Solusionario

Engineering Circuit Analysis, 7th Edition

Chapter Two Solutions

10 March 2006

1. (a)

12 μs (b) 750 mJ (c) 1.13 kΩ

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

(g) 39 pA (h) 49 kΩ (i) 11.73 pA

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Engineering Circuit Analysis, 7th Edition

Chapter Two Solutions

10 March 2006

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

(i) 32 mm

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Engineering Circuit Analysis, 7th Edition

Chapter Two Solutions

10 March 2006

3. (a)

( 400 Hp ) ⎜

⎛ 745.7 W ⎞ ⎟ = 298.3 ⎝ 1 hp ⎠

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 kJ

⎛ 1055 J ⎞ (d) ( 67 Btu ) ⎜ ⎟ = 70.69 ⎝ 1Btu ⎠ (e) 285.4´10-15 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.

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.

Engineering Circuit Analysis, 7thEdition

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 = (130.5×103 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 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

Chapter Two Solutions

10 March 2006

6.

The 400-mJ pulse lasts 20 ns. (a) To compute the peak power, we assume thepulse shape is square:
Energy (mJ) 400

20

t (ns)

Then

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 to teachers and educators for course preparation. If you are a student using this Manual, you areusing it without permission.

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:
Energy (mJ) 1

75

t (fs)

Then

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. 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

Chapter Two Solutions

10 March 2006

8.

The power drawn from the battery is (not quite drawn to scale):
P (W)...