Electrical Engineering: Principles & Applications (6th Edition)

Published by Prentice Hall
ISBN 10: 0133116646
ISBN 13: 978-0-13311-664-9

Chapter 1 - 1.7 - Problems - Introduction to Circuits - Page 40: P1.68


$I_R = -1.5A \\P_R = 18W \\ P_I = 48 W \\P_V=-66W$

Work Step by Step

First, we use Ohm's law to find the current: $I_R = \frac{V}{R} = \frac{-12V}{8 \Omega} = \fbox{-1.5A}$ Now, we use the equation for Power: $P=VI$. Recall, if the current goes from the negative to the positive, there is a negative sign in from of this equation. Recalling this, we obtain: $P_R = -(12V)(-1.5A) = \fbox{18W}$ $P_I = (12V)(4A) = \fbox{48W}$ $P_V = -(12V)(5.5A) = \fbox{-66W}$ Note, if the power is negative, the source is supplying energy, while if the power is positive, the source is absorbing energy.
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.