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: 90

Answer

$f=1.59\mu F$

Work Step by Step

We know that; $f=\frac{1}{2\pi \sqrt{LC}}$ This can be rearranged as: $C=\frac{1}{(2\pi f)^2 L}$ We plug in the known values to obtain: $C=\frac{1}{(2\times 3.1416\times 3500)^2\times 0.00130}=1.59\times 10^{-6}=1.59\mu F$
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