, 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
Copyright ©2018 McGraw-Hill Higher Education. Permission required for reproduction or display. All rights reserved.
, 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 = 1.3110−15 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.
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
Copyright ©2018 McGraw-Hill Higher Education. Permission required for reproduction or display. All rights reserved.
, 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 = 1.3110−15 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.
