## Electrical Engineering: Principles & Applications (6th Edition)

Voltage-controlled current source. $I_x=42mA$
$v_o=\sqrt{P_o16}=16V$ 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 $I_x=\frac{v_{in}}2+\frac{v_{in}}5=42mA$