I think Thevenin (or Norton) equivalent circuits do not consider variable sources. The same refers to non-linear resistors (and other elements in AC scope). But I understand what you mean: you would like to have something like these.

In your case you should first select all the elements that are not dependent on other and do not alter other elements, and simplify them. The next step is to find all independent voltage/current sources.

Now combine non-linear static elements, like resistors. The combination of a linear object and a non-linear object is also non-linear object (but there is a theoretical possibility that two non-linear functions make a linear one).

At this moment you get: combined resistances that are (generally speaking) non-linear and do not alter anything and independent and dependent sources, and the elements that alter sources. If possible, combine independent sources.

That's the hardest task now: to combine independent sources with dependent. The Kirchoff's laws might be necessary here.

**UPDATE**

According to your circuit, this is not that difficult as it seems on the first sight. Please forgive me there are no exact calculations as I did them last time almost 20 years ago...

First of all, take a look at the non-ideal current source `I1`

. Because it has `R1`

in parallel you can convert it to a non-ideal voltage source, which has resistance in series. This voltage source would have internal resistance 1 Ohm too and voltage `R1 * 4Ix`

that is `4*Ix`

volts as `R1 = 1 Ohm`

. I will name this new source as `V2`

.

At the moment on the left side of the circuit you have non-ideal voltage source `V2`

(equivalent to `I1`

current source), its internal resistance (equivalent to `R1`

), than voltage source `V1`

. The `R1`

resistance is gone as it became internal load of voltage source. More reading about source transformation.

Because in the same branch there are two voltage sources you can combine them. So it is `E = V1 + V2`

which leads to `(4 Ix - 10) V`

(`-`

because `V1`

is in opposition to `V2`

).

Now we have the first part of our task, the source. Now we're going to find equivalent resistance, and, moreover, we need to drive out `Ix`

from source equation, because after combining resistances to one there will be no `Ix`

.

As we know from Mr. Kirchoff, the load current (the one in `R3`

), say `I`

, divides in two: `Ix`

and `IL`

(`IL`

flows through `R3`

). The `Ix`

is `U2 / R2`

and `IL`

is `U2 / (R3 + RL)`

. You can write down proper equations yourself :).

Now you can find relation between `Ix`

and `IL`

(you need `IL`

in equation of voltage source) and make `E`

function of `IL`

. If this source is no more function of `Ix`

, you can combine other resistances to one equivalent. Do not forget source `E`

internal resistance (the one driven from `R1`

).

Please note that this method will lead you to have voltage source that is a function of load current (so in fact load resistance `RL`

). This is normal as `U2`

depends on this load (that's why I've written at the very beginning it is not true Thevenin method).

## Best Answer

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

The equivalent impedance looking into the port

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

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.