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 936: 20a

Answer

$L=3.6mH$

Work Step by Step

In the given scenario; $\frac{1}{2}CV^2=\frac{1}{2}LI^2$ This can be rearranged as: $CV^2=LI^2$ $L=\frac{CV^2}{I^2}$ We plug in the known values to obtain: $L=\frac{4\times 10^{-6}(1.5)^2}{(0.050)^2}=3.6\times 10^{-3}=3.6mH$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.