Answer
$Z = 218~\Omega$
Work Step by Step
We can find $Z$:
$Z = \sqrt{R^2+X_L^2}$
$Z = \sqrt{R^2+(\omega_d~L)^2}$
$Z = \sqrt{R^2+(2\pi~f_d~L)^2}$
$Z = \sqrt{(200~\Omega)^2+[(2\pi)~(60.0~Hz)~(0.230~H)]^2}$
$Z = 218~\Omega$
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 938: 43a
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