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

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- 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. Engineering Circuit Analysis, 7th Edition Chapter Two So