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: 26

Answer

(a) $L = 2.36\times 10^{-5}~H$ (b) $I_L = 165~\mu A$

Work Step by Step

(a) We can find the inductance: $I_L = \frac{V_L}{\omega~L}$ $L = \frac{V_L}{\omega~I_L}$ $L = \frac{V_L}{2\pi ~f~I_L}$ $L = \frac{2.2~V}{(2\pi) (45\times 10^6~Hz)~(330\times 10^{-6}~A)}$ $L = 2.36\times 10^{-5}~H$ (b) We can find the peak current: $I_L = \frac{V_L}{\omega~L}$ $I_L = \frac{V_L}{2\pi ~f~L}$ $I_L = \frac{2.2~V}{(2\pi) (90\times 10^6~Hz)~(2.36\times 10^{-5}~H)}$ $I_L = 1.65\times 10^{-4}~A$ $I_L = 165~\mu A$
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.