# Circuit analysis question?

1. Jun 30, 2013

### Questions999

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

We have the circuit.http://i.imgur.com/X4IYi4j.jpg
Find the equivalent thevenin resistance ..the result in my book is 8 ohm but I alway get 8/6 ..how so?

2. Relevant equations
Thevenin Teorem

3. The attempt at a solution

1/Re=1/4+1/2 and so Re=8/6 ..

2. Jun 30, 2013

### Staff: Mentor

You'll have to show your work so we can see where you are going wrong. Is R4 part of the circuit for which you want to find the Thevenin equivalent, or does it represent the load?

3. Jun 30, 2013

### Questions999

I dont calculate R4 because I have to use the Tevenin's transformation between the points a and b..to find equivalent resistance, I took the R3 and R2 in parallel and found 1/Re=1/R3+ 1/R2

4. Jun 30, 2013

### rude man

I get R thevenin = 80/21 ohms. I included R4.

Without R4 I get R thevenin = 5 ohms.

Last edited: Jun 30, 2013
5. Jun 30, 2013

### Staff: Mentor

So R4 is not part of the circuit under analysis. Fine.

What makes you think that R2 and R3 are in parallel? Even if you could suppress the two current sources (which you can't because one of them is an active source), wouldn't that leave the resistors in series?

You'll have to use another method to find the Thevenin equivalent since you have an active source in the circuit.

6. Jun 30, 2013

### Questions999

I they are in series,than the result is 6 ,still different from my book..

7. Jun 30, 2013

### Staff: Mentor

The controlled source is going to muck with the Thevenin resistance. You need another analysis method.

@rude man: I'm seeing the book's answer of 8Ω.

8. Jun 30, 2013

### rude man

As gneill pointed out, you have to include the effects of the voltage-dependent voltage source. You can't just parallel resistors etc.

The way I do it is to apply 1V to the input, then compute the input current i using KVL or whatever, then R thevenin = 1/i.

9. Jun 30, 2013

### rude man

Looks like I struck out on this one!

10. Jun 30, 2013

### Staff: Mentor

Hey, you do good work. A small slip every now and then can't be avoided. Besides, they tend to rejuvenate concentration

11. Jun 30, 2013

### rude man

Still bugs me. Want to send me your analysis privately?

12. Jun 30, 2013

### Staff: Mentor

Sure, I'll see if I can slip it into a PM.

13. Jul 5, 2013

### Staff: Mentor

i'm disappointed that you appear to have given up on this question. I seem to have worked it out needlessly, then.

Perhaps someone else would like clues on how to do it? It's a good exam-type question.

14. Jul 6, 2013

### gary32

The circuit isnt very clear but if there's 2 power sources then Kirchhoff's voltage law and the use of simultaneous equations would be helpful