University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 31 - Alternating Current - Problems - Exercises - Page 1044: 31.10

Answer

2.36 MHz.

Work Step by Step

$X_L=\frac{V}{I}=\frac{12.0V}{1.80\times10^{-3}A}=6667\Omega$ Solve $X_L=2 \pi fL$ for the frequency. $$f=\frac{X_L}{2 \pi L}=\frac{6667\Omega }{2 \pi (0.450\times10^{-3}H)}=2.36\times10^6 Hz$$
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