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: P1.67


$P_I=-120W$ $P_V=120W$

Work Step by Step

We use the equation for power to find: $$P_I = -VI \\ P_I = -(10A)(12V) \\ P_I = \fbox{-120W}$$ We use the equation for power again. In this case, however, there is no negative sign, for the current goes from higher (the positive) to lower (the negative) potential, meaning that the battery is being charged: $$P_V = VI \\ P_V = (10A)(12V) \\ P_V = \fbox{120W}$$ Note, in the case of the current generator, the power is negative, so energy is given off, but in the case of the resistor, the power is positive, so energy is taken in.
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