Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 24 - Alternating-Current Circuits - Problems and Conceptual Exercises - Page 871: 87

Answer

$4.16W$

Work Step by Step

We know that $Z=\sqrt{R^2+(\omega C)^{-2}}$ We plug in the known values to obtain: $Z=\sqrt{(3.30K\Omega)^2+(2\pi(60.0)(2.75\mu F))^{-2}}=3.44K\Omega$ Now $I_{rms}=\frac{V_{rms}}{Z}=\frac{122}{3.44K\Omega}=35.5mA$ We can find the average power consumed as $P_{avg}=I_{rms}^2R$ We plug in the known values to obtain: $P_{avg}=(0.0355)^2(3.30K\Omega)=4.16W$
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