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: P1.75

Answer

$U_{x}$=2V $i_{y}$=0.333A
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Work Step by Step

Applying KVL around of the circuit, we have -10+$V_{x}$+$4V_{x}$=0 then $V_{x}$=2V Using Ohm's law $i_{12}$ = $V_{12}$/12=0.667A and $i_{x}$=$v_{x}$/2=1A Then KCL applied to the node at the top of the 12 omega gives $i_{x}$=$i_{12}$+$i_{y}$ which yields $i_{y}$=0.333A
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