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: 17b

Answer

$I=0.365A$

Work Step by Step

We know that; $I=\omega Q$ This can be written as: $I=2\pi fQ$ We plug in the known values to obtain: $I=2(3.1416)(275)(2.11\times 10^{-4})$ $I=0.365A$
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