Solucionario de Análisis de Circuitos en Ingeniería 7ma edicion Hayt&Kemmerly

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Transcript of Solucionario de Análisis de Circuitos en Ingeniería 7ma edicion Hayt&Kemmerly

  • 1.Engineering Circuit Analysis, 7th Edition1. (a)12 s (b) 750 mJ (c) 1.13 k(d) 3.5 Gbits (e) 6.5 nm (f) 13.56 MHzChapter Two Solutions10 March 2006(g) 39 pA (h) 49 k (i) 11.73 pAPROPRIETARY 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 Edition2.(a) 1 MW (b) 12.35 mm (c) 47. kW (d) 5.46 mA(e) 33 J (f) 5.33 nW (g) 1 ns (h) 5.555 MWChapter Two Solutions10 March 2006(i) 32 mmPROPRIETARY 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 Edition3. (a)Chapter Two Solutions 745.7 W = 298.3 1 hp ( 400 Hp ) 10 March 2006kW 12 in 2.54 cm 1 m (b) 12 ft = (12 ft) = 3.658 m 1 ft 1 in 100 cm (c) 2.54 cm =25.4 mm 1055 J (d) ( 67 Btu ) = 70.69 1 Btu (e) 285.410-15 s =kJ285.4 fsPROPRIETARY 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 Edition4.Chapter Two Solutions10 March 2006(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 kJPROPRIETARY 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 Edition5.Chapter Two Solutions10 March 2006Motor 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 Edition6.Chapter Two Solutions10 March 2006The 400-mJ pulse lasts 20 ns. (a) To compute the peak power, we assume the pulse shape is square: Energy (mJ) 40020Thent (ns)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 Edition7.Chapter Two Solutions10 March 2006The 1-mJ pulse lasts 75 fs. (a) To compute the peak power, we assume the pulse shape is square: Energy (mJ) 175Thent (fs)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 Edition8.Chapter Two Solutions10 March 2006The power drawn from the battery is (not quite drawn to scale): P (W)106 t (min) 571724(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 Edition9.Chapter Two Solutions10 March 2006The 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 by300 s010 2 10 300 t + 10 dt = 600 t + 10t 300= 1.5 kJ 0(a) The total energy transferred is 288 + 1.5 = 289.5 kJ (b) The energy transferred in the last five minutes is 1.5 kJPROPRIETARY 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 EditionChapter Two Solutions10 March 2006charge q = 18t2 2t4 C.10. Total(a) q(2 s) = 40 C. (b) To find the m aximum charge within 0 t 3 s, we need to take the second derivitives:f irst anddq/dt = 36t 8t3 = 0, leading to roots at 0, 2.121 s d2q/dt2 = 36 24t2substituting t = 2.121 s into the expression for d2q/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). 50 70(b)(a)60 50030tim e (t)q (C)4020-5010 0-100-10 -20 00.511.5 tim e (s )22.53-150 00.511.5 i (A )22.5PROPRIETARY 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 11. Engineering Circuit Analysis, 7th Edition11.Chapter Two Solutions10 March 2006Referring to Fig. 2.6c,- 2 + 3e 5t A, i1 (t ) = 3t - 2 + 3e A,t0Thus, (a) i1(-0.2) = 6.155 A (b) i1 (0.2) = 3.466 A (c) To determine the instants at which i1 = 0, we must consider t < 0 and t > 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)= =0.1i (t )dt0.8 105 t 0.8 2 + 3e dt +03 = 2t e5t 5 -0.8+0.10 2 + 3e3t dt ( 2t + e ) 3t0.1 0= 33.91 CPROPRIETARY 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 Edition12.Chapter Two Solutions10 March 2006Referring 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 qtotal =121i (t )dt = -4(1) + 2(2) + 6(2) + 0(4) 4(2) = 4 C(c) See Fig. belowq (C) 8 -816 24106 812 14t(s) 16-16PROPRIETARY 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 Edition13. (a)VBA = (b) VED =2 pJ -1.602 10-19 CChapter Two Solutions=12.48 MV0 -1.602 10-19 C=03 pJ 1.602 10-19 C=10 March 200618.73 MV(c) VDC = (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 19 C from B to C, or +2 pJ to move +1.602x1019 C from C to B. +1.602x10 Thus, it requires 3 pJ + 2 pJ = 1 pJ to m ove +1.602x10 19 C from D to C to B. Hence,VDB =1 pJ = 6.242 MV. 1.602 10-19 CPROPRIETARY 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 EditionChapter Two Solutions10 March 200614.+ V1 V2 ++ Voltmeter+VoltmeterFrom 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 EditionChapter Two Solutions15. (a)Pabs = (+3.2 V)(-2 mA) = 6.4 mW(b)Pabs = (+6 V)(-20 A) = 120 W10 March 2006(or +6.4 mW supplied)(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 WPROPRIETARY 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 Edition16.Chapter Two Solutions10 March 2006i = 3te-100t mA and v = [6 600t] e-100t mV (a) The power absorbed at t = 5 ms is[Pabs = (6 600t ) e 100t 3te 100t =]t = 5 msW0.01655 W = 16.55 nW(b) The energy delivered over the interval 0 < t < is0Pabs dt=03t (6 600t ) e 200t dtJMaking use of the relationship0x n e ax dx=n! a n +1where n is a positive integer and a > 0,we find the energy delivered to be = 18/(200)2 - 1800/(200)3 = 0PROPRIETARY 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 Edition17.Chapter Two Solutions(a) Pabs = (40i)(3e-100t)| t = 8 ms =[di (b) Pabs = 0.2 i = - 180 e 100t dt (t (c) Pabs = 30 idt + 20 3e 100t 0 )[360 e 100t]2t = 8 ms]2t = 8 ms10 March 2006= 72.68 W= - 36.34 Wt = 8 mst = 90e 100t 3e 100t dt + 60e 100t 0 = 27.63 W t = 8 msP