solucionario Análisis de circuitos en ingeniería 7ed (hayt, kemmerly)

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solucionario Análisis de circuitos en ingeniería 7ed (hayt, kemmerly), las respuestas pero no el procedimiento

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  • 1.CHAPTER 2 ENGINEERING CIRCUIT ANALYSIS SELECTED ANSWERS 1. (a) 12 s (d) 3.5 Gbits (g) 39 pA (b) 750 mJ (e) 6.5 nm (h) 49 k (c) 1.13 k (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 = I2 R1 + I2 R2. Now invoke Ohms law. 33. 500 A, 2.5 mW; -500 A, 2.5 mW; -500 A, 2.5 mW; 500 A, 2.5 mW 35. -2 V (at t = 0.324 s) 37. 2 km. Hmmmm. 39. 1.7 . cm 41. 560 m, 1.3 W 43. 266 m; 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.

2. 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 ; 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 k; (b) R = 0 33. -250 cos 5t mV Copyright 2007 The McGraw-Hill Companies, Inc. All Rights Reserved. 3. CHAPTER 3 ENGINEERING CIRCUIT ANALYSIS SELECTED ANSWERS 35. (a) P5A = 5 vx = 1.389 kW P100 = (vx)2 / 100 = 771.7 W P25 = (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 P100 = (vx)2 / 100 = 240.9 W P25 = (vx)2 / 25 = 963.5 W Pdep = vx(0.8 iy) = 428.1 W 37. P8A = 8 vx = 240 W P6 = (vx)2 / 6 = 150 W P8A = 7 vx = 210 W P12 = (vx)2 / 12 = 75 W P4 = (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 k 55. (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 57. 5.5 k 59. 60 ; 213 ; 52 Copyright 2007 The McGraw-Hill Companies, Inc. All Rights Reserved. 4. 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 k, 1 k 73. 139 A; 868 W 75. 18 W 77. (a) ( ) )R(RRRRRR )R(RR V 4324321 432 S ++++ + ; (b) ( ) )R(RRRRRR )RR(RR V 4324321 4321 S ++++ ++ ; (c) ( ) )R(RRRRRR R V 4324321 2 S ++++ . 79. (a) 42 A; (b) 11.9 V; (c) 0.238 81. ++++ )R(RR)RR(RR RR V 5435432 53 S 83. vout = -56.02 sin 10t V Copyright 2007 The McGraw-Hill Companies, Inc. All Rights Reserved. 5. 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 V 7.1 V 7.5 V 4.9 V v v v v = = = = 5 6 7 8 1.7 V 3.8 V 3.5 V 2.4 V v v v v = = = = 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 A, i2 = 1.08 mA, i3 = -1.20 mA and i4 = -480 A Copyright 2007 The McGraw-Hill Companies, Inc. All Rights Reserved. 6. CHAPTER 4 ENGINEERING CIRCUIT ANALYSIS SELECTED ANSWERS 37. (a) -5700 ; (b) this value is unique. 39. (a) 330 A; (b) 330 A; (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 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 = 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. Copyright 2007 The McGraw-Hill Companies, Inc. All Rights Reserved. 7. 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 W 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 and in series with 10 ; (b) 26 V; (c) Cannot remove the resistor across which v appears or v may become lost. 23. 10 mW 25. 33 W 27. (a) 12.8 mV 29. 764 nA 31. Current source is 7.25 A, resistor is 2 ohms. 33. 1.57 V, 811 m 35. The final circuit is an 8.5 V voltage source in series with a 2.0 M resistor. 37. (a) An 8/5 A current source in parallel with 5 , in parallel with RL. Copyright 2007 The McGraw-Hill Companies, Inc. All Rights Reserved. 8. CHAPTER 5 ENGINEERING CIRCUIT ANALYSIS SELECTED ANSWERS 39. -2 V 41. (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. 43. (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. 45. (a) 6.7 , -300 mA, arrow upwards; (b) 6.7 , -150 mA, arrow upwards. 47. (a) 38.9 V, 178 ; (b) 1.96 W. 49. VTH = 0, RTH = 192 . 51. 15 , 15 53. VTH = 0; The Norton equivalent is 0 A in parallel with 1.3 . 55. VTH (and hence IN) = 0; RTH = RN = 198 m. 57. 2 M 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 , 6.3 W 63. 65 V, 15 , 70 W 65. (a) 200 V; (b) 125 W; (c) 80 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 i2 RLOAD = 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 k Copyright 2007 The McGraw-Hill Companies, Inc. All Rights Reserved. 9. CHAPTER 5 ENGINEERING CIRCUIT ANALYSIS SELECTED ANSWERS 71. 1.2 , 0.54 , 4.9 73. 9.9 75. 5.5 V, 1.0 77. -13 V, 27 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 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 . Copyright 2007 The McGraw-Hill Companies, Inc. All Rights Reserved. 10. 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 k and Rin = 5.1 k. 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 k and Rin = 5.1 k 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 , and a resistor R1 = 1 . 17. V1 = 21 V 19. ; -5.6 V(out -4 1 sin 3 Vv t= + ) 21. Rf = 236 k and R1 = 1 k. 23. (a) B must be the non-inverting input; (b) Choose R2 = RB = 1 ; (c) A is the inverting input. 25. vout(0.25 s) = 0.93 V 27. 4.2 V 29. = N 1 f R R- i i iv 31. Pick R1 = 10 k. Then vS = -0.21 V. Copyright 2007 The McGraw-Hill Companies, Inc. All Rights Reserved. 11. CHAPTER 6 ENGINEERING CIRCUIT ANALYSIS SELECTED ANSWERS 33. One possible solution of many: 35. Set R = 10 k: Then connect several into: after setting Rf2 = Rf1 = Rin = R =10 k. 37. 1 kV 39. -179 kV 41. 1.7 V 43. Rf = 0, Rin = 100 k, R2 = 51 . Copyright 2007 The McGraw-Hill Companies, Inc. All Rights Reserved. 12. CHAPTER 6 ENGINEERING CIRCUIT ANALYSIS SELECTED ANSWERS 45. Rf = 120 k and Rin = 200 k, R = 560 . 47. R = 400 , R1 = 82 . I Is 49. R = 91 , R1 = 560 , 467 > RL > 67 . 51. (a) 3.7 mV; (b) 28 mV; (c) 3.7 V. 53. A101 100A- in out + = v v ; A = 9999. 55. vout = -16 mV Copyright 2007 The McGraw-Hill Companies, Inc. All Rights Reserved. 13. CHAPTER 6 ENGINEERING CIRCUIT ANALYSIS SELECTED ANSWERS 57. (a) (b) vout = 105 (-0.00004v2 - 9.9998010-6 v1)+5v2 = 1.00008