Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 30 - Electromagnetic Induction - Exercises and Problems - Page 874: 72


$I_{max} = 0.050~A$

Work Step by Step

We can find the maximum current: $\frac{1}{2}LI^2 = \frac{1}{2}C(\Delta V)^2$ $I = (\Delta V)~\sqrt{\frac{C}{L}}$ $I = (5.0~V)~\sqrt{\frac{0.10\times 10^{-6}~F}{1.0\times 10^{-3}~H}}$ $I = 0.050~A$
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