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 32 - AC Circuits - Exercises and Problems - Page 925: 31


$f = 1.59\times 10^5~Hz$ $X_L = X_C = 1.0~\Omega$

Work Step by Step

We can equate the reactances to find the required frequency: $X_L = X_C$ $\omega ~L = \frac{1}{\omega~C}$ $\omega^2 = \frac{1}{L~C}$ $(2\pi~f)^2 = \frac{1}{L~C}$ $f^2 = \frac{1}{4~\pi^2~L~C}$ $f = \frac{1}{2\pi~\sqrt{L~C}}$ $f = \frac{1}{(2\pi)~\sqrt{(1.0\times 10^{=6}~H)(1.0\times 10^{-6}~F)}}$ $f = 1.59\times 10^5~Hz$ We can find the value of the reactance: $X_L = \omega ~L$ $X_L = 2\pi~f ~L$ $X_L = (2\pi)~(1.59\times 10^5~Hz) ~(1.0\times 10^{-6}~H)$ $X_L = 1.0~\Omega$ Therefore, $X_L = X_C = 1.0~\Omega$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.