Problem 4.5-3

The circuit is as follow:

We apply again the KVL for the 2 loops, and we get:

va = R1*ia + 6

va = R3*ib + 4*va

Thus we get the following system:

- 2.10³ *(ia +ib) = 10³ *ia + 6

6. 10³ *(ia +ib) = 3. 10³ *ib

We have 2 equations for 2 unknowns so we can solve the system. We get

ia =  0.006 A and ib = - 0.012 = - 12 mA