Electrical Engineering: Principles & Applications (6th Edition)

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

Chapter 1 - Practice Test - Practice Test - Page 44: T1.2


$V_R = -6V$ $P_I=12W$

Work Step by Step

The current goes from the negative to the positive for the resistor. Thus, we find: $V=-IR = - (3A)(2\Omega) =\fbox{ -6V}$ We first need to find the voltage of the current source using Kirchhoff's Voltage Law: $-10+6+v=0 \\ v-4=0 \\ v=4V$ Using this, we find power: $P = VI = (4V)(3A) = \fbox{12 W}$
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