Circuitos
Páginas: 33 (8080 palabras)
Publicado: 26 de febrero de 2013
CHAPTER 2 ENGINEERING CIRCUIT ANALYSIS 1. (a) 12 μs (b) 750 mJ (c) 1.13 kΩ (d) 3.5 Gbits (e) 6.5 nm (f) 13.56 MHz (g) 39 pA (h) 49 kΩ (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 C12 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 Ohm’s law. 500 μA, 2.5 mW; -500 μA, 2.5 mW; -500 μA, 2.5 mW; 500μA, 2.5 mW -2 V (at t = 0.324 s) 2 km. Hmmmm…. 1.7 μΩ.cm 560 mΩ, 1.3 W 266 mΩ; 514 mA Design. Many possible solutions. Hint: Start with finding resistivity, then choose geometry.
Copyright ©2007 The McGraw-Hill Companies, Inc. All Rights Reserved.
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 Ω; 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 Vv5 = 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 kΩ; (b) R = 0 -250 cos 5tmV
Copyright ©2007 The McGraw-Hill Companies, Inc. All Rights Reserved.
CHAPTER 3 ENGINEERING CIRCUIT ANALYSIS 35. (a) P5A P100Ω P25Ω Pdep (b) P5A P100Ω P25Ω Pdep 37. P8A P6Ω P8A P12Ω P4Ω 39. 41. 43. 45. 47. 49. 51. 53. 55. = –8 vx = (vx)2 / 6 = –7 vx = (vx) / 12 = (vx) / 4
2 2
SELECTED ANSWERS
= –5 vx = (vx) / 100 = (vx) / 25 = –vx(0.8 ix) = –0.8 (vx)2 / 25
2 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 kΩ (a) 10 kΩ || 10 kΩ; (b) 47 kΩ + 10 kΩ + 1 kΩ || 1kΩ || 1kΩ; (c) 47 kΩ || 47kΩ + 10 kΩ || 10 kΩ + 1 kΩ 5.5 kΩ 60 Ω; 213 Ω; 52 Ω
57. 59.
Copyright ©2007 The McGraw-Hill Companies, Inc. All Rights Reserved.
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 kΩ, 1 kΩ 139 μA; 868 μW 18 μW
R 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 3
vout = -56.02 sin 10t V
83.
Copyright ©2007 The McGraw-Hill Companies, Inc. All Rights Reserved. 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 =...
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 C12 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 Ohm’s law. 500 μA, 2.5 mW; -500 μA, 2.5 mW; -500 μA, 2.5 mW; 500μA, 2.5 mW -2 V (at t = 0.324 s) 2 km. Hmmmm…. 1.7 μΩ.cm 560 mΩ, 1.3 W 266 mΩ; 514 mA Design. Many possible solutions. Hint: Start with finding resistivity, then choose geometry.
Copyright ©2007 The McGraw-Hill Companies, Inc. All Rights Reserved.
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 Ω; 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 Vv5 = 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 kΩ; (b) R = 0 -250 cos 5tmV
Copyright ©2007 The McGraw-Hill Companies, Inc. All Rights Reserved.
CHAPTER 3 ENGINEERING CIRCUIT ANALYSIS 35. (a) P5A P100Ω P25Ω Pdep (b) P5A P100Ω P25Ω Pdep 37. P8A P6Ω P8A P12Ω P4Ω 39. 41. 43. 45. 47. 49. 51. 53. 55. = –8 vx = (vx)2 / 6 = –7 vx = (vx) / 12 = (vx) / 4
2 2
SELECTED ANSWERS
= –5 vx = (vx) / 100 = (vx) / 25 = –vx(0.8 ix) = –0.8 (vx)2 / 25
2 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 kΩ (a) 10 kΩ || 10 kΩ; (b) 47 kΩ + 10 kΩ + 1 kΩ || 1kΩ || 1kΩ; (c) 47 kΩ || 47kΩ + 10 kΩ || 10 kΩ + 1 kΩ 5.5 kΩ 60 Ω; 213 Ω; 52 Ω
57. 59.
Copyright ©2007 The McGraw-Hill Companies, Inc. All Rights Reserved.
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 kΩ, 1 kΩ 139 μA; 868 μW 18 μW
R 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 3
vout = -56.02 sin 10t V
83.
Copyright ©2007 The McGraw-Hill Companies, Inc. All Rights Reserved. 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 =...
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