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

1. A voltage controlled current source 2.$v_{x}$=15v
v x = (4 Ω) × (1 A) = 4 V $i_{s} = (v_{x} /2) + 1 = 3$ $i_{s}=2+1$i_{s}=3A Applying KVL around the outside of the circuit: $v_{s} = 3i_{s} + 4 + 2 v_{s} =15v$