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 868: 20

Answer

$I_{rms}=1.92mA$

Work Step by Step

We know that $Z=\sqrt{R^2+(\frac{1}{\omega C})^2}$ We plug in the known values to obtain: $Z=\sqrt{(10.0)^2+[\frac{1}{2\pi (105)(0.250)}]^2}$ $Z=11.7\Omega$ Now we can find the rms current as $I_{rms}=\frac{rms}{Z}$ We plug in the known values to obtain: $I_{rms}=\frac{22.5}{11.7}=1.92mA$
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