Anlalisis de circuitos en ingeniería. SOLUCIONARIO HAYT 7ma ed.

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Transcript of Anlalisis de circuitos en ingeniería. SOLUCIONARIO HAYT 7ma ed.

  1. 1. Engineering Circuit Analysis, 7th Edition Chapter Two Solutions 10 March 2006 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 PROPRIETARY MATERIAL . 2007 Th e McGraw-Hill Companies, Inc. Lim ited distr ibution perm itted onl y to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
  2. 2. Engineering Circuit Analysis, 7th Edition Chapter Two Solutions 10 March 2006 2. (a) 1 MW (e) 33 J (i) 32 mm (b) 12.35 mm (f) 5.33 nW (c) 47. kW (g) 1 ns (d) 5.46 mA (h) 5.555 MW PROPRIETARY MATERIAL . 2007 Th e McGraw-Hill Companies, Inc. Lim ited distr ibution perm itted onl y to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
  3. 3. Engineering Circuit Analysis, 7th Edition Chapter Two Solutions 10 March 2006 3. (a) ( ) 745.7 W 400 Hp = 1 hp 298.3 kW (b) 12 ft = 12 in 2.54 cm 1 m (12 ft) 3.658 m 1 ft 1 in 100 cm = (c) 2.54 cm = 25.4 mm (d) ( ) 1055 J 67 Btu = 1 Btu 70.69 kJ (e) 285.410-15 s = 285.4 fs PROPRIETARY MATERIAL . 2007 Th e McGraw-Hill Companies, Inc. Lim ited distr ibution perm itted onl y to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
  4. 4. Engineering Circuit Analysis, 7th Edition Chapter Two Solutions 10 March 2006 4. (15 V)(0.1 A) = 1.5 W = 1.5 J/s. 3 hrs running at this power level equates to a transfer of energy equal to (1.5 J/s)(3 hr)(60 min/ hr)(60 s/ min) = 16.2 kJ PROPRIETARY MATERIAL . 2007 Th e McGraw-Hill Companies, Inc. Lim ited distr ibution perm itted onl y to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
  5. 5. Engineering Circuit Analysis, 7th Edition Chapter Two Solutions 10 March 2006 5. Motor power = 175 Hp (a) With 100% efficient mechanical to electrical power conversion, (175 Hp)[1 W/ (1/745.7 Hp)] = 130.5 kW (b) Running for 3 hours, Energy = (130.5103 W)(3 hr)(60 min/hr)(60 s/min) = 1.409 GJ (c) A single battery has 430 kW-hr capacity. We require (130.5 kW)(3 hr) = 391.5 kW-hr therefore one battery is sufficient. PROPRIETARY MATERIAL . 2007 Th e McGraw-Hill Companies, Inc. Lim ited distr ibution perm itted onl y to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
  6. 6. Engineering Circuit Analysis, 7th Edition Chapter Two Solutions 10 March 2006 6. The 400-mJ pulse lasts 20 ns. (a) To compute the peak power, we assume the pulse shape is square: 400 Energy (mJ) t (ns) 20 Then P = 40010-3 /2010-9 = 20 MW. (b) At 20 pulses per second, the average power is Pavg = (20 pulses)(400 mJ/pulse)/(1 s) = 8 W. PROPRIETARY MATERIAL . 2007 Th e McGraw-Hill Companies, Inc. Lim ited distr ibution perm itted onl y to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
  7. 7. Engineering Circuit Analysis, 7th Edition Chapter Two Solutions 10 March 2006 7. The 1-mJ pulse lasts 75 fs. (a) To compute the peak power, we assume the pulse shape is square: 1 Energy (mJ) t (fs) 75 Then P = 110-3 /7510-15 = 13.33 GW. (b) At 100 pulses per second, the average power is Pavg = (100 pulses)(1 mJ/pulse)/(1 s) = 100 mW. PROPRIETARY MATERIAL . 2007 Th e McGraw-Hill Companies, Inc. Lim ited distr ibution perm itted onl y to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
  8. 8. Engineering Circuit Analysis, 7th Edition Chapter Two Solutions 10 March 2006 8. The power drawn from the battery is (not quite drawn to scale): 5 7 17 24 P (W) 10 6 t (min) (a) Total energy (in J) expended is [6(5) + 0(2) + 0.5(10)(10) + 0.5(10)(7)]60 = 6.9 kJ. (b) The average power in Btu/hr is (6900 J/24 min)(60 min/1 hr)(1 Btu/1055 J) = 16.35 Btu/hr. PROPRIETARY MATERIAL . 2007 Th e McGraw-Hill Companies, Inc. Lim ited distr ibution perm itted onl y to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
  9. 9. Engineering Circuit Analysis, 7th Edition Chapter Two Solutions 10 March 2006 9. The total energy transferred during the first 8 hr is given by (10 W)(8 hr)(60 min/ hr)(60 s/ min) = 288 kJ The total energy transferred during the last five minutes is given by 300 s 0 10 + 10 300 t dt = 300 2 0 10 + 10 = 600 t t 1.5 kJ (a) The total energy transferred is 288 + 1.5 = 289.5 kJ (b) The energy transferred in the last five minutes is 1.5 kJ PROPRIETARY MATERIAL . 2007 Th e McGraw-Hill Companies, Inc. Lim ited distr ibution perm itted onl y to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
  10. 10. Engineering Circuit Analysis, 7th Edition Chapter Two Solutions 10 March 2006 10. Total charge q = 18t2 2t4 C. (a) q(2 s) = 40 C. (b) To find the m aximum charge within 0 t 3 s, we need to take the first and second derivitives: dq/dt = 36t 8t3 = 0, leading to roots at 0, 2.121 s d2 q/dt2 = 36 24t2 substituting t = 2.121 s into the expression for d2 q/dt2 , we obtain a value of 14.9, so that this root represents a maximum. Thus, we find a maximum charge q = 40.5 C at t = 2.121 s. (c) The rate of charge accumulation at t = 8 s is dq/dt|t = 0.8 = 36(0.8) 8(0.8)3 = 24.7 C/s. (d) See Fig. (a) and (b). 0 0.5 1 1.5 2 2.5 3 -20 -10 0 10 20 30 40 50 60 70 time (s) q(C) 0 0.5 1 1.5 2 2.5 3 -150 -100 -50 0 50 i (A) time(t) (a) (b) PROPRIETARY MATERIAL . 2007 Th e McGraw-Hill Companies, Inc. Lim ited distr ibution perm itted onl y to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
  11. 11. Engineering Circuit Analysis, 7th Edition Chapter Two Solutions 10 March 2006 11. Referring to Fig. 2.6c, 0A,32- 0A,32- )( 3 5 1 >+ 0 separately: for t < 0, - 2 + 3e-5t = 0 leads to t = -0.2 ln (2/3) = +0.0811 s (impossible) for t > 0, -2 + 3e3t = 0 leads to t = (1/3) ln (2/3) = 0.135 s (impossible) Therefore, the current is never negative. (d) The total charge passed left to right in the interval 0. 8 < t < 0.1 s is q(t) = 0.1 10.8 ( )i t dt = 0 0.1 5 3 0.8 0 2 3 2 3t t e dt e d + + + t = ( ) 0 0.1 5 3 0 -0.8 3 2 2 5 t t t e t e + + = 33.91 C PROPRIETARY MATERIAL . 2007 Th e McGraw-Hill Companies, Inc. Lim ited distr ibution perm itted onl y to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
  12. 12. Engineering Circuit Analysis, 7th Edition Chapter Two Solutions 10 March 2006 12. Referring to Fig. 2.28, (a) The average current over one period (10 s) is iavg = [-4(2) + 2(2) + 6(2) + 0(4)]/10 = 800 mA (b) The total charge transferred over the interval 1 < t < 12 s is = 12 1 total )( dttiq = -4(1) + 2(2) + 6(2) + 0(4) 4(2) = 4 C (c) See Fig. below 2 4 6 8 10 12 16 q (C) t(s) -16 8 16 -8 14 PROPRIETARY MATERIAL . 2007 Th e McGraw-Hill Companies, Inc. Lim ited distr ibution perm itted onl y to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
  13. 13. Engineering Circuit Analysis, 7th Edition Chapter Two Solutions 10 March 2006 13. (a) VBA = -19 2 pJ -1.602 10 C = 12.48 MV (b) VED = -19 0 -1.602 10 C = 0 (c) VDC = -19 3 pJ 1.602 10 C = 18.73 MV (d) It takes 3 pJ to move +1.602x1019 C from D to C. It takes 2 pJ to move 1.602x1019 C from B to C, or 2 pJ to move +1.602x10 19 C from B to C, or +2 pJ to move +1.602x1019 C from C to B. Thus, it requires 3 pJ + 2 pJ = 1 pJ to m ove +1.602x1019 C from D to C to B. Hence, VDB = -19 1 pJ 1.602 10 C = 6.242 MV. PROPRIETARY MATERIAL . 2007 Th e McGraw-Hill Companies, Inc. Lim ited distr ibution perm itted onl y to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
  14. 14. Engineering Circuit Analysis, 7th Edition Chapter Two Solutions 10 March 2006 14. + Voltmeter + V1 - Voltmeter + V2 + From the diagram, we see that V2 = V1 = +2.86 V. PROPRIETARY MATERIAL . 2007 Th e McGraw-Hill Companies, Inc. Lim ited distr ibution perm itted onl y to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
  15. 15. Engineering Circuit Analysis, 7th Edition Chapter Two Solutions 10 March 2006 15. (a) Pabs = (+3.2 V)(-2 mA) = 6.4 mW (or +6.4 mW supplied) (b) Pabs = (+6 V)(-20 A) = 120 W (or +120 W supplied) (d) Pabs = (+6 V)(2 ix) = (+6 V)[(2)(5 A)] = +60 W (e) Pabs = (4 sin 1000t V)(-8 cos 1000t mA)| t = 2 ms = +12.11 W PROPRIETARY MATERIAL . 2007 Th e McGraw-Hill Companies, Inc. Lim ited distr ibution perm itted onl y to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
  16. 16. Engineering Circuit Analysis, 7th Edition Chapter Two Solutions 10 March 2006 16. i = 3te-100t mA and v = [6 600t] e-100t mV (a) The power absorbed at t = 5 ms is Pabs = ( )[ ] W36006 5 100100 mst tt teet = = 0.01655 W = 16.55 nW (b) The energy delivered over the interval 0 < t < is ( ) = 0 0 200 J60063 dtettdtP t abs Making use of the relationship 10 ! + = n axn a n dxex where n is a positive integer and a > 0, we find the energy delivered to be = 18/(200)2 - 1800/(200)3 = 0 PROPRIETARY MATERIAL . 2007 Th e McGraw-Hill Companies, Inc. Lim ited distr ibution perm itted onl y to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
  17. 17. Engineering Circuit Analysis, 7th Edition Chapter Two So