# Thevenin Equivalent

1. Apr 16, 2007

### cepheid

Staff Emeritus
1. The problem statement, all variables and given/known data

Find the Thevenin equivalent of the circuit shown below

2. Relevant equations

Nothing but basic principles

3. The attempt at a solution

Alright, I realize that it will take some time for the attachement to be approved, but it's a really simple circuit. It has just one loop. Starting from ground, you go up through a 10 V voltage source, across a 1k resistor in series with it, and then back down to ground through a "controlled" current source, which means going in the opposite direction of the current indicated by the arrow ('up'). Note that it is a controlled source, even though it doesn't look like it in the diagram. This is because Circuit Maker is mildly retarded and doesn't represent controlled sources using diamonds, but using some other much more complicated circuit symbol. So I just threw an independent current source in there. Anyway, the output voltage is taken to be across this branch (the current source). Anyway, the current across the 2k is ix, and the current supplied by the dependent source is 3ix. I realize I haven't done anything towards a solution yet, but that's because I'm having conceputal difficulty with the controlled source here. So to be fair, for the time being I won't ask for help with how to solve it except to ask, "what do I do with the controlled source when I am trying to calculate the Thevenin equivalent voltage and resistance?"

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Last edited: Apr 16, 2007
2. Apr 17, 2007

### SGT

Applying KCL to the circuit you get
$$i_x = -3i_x$$. So, $$i_x = 0$$ and $$e_{oc} = V_s = 10V$$

3. Apr 17, 2007

### cepheid

Staff Emeritus
Hmm, that's kind of weird. So this configuration is not really possible with non-zero current unless a load resistor is attached?

Also, I know that you're supposed to find the "Thevenin equivalent" resistance of the network by hooking up a "test source" across the terminals and calculating the current drawn from it. When you do so, you are supposed to "turn off" the network's internal sources. Again, how does this apply to the controlled current source? Do I leave it as is, and add its current to that drawn by the test source?

4. Apr 17, 2007

### SGT

Yes.
Only the independent sources are cancelled. The controlled ones remain in the circuit.
There is an alternative way to calculate the equivalent resistance. You can calculate the short circuit current for the Norton equivalent. Since the Thevenin and Norton equivalents use the same resistance, you can calculate it as the quocient of the open circuit voltage and the short circuit current.

5. Apr 18, 2007

### cepheid

Staff Emeritus
I'll try all that, thanks for the tips.