Problem 4.4-4

We have three currents as below:

We have the three KVL for the 3 loops:

8 = - 125*(i1 + i0) + 250*(i2 – i0)

12 = 500*i1 + 125*(i0 + i1)

12 = 250*(i3 – i0) + 500*i3

We have 3 equations for 3 unknowns thus we can solve the system.

We get i0 = - 0.024 A

i1 = 0.024 A

i2 = 0.008 A

So finally we get Va = R*i2 = 500*i2 = 4 V