From the circuit, notice the closed path containing the voltage
source, the 5 ? resistor, and the 20 ? resistor. We can apply
Kirchhoff s voltage law around this closed path
500 = 5 $i_\Delta$ + 20$i_0$
Now we need to generate a second equation containing these two currents. There are three
nodes in the circuit, so we turn to Kirchhoffs current law KCL to generate the second
equation. We select node b to produce the following equation:
$i_0$ = 5 $i_\Delta$ + $i_\Delta$ = 6 $i_\Delta$
From the two equations we will find that
$i_0$ = 24 A
$i_\Delta$ = 4 A
$v_0$ = 20$i_0$ = 20 X 24 = 480V