# Thevenin equivalent with dependent source

Well, I'm trying to find the Thevenin equivalent of this circuit: As you can see, the only current source is a dependent source. However, the control variable is in a wire where, if nothing is connected between A and B, is $i_\Delta = 0$ A, and that makes the source 0 A. However, if I connect an impedance $Z$ to the circuit, I will have current, and a Thevenin equivalent. I don't know how to solve this problem. I've bee searching here for a while, and I found this: Thévenin equivalent with a dependent source . I can see how is solved that case. However, I still don't know how to apply that method to my circuit.

Any help about finding the Thevenin equivalent is appreciated n.n

It's actually quite straightforward to find the Thevenin impedance of this circuit.

The equivalent impedance looking into the port ab is defined by:

$$Z_{ab} \equiv \frac{V_{ab}}{I_{ab}} = \frac{V_{ab}}{i_{\Delta}} = Z_{th}$$

But you can write by inspection a simple KVL equation for $V_{ab}$ in terms of $i_{\Delta}$:

$$V_{ab} = i_{\Delta}(-jX_{1uF} + 10k\Omega) + 200 i_{\Delta} \cdot 100 \Omega) = i_{\Delta}(-jX_{1uF} + 30k)\Omega$$

Generally speaking, to find the Thevenin equivalent of a circuit with only dependent source(s), you must be sure to "activate" the dependent source(s) with a test source.

This is what was done above. We solved for the voltage across the port due to a test current source, $I_{ab}$ which, in this case, equals the controlling variable thus making this problem particularly easy to solve.

• Great! Thank you! Jan 5, 2014 at 19:53