# Engineering Circuit Analysis 9th Edition Solution Manual

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Engineering circuit-analysis-solutions-7ed-hayt • 1. Engineering Circuit Analysis, 7th Edition Chapter Two Solutions 10 March 20061. (a) 12 μs (d) 3.5 Gbits (g) 39 pA (b) 750 mJ (e) 6.5 nm (h) 49 kΩ (c) 1.13 kΩ (f) 13.56 MHz (i) 11.73 pAPROPRIETARY MATERIAL. © 2007 Th e McGraw-Hill Companies, Inc. Lim ited distr ibution perm itted onl y toteachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. • Engineering Circuit Analysis, 7th Edition Chapter Two Solutions 10 March 20062.

(a) 1 MW (e) 33 μJ (i) 32 mm (b) 12.35 mm (f) 5.33 nW (c) 47. KW (g) 1 ns (d) 5.46 mA (h) 5.555 MWPROPRIETARY MATERIAL.

© 2007 Th e McGraw-Hill Companies, Inc. Gto 52 Zp Manual. Lim ited distr ibution perm itted onl y toteachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. • Engineering Circuit Analysis, 7th Edition Chapter Two Solutions 10 March 2006 ⎛ 745.7 W ⎞3. (a) ( 400 Hp ) ⎜ ⎟ = 298.3 kW ⎝ 1 hp ⎠ ⎛ 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 kJ ⎝ 1 Btu ⎠ (e) 285.4´10-15 s = 285.4 fsPROPRIETARY MATERIAL. © 2007 Th e McGraw-Hill Companies, Inc. Lim ited distr ibution perm itted onl y toteachers and educators for course preparation.

If you are a student using this Manual, you are using it without permission. • Engineering Circuit Analysis, 7th Edition Chapter Two Solutions 10 March 20064. (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 toteachers and educators for course preparation.

If you are a student using this Manual, you are using it without permission. • Engineering Circuit Analysis, 7th Edition Chapter Two Solutions 10 March 20065. 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 = (130.5×103 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 toteachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

• Engineering Circuit Analysis, 7th Edition Chapter Two Solutions 10 March 20066. The 400-mJ pulse lasts 20 ns. (a) To compute the peak power, we assume the pulse shape is square: Energy (mJ) 400 t (ns) 20Then P = 400×10-3/20×10-9 = 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 toteachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. • Engineering Circuit Analysis, 7th Edition Chapter Two Solutions 10 March 20067. The 1-mJ pulse lasts 75 fs. (a) To compute the peak power, we assume the pulse shape is square: Energy (mJ) 1 t (fs) 75Then P = 1×10-3/75×10-15 = 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.