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

a. $v_x=1.667V$ b. $i_x=0.556A$ c. $P_{voltage\ source}=5.556W$
a. $10=v_x+5i_x$ so $v_x=10/6=1.667V$ b. $i_x=v_x/3=0.556A$ c. $P_{voltage\ source}=-10i_x=5.556W$