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: P1.64


$I_R= 2A$ $P_V = 20W$ $P_I = -40W$

Work Step by Step

This is a series circuit, so the current across the entire thing is equal. Thus, we find that the current through the resistor is: $I_R = \fbox{2A}$. We know that power through a resistor is given by: $P=I^2R$. Thus, we find: $P_R = 2^2\times(5) = 4\times 5=\fbox{20W}$ We now use Kirchhoff's law to find the voltage of the voltage source: $V +10+10 = 0 \\ V=-20V$ From this, we find power: $P_V = VI = -20 \times 2 = \fbox{-40W}$
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