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

$I_x=4.25A$
Using Ohm's law for $v_1=1A\times4\Omega=4V$ And then $i_1=v_1\div2\Omega=2A$ Then KCL yields $i_2=i_1+1A=3A$ and back to Ohm's law for $v_2=2\Omega\times i_2=6V$ Use KVL with $v_1+v_2-v_3=0, v_3=4V+6V=10V$ and then Ohm's law again for $i_3=v_3\div8\Omega=1.25A$ Finally, use KCL to find $I_x=i_2+i_3=4.25A$