Problem 5.5-8

As we have only one dependent source and no independent one, then we have i = 0, when terminals a-b are open. Therefore v_oc = 0 and we also find that i_sc = 0.

The equivalent Thévenin circuit is then a resistor alone. Let’s find its value.

For this, we impose a 1A current source between a and b and the Rt, the Thévenin equivalent resistor will be such that Rt = Vab/1.

We have the following circuit:

We have i = i1 – i2 = 1A (1)

Moreover we have the KVL in the left loop, which gives:

1.5*v1 – 4*i2 = 6*i1 (2)

But we know that v1 = – 6*i = – 6V

Thus we get from (2)

– 9 – 4*i2 = 6*i1

So we have two equations and two unknowns, thus we solve for i1 and i2 and we get:

i2 = – 1.5 A and i1 = – 0.5 A

We have Vab  = 6 + 6*i1 = 6 – 3 = 3 V

so finally we have the Thévenin resistor Rt = 3 Ω