Problem 4.4-8

Again, here is the circuit we have:

We can apply the KVL for the 3 loops by expressing everything in terms of i0, i1 and i2.

We have:

12 = 2* (i1 – i0) + 4*i1

12 = 3*i2 + 2*(i0 + i2)

i0 = 1 A

 

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

i0 = 1 A

i1 = 2.33 A

i2 = 2 A

 

Thus the voltage we are looking for is v = - 2* (i1 – i0) + 2*(i0 + i2)

So v = 3.33 V