# 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

## 1 Answer

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