Engineering Circuit Analysis Solutions

Engineering Circuit Analysis, 7th Edition

1. (a)

12 μs
(b) 750 mJ
(c) 1.13 kΩ

(d) 3.5 Gbits
(e) 6.5 nm
(f) 13.56 MHz

Chapter Two Solutions

10 March 2006

(g) 39 pA
(h) 49 kΩ
(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.

Engineering Circuit Analysis, 7th Edition

2.

(a) 1 MW
(b) 12.35 mm
(c) 47. kW
(d) 5.46 mA

(e) 33 μJ
(f) 5.33 nW
(g) 1 ns
(h) 5.555 MW

Chapter Two Solutions

10 March 2006

(i) 32 mm

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.

Engineering Circuit Analysis, 7th Edition

3. (a)

Chapter Two Solutions

⎛ 745.7 W ⎞
⎟ = 298.3
⎝ 1 hp ⎠

( 400 Hp ) ⎜

10 March 2006

kW

⎛ 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) 285.4´10-15 s =

kJ

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.

Engineering Circuit Analysis, 7th Edition

4.

Chapter Two Solutions

10 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 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.

Engineering Circuit Analysis, 7th Edition

5.

Chapter Two Solutions

10 March 2006

Motor power = 175