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