′′

⎟⎟

⎜⎜

′+′+

+−−

πμμ

μπμμ

ω

ω

CC 2C R R

1-R

C

R

CC C

tan22

LS

SL1

mg

CHAPTER 10 ENGINEERING CIRCUIT ANALYSIS SELECTED ANSWERS

77. ( )12 2 −+ ωωω

j, 2 Ω, 2 H

79. 2.5 Ω, 1.25 H, 0.89 63 A∠ − ° 81. , j150 Ω 158 108 V∠ ° 83. (a) 88 cos (t – 107o) mV 85.

87. (a) (b) out24 2

S

0.802 1 6.4 10 ω−

=+ ×

VV

89. v1(t) = 3.2×10-3 cos (2×104t – 87o) + 310×10-12 cos (2×105t + 177o) V and

v2(t) = 31×10-9 cos(2×104t – 177o) + 116×10-12 cos(2×105t – 93o) V 91. , , 57 77∠ − ° 26 140∠ − ° 51 50∠ − ° , 143 13∠ ° , 51 140∠ − ° , 51 140∠ − ° 93. -40.5o, 28o

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 Ω; 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 μF; 40 μF 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

55. ( )old new

2rms

P tan - tan C =

Vθ θω

57.5.1 Arms; 1200 W; -1200 W; 1200 VA; 1200 VA; j1200 VA 59. 520∠3o kVA, 38 kVA, -j49.6 kVA, j77 kVA, 480 + j0 kVA; 520∠3o kVA; NO!; 520 kW; 28 kVAR

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

CHAPTER 12 ENGINEERING CIRCUIT ANALYSIS SELECTED ANSWERS

1. -9.3 V; -0.7 V; 9.3 V 3. Van = |Vp| ∠ 0o Vdn = |Vp| ∠ -180o

Vbn = |Vp| ∠ -60o Ven = |Vp| ∠ -240o

Vcn = |Vp| ∠ -120o Vfn = |Vp| ∠ -300o

Van = |Vp| ∠ 0o Vdn = |Vp| ∠ 180o

Vbn = |Vp| ∠ 60o Ven = |Vp| ∠ 240o

Vcn = |Vp| ∠ 120o Vfn = |Vp| ∠ 300o

5. 56.7 ∠ -11.5

o V; 190 35.0∠

o V

7.

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

o A; 34.4 -12.1∠

o A, 7.60 ∠ -109

o A, 36 ∠ 180

o A

13. 91.5 Fμ ; 6.68 kVA 15. 1040 W; 81.3 144∠ o V 17. 1.18Ω ; 282∠ 20.8 ; 450 V° 173 V∠ ° ; 15.8 – j6.00 kVA 19. 6.80 96.1 A rms∠− ° 21. 2.97 ; 52.8 W; 1990 W; 0.96 leading 17 A∠ °

CHAPTER 12 ENGINEERING CIRCUIT ANALYSIS SELECTED ANSWERS

23. 0.894; 22 μF; 541 VAR 25. 346 ; 48 ; 11.230 V∠− ° 24 j− Ω 86.6 A rms∠ ° 27. 5.48 A rms; 3.16 A rms; 240 V rms 29. 40.2 45 A rms; 60.5 170 A rms; 36 30 A rms; 4320 4320 VAj∠ ° ∠− ° ∠− ° + 31. 243 ; 24 ; 41.730 V∠ ° 1 A rms∠− ° ∠ 31 A rms− ° 33. 33.9 ; 25.2 ; 5345∠ ° 7.6 A∠− ° 157 A rms∠− ° ; 6100 3300 VAj+ 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.

CHAPTER 13 ENGINEERING CIRCUIT ANALYSIS SELECTED ANSWERS

1. M 21 663 Hμ= 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 7. ; ; 32300 3400 A/st te e− −− + 31700 4600 A/st te e− −− + 3

2 1700 4600 At ti e e− −= + 9. 4.8 W; 0 each; 0 10.4 W; − 11. 106 ; 25 W 76 j+ Ω

13. =)(tic A,)01.0(

3022 μ

+tt >0 t

15. (6 + j5ω) I

1 – j2ω I

2 – 6 I

3 = 100, -j2ω I

1 + (4 + j5ω) I

2 – j4ω I

3 = 0, -6 I

1 - j4ω I

2 +

(11 + j6ω) I3 = 0 ; I

3 = 4.32 ∠ -54

o A

17. 2 2

2 2

0.2 0.020.125 0.25 25 0.25

j⎡ ⎤ω ω

+ + ω −⎢ ⎥+ ω + ω⎣ ⎦; 2.8 1.2 j+ Ω

19. 27.3cos(10 ) V; 23.6cos(1069t + ° 66t + ° ) V; 9.6 W, 5.76 W 21. 1.3 60 A∠ − °

23. W1881.1K82.1K

K16.224

2

+−

25. 0.84 W; 0.26 W; 1.1 W 27. 1.7 ; 0.39 ∠ -8042∠ ° o; 2.2 0.05∠ °

29. 21 2

1.7L L 1j

k= −

+V

31. 4.56 ; 10 4 n j− Ω 63 j+ Ω 33. M = 5 H, L1 = 9 H, L = 11 H 2

35. 600 mH; 880 mH; 750 mH

CHAPTER 13 ENGINEERING CIRCUIT ANALYSIS SELECTED ANSWERS

37. OC SC 4

4

×

×

= ω Ω

= ω Ω

T AocT Boc

j M

j M

Z

Z; 4 8 10× ×= = − ω Ω + ω ωT A T B

SS SS Ωj M j j MZ Z ;

4 10 8

26 12 8

×

×

= − ω − ω + ω Ω

= ω ω − ω Ω

T Ain

T Bin

j j j M

j j j M

Z

Z

39. 4.9 1 0.5

jj

ωω

Ω+

41. 25 ; ; 0.62 j+ Ω 24 j Ω 25 j− Ω 43. 20 ; 20 ; 2031 j+ Ω Ω28 j+ 25 j+ Ω ; 21 24 j+ Ω 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

57. Age576Age576108.28

1201000

1IQ2

3 ××⎟⎟⎠

⎞⎜⎜⎝

⎛×+×

= ; half a century

59. You need to purchase (and wire in) a three-phase transformer rated at ( )( )( ) 6.3102083 = kVA.

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. 8e–t; 19; ; ; ; o11 38∠ °∠01 °∠01 °∠01 ; °∠988 5. 6.6 Cμ ; C 9 μ ; No. 7. ; 38.1 cos(15 60 )te t− − ° 38.1 cos(15 60 )te t− − ° ; 4.1− ; 4.1− 9. 30 ; cos (5 ; -19 V; 230 V∠ ° 236 te− 0 56 ) Vt − ° 12 50 j s−= − +s ; 2 1s 50 −− j 11. impedance R; an impedance Z L (L σ= = +s )ωj L , an impedance

Z c = 1( )

1j C Cσ ω

=+ s

; Z =100R . Ω Z L ( 2 10)(0.002) 20 101 j= − + = ∠ ° Ω ; Yes.

13. 0.35 ; 105 A∠− ° 2350 cos(10 105 ) mAte t− − ° 15. 185 ; 48 V∠− ° 3185 cos(4 48 ) Vte t− − °

17. Ks

19. 5s

; 38+s

; 0 ; Ks

21. ( )25 1 e−− s

s ; ( )25 1 e−− s

s

23. 6 38 12

e− −⎡ ⎤−⎣ ⎦+s

s ; 8

2−s6 31 e e−⎡ ⎤−⎣ ⎦

s ; 6 38 12

e− −⎡ ⎤−⎣ ⎦+s

s

25. u(t); te 3− )(tδ ; t u(t) ; 275 )(tδ ; u(t) 27. 0 ; ; .047 0.11j+ 0.18 0.20j− + 3(0.47 6.5) 10j −+ ×

29. 2 51 ( )e e− −−s s

s ; 24 e− s

s; 2 64( )

3e− −=

+sF s

s ; 24e− s ; 2.9

31. ( )s3s

s2 −− − ee ; se

s42 − ; 84

23 −−

+se

s ; ; se 53 − 4e−− s

33. 90 ( ) 4.5 ( )t u t− ( ) 2 ( )t u t; 11 + ; ; ( )tte u t− 2 31 1 ( )2 2

t t te e e u t− − −⎡ ⎤− +⎢ ⎥⎣ ⎦ δ δ

CHAPTER 14 ENGINEERING CIRCUIT ANALYSIS SELECTED ANSWERS

35. 2.5 mA 37. ; )(2)()( tuetut t−++δ )()1(2)2( ttt δδδ +−+− ; ; )1(2 1 −− te δ )5()1( −+− tt δδ 39. ; ; ; ;

)(5 tue t− )()25( 4 tuee tt −− − )()(6 4 tuee tt −− − )()4(6 4 tuee tt −− −

)()16(6)(18 4 tueet tt −− −+δ 41. 2 ; ; )( 3)( tuetu t−− )(4)(2 3 tuet t−+δ )8.0(3 −tδ ; ;

)()(3 62 tuee tt −− −

)()75.075.03( 622 tueete ttt −−− +−

43. =)(tf )(32)(

32)( 3 tuetut t−−+δ ; ( ) 0.5 ( ) 0.25 ( ) 0.35cos(2 135 )f t tu t u t t= + + + °

45. ; )()()( 2 tuetth t−−= δ [ ] )(2)( 2 tueeth tt −− −= ;

)(481)(

49)(

21)(6)(2)( 3 tuetuetutett

dtdth ttt −−− −+−+= δδ

47. 4 /3( ) (1.9 5.59 ) ( )tf t e−= − u t

49. 50 V; ; 0)20(1.02.01.0 ' =−++ ccc vvv V )(3

1303

20)( 3 tuetv tc ⎟

⎠⎞

⎜⎝⎛ += −

51. )()24( 15.0 tue t−− 53. ; )()62()( tuety t−+= )()16( tue t −−

55. -600 mA; ; 5010040 += ci ∫∞

−+

0100dtic )(6.0 5.0 tue t−−

57. 1R 250 m , C 1 F, L H3

= Ω = = ; V )()5.625.1275()( 53 tueeetv ttt −−− −−=

59. A )(6.1)(4.0)( 2 tuetti tc

−−+ δ 61. STABLE; UNSTABLE. 63. STABLE; STABLE. 65. 7 V ; 7 V 67. 2, 3, 0, 0, ∞

69. asbsn

++ ; ,ab − ab −

CHAPTER 15 ENGINEERING CIRCUIT ANALYSIS SELECTED ANSWERS

1.

= 0.032s Ω

= 384 μV

0.032s Ω 12s

mA

2

2

20 11,000 200,000700 100,000

+ ++ +

s ss s

2

2

20 11,000 200,000700 100,000

+ ++ +

s ss s

20020+s

s500

1000+s; ; 8.1 54 ∠ ° Ω3. ; ; ; 11 − Ω

2

2

16 50 400080

+ ++

s ss s

5. ; 0.16 ; 4.7 j− Ω 6.85 114 ∠− ° Ω 910 mΩ; ; 1 Ω

( )B L

2L B m L B B B L L B B

r R 1 + Cr R C C + (g r R C + r R C + r R C + r C + R C ) + r + R

π μ

π π μ π μ π π π μ π μ μ π

Z sZ s Z Z Z s

7.

-0.28t u(t) μA 9. 4.5 e

11. v1(t) = –5.6e–6.67t + 3.6 V, t ≥ 0

–6.67t v2(t) = –3.73e + 4.4 V, t ≥ 0

41

2 1( )3 6

ti t e−= − 4

22 1( )3 12

ti t e−= +A, t ≥ 0 and A, t ≥ 0 13.

( ) ( )2 2

35 - 131

2 100 6 + 100⎡ ⎤ ⎡+ + +⎣ ⎦ ⎣

s

s s15.

⎤⎦

;

e-6t -2t [0.092cos 2t - 1.5 sin 2t] - e [0.092 cos10t - 0.34 sin 10t] A

-0.61t -0.55t cos(0.34t + 99o)] u(t) + 0.79 e17. [0.63e

2

4 3 2

200 ( + 9 + 12)2 + 17 + 90 + 185 + 250

s s ss s s s

-3t19. ; [185 e cos (4t - 48o) + 86e-1.25t cos (1.9t + 107o)] u(t)

21. I1 = 271.7∠-96.5o A and I = 272∠-96.5o A 2

CHAPTER 15 ENGINEERING CIRCUIT ANALYSIS SELECTED ANSWERS

23. 2[1.301 e-142.8t cos (742.3t + 12.54o) + 0.00202 cos (2t – 6.538o -5 δ(t) ) – 6.601×10 - 1.564 e-142.8t cos (742.3t – 33.56o) - 2.998 cos (2t + 179.9o)]2 W 25. (a)

(b)

( )6

2 5

7.5 10 + 1500 V + 5000 + 5 10×

×s

s s s2

2500 + 0.5 0.001 + 5 + 500

Ωs

s s,

(c) [-3 e-2.5×106t -0.2t+ 3 e + 3×10-3 + 21 cos(711t + 89.9o)] u(t) 27.

4 3 2

2

420 133 21 60 960 19 3

+ + + +Ω

+ +s s s s

s s2

70 V60 19 3+ +s s 4 3 2

70 A420 133 21 60 9+ + + +s s s s

; ;

29.

13 13 2

1 10 3 15 2 18 18

30303(0.2239 10 + 0.1613 10 + 98700 ) = (0.4639 10 + 0.7732 10 + 0.5691 10 + 0.1936 10 )

× ×× × ×

s sVs s s s ×

,

3 12 2 14 14

2 10 3 15 2 18 18

7609(705000 + 0.1175 10 + 0.6359 10 + 0.8897 10 ) = (0.4639 10 + 0.7732 10 + 0.5691 10 + 0.1936 10 )

× ×× × ×

s s sVs s s s

××

;

-165928t v1(t) = [3.504 + 0.3805×10-2 e – 0.8618 e-739t -0.3404t] u(t) V, – 2.646 e

-165928t v2(t) = [3.496 – 0.1365×10-2 e + 0.309 e-739t – 2.647 e-0.3404t] u(t) V

31.

3 14 2 17

3 10 2 13

174.639 109 + 7.732 10 + 5.691 10 + 1.936 1098700 + 1.645 10 + 1.21 10 + 2.059 10

× × ×× × ×

s s ss s s 12

×Ω ,

12

9 3 14 2 17 17

2.059 10 (5 - 3)(4.639 10 + 7.732 10 + 5.691 10 + 1.936 10 )

×× × × ×

ss s s s

, 201 μA

CHAPTER 15 ENGINEERING CIRCUIT ANALYSIS SELECTED ANSWERS

( )( )2 6

5000 + 10 + 6385

ss s

33. ; [-0.76 e-6385t + 0.77 cos (103 ot – 8.9 )] u(t) A;

0.77 cos (103t – 8.9o) A.

1 31, , double at = 04 4

j± − ± s35. Poles at ; zeroes at s = 0, .; Poles at s = , Zeroes at 2, 1j± ∞

1 2, j− ± ∞ 37. (a.)

-1 zeros at s = -25 and -12.5 s , and -1 poles at s = 0 and s = -1.7 s .

(b.) zeros at s = -9. 1 and -105 -1 s , and poles at s = -1.55×105 and s = -3.2 s-1.

)5.1(6)4)(1(5

+++

=s

ssZ in39. ; ; 1s 4 ,1 −−−∞− ,5.1

)cos1(10 tππ

−π20 , (10/ π) (1 + cos πt), 0 41. 0, ,

43. V, 16 V, zero 88 −t 45.

ss ee 422 153015 −− +−

ss-215 - 15e s

sse 2

2

1515 −−ss

, 15 u(t) – 15 u(t - 2), , 15 t u(t) – 15 u (t - 2), , 15 t

u(t) – 30 u (t - 2) + 15 u(t), -22 2

15 15 - e + 9 + 9

sss s

, 5 sin 3t u(t) – 15 cos [3(t – 2)] u(t - 2),

47. h(t) = δ(t); V )(8 tue t−

49. (a) (b)

(c) 5

CHAPTER 15 ENGINEERING CIRCUIT ANALYSIS SELECTED ANSWERS

51.

; 4.7, 10; 15 cm

3 2

2

10 30 10 306 13

+ − −+ +

s s ss s

117∠ ° 3.6 34∠ ° 53∠ °53. ; -2.3, ∞ ; 2.2 , 2.2 63 ,∠ ° , 5 , °∠03

55.

256 +− ωω

424

2 +ω; ωω

ω2)5( j+−)2(100

2

j+ ; 100 ; graph; 2 rad/s, 69

57. A (all t); A )(]4sin34cos5(5[ 26 tuttee tt ++− −−te 65 −−

-24t -1.7t59. ; i (t) = [10 – 2.1e – 0.885 e ] u(t) A -11.7 and 24 s− − in 61.

5.275.65.2

2 ++ ss ; [1 + 0.066 e-6.4t -0.39t ] u(t) V – 1.1e

s510R 5+

5

5

101010

++

ss

5105)(

+−

=s

ss ; ; 63. H

65. 0 F, 400 ; 5 nF, R1 = 200 kΩ ; 50 nF, 200 kΩ ; RΩ 1=fA kΩ , C nF, R , R R 10k

10=fA

=B1 = Ω Ω= 100fB A1

67. One possible design: If we use a 1-μF capacitor, then R = 159 Ω. To complete the design, select Rf = 2 kΩ and R = 1 kΩ. 1 69. One possible design: If we use 100-nF capacitors, then R = 3.167 kΩ.

CHAPTER 17 ENGINEERING CIRCUIT ANALYSIS SELECTED ANSWERS

1. (a) ; (b) 651; (c) 21; (d) 600 mA; (e) -141 mA 1

2

3

4 8 9 125 0 7 47 3 1 0

−⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥− =⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

III

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

3. (a) 390 W; (b) 250 W; (c) 180 W

5. 22 15 2

2 5+ +

+s s

s0

7. 15 25( 4)++

ss s

9. (a) 1 3 4

2

R R RCR

jω ; (b) 30.8 10 (L 0.8 mH)injω × Ω =

11. 142 mS, -77 mS 13. 40 mS, -40 mS, 40 mS, -30 mS 15. (a) 50 Ω; (b) 60 Ω 17. 200 mS, -300 mS, -400 mS, 150 mS 19. (a) Input is applied between g-s and output taken from d-s;

(b) ( )gs gdj C Cω + , gdj Cω− , m gdg j Cω− , ( )1gs gd

d

j C Cr

ω+ + ;

(c) 4.8 pSj ω , 1.4 pSj ω− , , 3 14.7 10 (1.4) 10 Sjω− −× − × 2

2 4 110 (0.4 1.4) 10 Sjω− −+ + × 21. 784 Ω, 1.72 kΩ, 367 Ω 23. 9.9 Ω 25. (a) 56; (b) -9.6; (c) 530; (d) 3.4 Ω; (e) 35 Ω 27. (a) (b)

CHAPTER 17 ENGINEERING CIRCUIT ANALYSIS SELECTED ANSWERS

29. (a) 1.55 V; (b) -17.5 μA

31. 7.5 1.1

( )4.5 11

⎡ ⎤Ω⎢ ⎥−⎣ ⎦

33. (a) -2; (b) 4; (c) 8; (d) 1 Ω; (e) 1.3 Ω

35. 11

12

133 4894 2.6

zz

= ∠− °Ω= ∠− °Ω

21

22

9400 87565 3.6

zz

= ∠ °Ω= ∠− °Ω

37. (a) ; (b) 10 220 0.2 SΩ −⎡ ⎤

⎢ ⎥⎣ ⎦

42 1.717 0.17 SΩ −⎡ ⎤

⎢ ⎥⎣ ⎦

39. (a) 1.2; (b) 9.6 Ω; (c) 240 mS

41. (a) ; (b) 8.6 kΩ 4

1000 0.0110 2 10 S−

Ω⎡ ⎤⎢ ⎥×⎣ ⎦

43. (a) ( )( )

( ) ( )1 1

1 1md

j C j r C j r Cg

rj r C C j r C Cμ π π π μ

π π μ π π μ

ω ω ωω ω

++ +

+ + + +; (b)

( )( )1

mg j C r

j r C Cμ π

π π μ

ω

ω

+ +

(c) ( )1xrr

j r C Cπ

π π μω+

+ +; (d) ( )1

j C rj r C C

μ π

π π μ

ωω+ +

45. (a) ; (b) 11 Ω 0.61 3.3

0.053 S 0.81Ω⎡ ⎤

⎢ ⎥⎣ ⎦

47. (a) , 1.4 2

0.2 S 1Ω⎡ ⎤

⎢ ⎥⎣ ⎦

1.5 31 S 16

Ω⎡ ⎤⎢ ⎥⎢ ⎥⎣ ⎦

, 11/ 7 41/ 7 S 1

Ω⎡ ⎤⎢ ⎥⎣ ⎦

; (b) 4.71 15.9

0.962 S 3.47Ω⎡ ⎤

⎢ ⎥⎣ ⎦

49. (a) , , 1 0

1/ R 1⎡ ⎤⎢ ⎥⎣ ⎦

1 R0 1⎡ ⎤⎢ ⎥⎣ ⎦

1/ 00a

a⎡ ⎤⎢ ⎥⎣ ⎦

; (b) 0.58 14

0.115 S 4.5Ω⎡ ⎤

⎢ ⎥⎣ ⎦