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: T1.3


$V_{ab}=-8V$ $P_{I_1}=-24W$ $P_{I_2}=8W$ $P_{R_1}=5.33W$ $P_{R_2}=10.67W$

Work Step by Step

We first apply Kirchhoff's Current Law and Ohm's law, which states that $I=\frac{V}{R}$, to find: $3-1+\frac{V_{ab}}{12} + \frac{V_{ab}}{6} = 0 \\ 2 + \frac{3V_{ab}}{12} = 0 \\ 3V_{ab} = -24 \\ V_{ab} = \fbox{-8V} $ We use the equation for power, recalling that if the current enters through the negative terminal, a negative sign is added in front of $VI$ in the equation. Thus, we find: $P_{I_1}=VI = -8 \times 3 = \fbox{-24W}$ $P_{I_2}=-VI = -(-8 \times 1) = \fbox{8W}$ $P_{R_1}= \frac{V^2}{R} = \frac{8^2}{12} = \fbox{5.33W}$ $P_{R_2}= \frac{V^2}{R} = \frac{8^2}{6} = \fbox{10.67W}$ Note, if the power is positive, the element is absorbing energy, while if the power is negative, the element is giving off energy.
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