Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 18 - Electric Currents - General Problems - Page 524: 71

Answer

a. 33.4 Hz b. 0.990 A c. V = 33.6 sin(210t) in volts

Work Step by Step

a.) I(t) = 1.40 sin(210t) $I(t) = I_{0} sin(2πft)$ $2πf = 210 \frac{rad}{s}$ and $I_{0} = 1.40 A$ $f = \frac{210 \frac{rad}{s}}{2π rad}$ = 33.4 $\frac{1}{s}$ = 33.4 Hz b.) $ I_{0} = 1.40 A$, so $I_{rms} = \frac{I_{0}}{\sqrt 2}$ $I_{rms} = \frac{1.40}{\sqrt 2} = 0.9899 A \approx 0.990 A$ c.) $V = IR = [1.40 sin(210t)] \times 24 Ω = 33.6 sin(210t)$ in volts
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