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Solution manual for Engineering Circuit Analysis 9th Edition by William H. Hayt

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Engineering Circuit Analysis 9 th Edition Chapter Two Exercise Solutions Copyright ©2018 McGraw-Hill Higher Education. Permission required for reproduction or display. All rights reserved. 1. Conv ... ert the following to engineering notation: (a) 0.045 W = 45 mW (b) 2000 pJ = 2 nJ (c) 0.1 ns = 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 km (g) 2,500,000,000,000 bits = 2.5 terabits (h) 3 15 2 21 3 3 10 atoms 10 cm 10 atoms/m cm 1 m       =    (it’s unclear what a “zeta atom” is) = 45´10-3 W = 2000 ´10-12 = 2´10-9 J = 0.1´10-9 =100 ´10-12 s =18´103m = 2.5´1012 bits Engineering Circuit Analysis 9 th Edition Chapter Two Exercise Solutions Copyright ©2018 McGraw-Hill Higher Education. Permission required for reproduction or display. All rights reserved. 2. Convert the following to engineering notation: (a) 1230 fs = 1.23 ps (b) 0.0001 decimeter = 10 m (c) 1400 mK (d) 32 nm = 32 nm (e) 13,560 kHz = 13.56 MHz (f) 2021 micromoles = 2.021 millimoles (g) 13 deciliters (h) 1 hectometer =1.23103 ´10-15 =1.23´10-12 s =1´10-4 ´10-1 =10´10-6 m =1.4 ´103 ´10-3 =1.4 K = 32 ´10-9m =1.356 ´104 ´103 =13.56 ´106 Hz = 2.021´103 ´10-6 = 2.021´10-3 moles =13´10-1 =1.3 liters =100 meter [Show More]

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