Problem 4.3-2

There are three currents i0, i2 and i3 where we have:

i0 = 1A

i1 + i3 = 2A

If we write down the voltage v2 we have on the left: v2 = v + v1, where v is the voltage of the resistor of 20Ω
we have on the right: v2 = v’ + v3, where v’ is the voltage of the resistor of 10Ω

Thus we have v + v1 = v’ + v3 which is

20*(i0+i1) +5*i1 = 10*(i3-i0) + 15*i3

Substituting the current i0 with 1A and i3 = 2 – i1 we get

20*(1 + i1) + 5*i1 = 10*(2 – i1 – 1) + 15*(2 – i1)

thus we find 50*i1 = 20 thus i1 = 2/5 A

Then we can get i3 = 2 – 2/5 = 8/5 A

Finally we get: v1 = R*i1 = 5*(2/5) = 2V
And v3 = 15*i3 = 15*(8/5) = 24 V
And v2 = v + v1 = 20*(i0+i1) +5*i1 = 30 V