Answer
$I_{max} = 0.050~A$
Work Step by Step
We can find the maximum current:
$\frac{1}{2}LI^2 = \frac{1}{2}C(\Delta V)^2$
$I = (\Delta V)~\sqrt{\frac{C}{L}}$
$I = (5.0~V)~\sqrt{\frac{0.10\times 10^{-6}~F}{1.0\times 10^{-3}~H}}$
$I = 0.050~A$
Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)
by Knight, Randall D.
Published by
Pearson
ISBN 10:
0133942651
ISBN 13:
978-0-13394-265-1
Chapter 30 - Electromagnetic Induction - Exercises and Problems - Page 874: 72
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