Answer
$I=95.5mA$
Work Step by Step
We know that;
$I=\frac{V}{\omega L}$
Since $\omega=2\pi f$,
$I=\frac{V}{2\pi f L}$
We plug in the known values to obtain:
$I=\frac{30}{2\times 3.1416\times 1000(0.050)}=95.5\times 10^{-3}=95.5mA$
Fundamentals of Physics Extended (10th Edition)
by Halliday, David; Resnick, Robert; Walker, Jearl
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 937: 29a
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