Solution Manual for Engineering Circuit Analysis (9th Edition) – William H. Hayt, Jack E. Kemmerly & Steven M. Durbin – Complete Problem Solutions

, Engineering Circuit Analysis 9th Edition Chapter Two Exercise Solutions


1. Convert the following to engineering notation:

(a) 0.045 W = 45´10-3 W = 45 mW

(b) 2000 pJ = 2000 ´10-12 = 2 ´10-9 J = 2 nJ

(c) 0.1 ns = 0.1´10-9 = 100 ´10-12 s = 100 ps

(d) 39,212 as = 3.9212×104×10-18 = 39.212×10-15 s = 39.212 fs

(e) 3 Ω

(f) 18,000 m = 18 ´103 m = 18 km

(g) 2,500,000,000,000 bits = 2.5´1012 bits = 2.5 terabits

3
 1015 atoms  102 cm 
(h)   = 10 atoms/m (it’s unclear what a “zeta atom” is)
21 3
3 
 cm  1 m 




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, Engineering Circuit Analysis 9th Edition Chapter Two Exercise Solutions


2. Convert the following to engineering notation:

(a) 1230 fs = 1.23103 ´10-15 = 1.23´10-12 s = 1.23 ps

(b) 0.0001 decimeter = 1´10-4 ´10-1 = 10 ´10-6 m = 10 m

(c) 1400 mK = 1.4 ´103 ´10-3 = 1.4 K

(d) 32 nm = 32 ´10-9 m = 32 nm

(e) 13,560 kHz = 1.356 ´104 ´103 = 13.56 ´106 Hz = 13.56 MHz

(f) 2021 micromoles = 2.021´103 ´10-6 = 2.021´10-3 moles = 2.021 millimoles

(g) 13 deciliters = 13´10-1 = 1.3 liters

(h) 1 hectometer = 100 meters




Copyright ©2018 McGraw-Hill Higher Education. Permission required for reproduction or display. All rights reserved.

, Engineering Circuit Analysis 9th Edition Chapter Two Exercise Solutions


3. Express the following in engineering units:

(a) 1212 mV = 1.121 V

(b) 1011 pA = 1011×10-12= 100 mA

(c) 1000 yoctoseconds = 1´103 ´10-24 = 1´10-21 seconds = 1 zs

(d) 33.9997 zeptoseconds

(e) 13,100 attoseconds = 1.3110−15 s = 1.31 fs

(f) 10−14 zettasecond=10-14×1021=107=10×106 s = 10 Ms

(g) 10−5 second = 10 ´10-6 seconds = 10 s

(h) 10−9 Gs = 10-9 ´109 = 1 second




Copyright ©2018 McGraw-Hill Higher Education. Permission required for reproduction or display. All rights reserved.