University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 30 - Inductance - Problems - Exercises - Page 1014: 30.16


See explanation.

Work Step by Step

a. $U=\frac{1}{2}LI^2=\frac{1}{2}(12.0H)(0.500A)^2=1.50J$. b. $P=I^2R=(0.500A)^2(180\Omega)=45.0W$ c. No. the stored energy depends on the current, which is constant. The energy being dissipated by the inductor comes from the emf source (e.g., battery) that drives the current.
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.