Thevenin equivalent with dependet sources

[elamre]

Homework Statement

Hello, i was studying diodes for my exam on monday, but i had some struggle finding the Thevenin equivalent. Here is the circuit:
http://img411.imageshack.us/img411/5336/schakeling.png

Homework Equations

I have no clue what to put here?

The Attempt at a Solution

I have made 2 equations with which i should be able to solve it (i think?) But i have no clue if I am doing it right! So can somebody check it?

i_x = (5-Voc)/1000
and
i_x - i_x /2 = 5/1000

 

[gneill]

What is the motivation behind your second equation?

 

[elamre]

Well i tried to follow an example out of my book (electrical engineering fifth edition by allan r hambley) And i need my second equiation to solve the first equation. Its just that i can't think of 2 =/.

 

[gneill]

Well i tried to follow an example out of my book (electrical engineering fifth edition by allan r hambley) And i need my second equiation to solve the first equation. Its just that i can't think of 2 =/.

I see. Well, it doesn't look like a valid equation to me.

Here's a trick that often helps with these types of circuits. If you consider a general Thevenin source it's a voltage supply Vth with a series resistance Rth. When it's connected to some load resistance R, the two resistances will form a voltage divider for the Thevenin voltage source and the voltage across the load will thus be
$$Vo = V_{th} \frac{R}{R + R_{th}}$$

Now, if you stick a resistance R onto your circuit on the right as a load, you should be able to apply nodal analysis at the output node and solve for Vo. One equation! If you can then put the result in the form above you can read off both the Thevenin voltage and resistance directly.

 

[elamre]

Well i know that, but what about the dependent source? :)

 

[gneill]

Well i know that, but what about the dependent source? :)

Solve for Vo with the dependent source in place, of course. You'll be finding the expression for the current Ix through the 1000Ω resistor as part of the node equation analysis, so you can substitute for Ix/2 easily.