Engineering Circuit Analysis 9th Edition Solution Manual by Hayt | Chapter 2 Verified Answers 2025/2026

Engineering Circuit Analysis 9th Edition Hayt
SOLUTION MANUAL

,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 103m= 18 km

(g) 2,500,000,000,000 bits  2.5 1012 bits = 2.5 terabits
3
 1015 atoms  102 cm 
(h)     1021 atoms/m3 (it’s unclear what a “zeta atom” is)
 1 m 
3
 cm

,
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3moles = 2.021 millimoles

(g) 13 deciliters  13101  1.3 liters

(h) 1 hectometer  100 meters

, 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