Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 31 - Electromagnetic Oscillations and Alternating Current - Problems - Page 940: 73a

Answer

$L = 2.41~\mu H$

Work Step by Step

We can find the inductance: $\omega = \frac{1}{\sqrt{LC}}$ $\sqrt{LC} = \frac{1}{\omega}$ $\sqrt{LC} = \frac{1}{2\pi~f}$ $LC = \frac{1}{4\pi^2~f^2}$ $L = \frac{1}{4\pi^2~f^2~C}$ $L = \frac{1}{(4\pi^2)~(8.15\times 10^3~Hz)^2~(158\times 10^{-6}~F)}$ $L = 2.41~\mu H$
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