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