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

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