Physics: Principles with Applications (7th Edition)

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

Chapter 21 - Electromagnetic Induction and Faraday's Law - Problems - Page 622: 66

Answer

a) $I_{rms}=4.3\times10^{-2}A$ b) $\phi=2.7^o$ c) $P=5.1W$ d) Resistor: $V_{rms}=120V$ Inductor: $V_{rms}=5.6V$

Work Step by Step

$I_{rms}=\frac{V_{rms}}{Z}$ $Z=\sqrt{R^2+X_L^2}=\sqrt{R^2+(2\pi fL)^2}=2803.1\Omega$ $I_{rms}=\frac{120V}{2803.1\Omega}=4.3\times10^{-2}A$ b) $\phi=\arctan(\frac{X_L-X_C}{R})=2.7^o$ c) $P=I_{rms}^2R=5.1W$ d) V_{rms}=I_{rms}Z Resistor: $Z=R$; $V_{rms}=(0.04281A)(2.80\times10^3\Omega)=120V$ Inductor: $Z=X_L=2\pi fL$ $V_{rms}=(0.04281A)(2\pi(60.0Hz)(350\times10^{-3}))=5.6V$
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