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: 85a

Answer

$L=0.689\mu H$

Work Step by Step

We know that; $f=\frac{1}{2\pi\sqrt{LC}}$ This can be rearranged as follows to make $L$ the subject of the formula: $L=(\frac{1}{C})(\frac{1}{2\pi f})^2$ We plug in the known values to obtain: $L=(\frac{1}{340\times 10^{-6}})(\frac{1}{2\times 3.1416\times 10400})^2=0.689\times 10^{-6}=0.689\mu H$
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