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: P1.74

Answer

Voltage-controlled current source. $I_x=42mA$

Work Step by Step

$v_o=\sqrt{P_o16}=16V$ Then use Ohm's law to find $i_1=v_o\div32\Omega=0.5A$ and $i_o=v_o\div 8\Omega=1A$ Use KCL to find $25v_{in}=i_1+i_o=1.5A$ and $v_{in}=1.5A\div25=60mA$ Then $I_x=\frac{v_{in}}2+\frac{v_{in}}5=42mA$
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