Problem 4.5-5

This is the circuit we have:

In the upper loop we have the following relationship:

3ix = R1*(i1 + i0) = 2*(i1 + i0)

And moreover ix = i1 + i2

thus  we get 3*(i1 + i2) = 2*(i1 + i0)

 

Applying the KVL to the left bottom loop, we have:

12 = 2*(i1 + i0) + R2 * ix

which is 12 = 2*(i1 + i0) + 2*(i1 + i2)

We finally have 3 equations

i2 = 1 A

3*(i1 + i2) = 2*(i1 + i0)

12 = 2*(i1 + i0) + 2*(i1 + i2)

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

i0 = 2.2 A

i1 = 1.4 A

i2 = 1 A

and thus ix = i1 +i2 = 2.4 A