# Homework Help: Thevenin & Norton Circuit with Current source???

1. Sep 10, 2016

### Marcin H

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

2. Relevant equations
V=IR
Thevenin/Norton

3. The attempt at a solution
I am trying to find the Voc, but this circuit isn't making sense to me. So to start we can ignore the 9ohm resistor and that branch for now when Finding Voc so you are basically left with the square circuit on the left and I labeled my ground in the bottom right of that square. Now to find the Voc I have to find the voltage across the 3ix current source, but I am not sure how. How can I apply KVL or KCL to the loop if I don't know the resistance or voltage of the current source? And if you do KCL at the Top middle node then you get ix = 3ix which would give you 1=3 . Should I find I norton first and then I can find Rt which would just be 11ohms and then do V= InRt?

2. Sep 10, 2016

### Staff: Mentor

As you've noticed, if ab is open then only the one loop is involved, hence only one current.

For what possible value of $i_x$ can $i_x = 3 i_x$? What potential at the central node will guarantee this value of current?

3. Sep 10, 2016

### Marcin H

Your text is a bit glitchy for me, but are you asking what value of ix will give you ix = 3ix? 0 is the only value, but I don't really get how that works because if you take that equation and divide by ix you get 1=3 which gives you nothing. and how does a current of 0 in this circuit make sense? You have a voltage source and a current source.

Also, if it is 0, I am not sure how to find the potential that will make that true. Do I just use KCL at the node?

4. Sep 10, 2016

### Staff: Mentor

I did fix the text, undoubtedly shortly after you opened the post!

Yes, the only value of $i_x$ that can satisfy the requirement is zero. What must be the potential difference across the 2 Ω resistor in that case? So what is the potential at the center node?

5. Sep 10, 2016

### Marcin H

It would also have to be 0

EDIT* So if ix is 0 then does that mean the 3ix current source is also at 0A meaning that the voltage across there has to be 0 so that would mean Voc is also 0. So the thevenin voltage is 0?

Last edited: Sep 10, 2016
6. Sep 10, 2016

### cnh1995

No.
Use KVL when Ix=0.

7. Sep 10, 2016

### Marcin H

Ok so how could I show that exaclty using some equations? Can I right Voc =V1 = 3(ix)R = 0. V1 is what I labeled my middle node.

Also, I am not sure how I could find my norton equivalent now. Since I have Vthevenin (0V) and Rthevenin (11ohms) can I just say Inorton = Vth/Rth = 0/11 = 0? This seems way to easy. I can get some node equations when I put a wire from a to b to find Isc but it looks like In would be 0 if you use KCL at the node. ix=3ix+In and if ix =0 then IN=0. I have never had a problem like this and it just doesn't seem right.

8. Sep 10, 2016

### cnh1995

Show what? Vth=0? It is not zero..I edited my post a couple of minutes after posting.

9. Sep 10, 2016

### cnh1995

Dependent sources can be a bit confusing in some cases. You got Ix=0 right. But since Ix=0, what is the voltage across the 2 ohm resistor? Apply KVL to the circuit and see.

10. Sep 10, 2016

### Marcin H

Oh. I didn't see your edit. If I use KVL (starting in the bottom left corner) I would get 0 = -8V +2ix + 3ix. But that doesn't make sense. how can I use KVL when there is a current source with an unkown resistance. I dont know how to find the voltage across the current source. That equation also does't make sense because if you plug in 0 for ix you will get -8V = 0 which is wrong.

EDIT*

If ix is 0 then V=IR = (0)*(2ohms) = 0

11. Sep 10, 2016

### cnh1995

Right.
Current through the dependent source is 0. In fact, no current is flowing im any branch.
What does this tell you about the current source?

12. Sep 10, 2016

### Marcin H

Is the current source off or something? I am not too sure what that tells me about the current source. Would the coltage across it be 8V then? Is it acting just like its shorted? I can just remove that component?

13. Sep 10, 2016

### Staff: Mentor

Your conclusion that if the current source is 0 Amps that the potential across it must be 0 Volts is not correct. A current source produces whatever potential difference is required in order to achieve and maintain its designated current. That can be any voltage, positive, negative, or zero.

In this case its goal is achieved if the current $i_x$ is zero, making $3i_x$ zero also. In order for that to happen the current through the 2 Ω resistor must be zero. So what what can you say about the potentials (with respect to your chosen reference point) at both ends of that resistor?

14. Sep 10, 2016

### Marcin H

Would it just be 8V then?

15. Sep 10, 2016

### Staff: Mentor

Well, confirm your hypothesis. Would KVL be satisfied for the loop?

16. Sep 10, 2016

### Marcin H

-8V+8V=0 so I guess it will. Hmm. So Vth is 8V.

SO if Vth is 8V I can just use Rth which is 11ohms and use that to find Inorton right? In = (8/11) A

17. Sep 10, 2016

### Staff: Mentor

How do you conclude that Rth is 11 Ω? Remember, you can't suppress a controlled source to find the Thevenin resistance: the actions of controlled sources will affect what the load "sees" as the impedance of the network, since they muck with the V-I characteristics that appear at the output.

Finding the open circuit voltage and the short circuit current is the way to finding the Thevenin or Norton equivalent. The Thevenin (or Norton) resistance is the ratio of Voc/Isc.

18. Sep 10, 2016

### Marcin H

I thought we could find Rt by shorting all the sources and just looking at the resistors. I remember that rule from somewhere, but I'm not sure it's right. But if we said that the current throught the 2 ohm and the current source is 0 then Wouldn't In also be 0? or would the voltage source be providing a current to IN now?

19. Sep 10, 2016

### Staff: Mentor

You can suppress fixed sources (shorting voltage sources, opening current sources), but NOT controlled sources. Controlled sources are active components and change their voltage and/or currents depending upon the conditions in the circuit (or externally applied conditions). In some ways they appear to act like variable resistors.

So your path forward right now is to determine the short circuit current.

20. Sep 10, 2016

### Marcin H

Would it not be 0? Can we use the KCL equation ix = 3ix + Isc and use the fact that ix=0 to show that In=0? Or is that not true once you put a wire from A to B. Here is what I came up with after shorting a-b. Is this correct?

21. Sep 10, 2016

### Staff: Mentor

No, Isc won't be zero. Once you short the output you have a new circuit, so your previous value of 8 V for V1 won't apply anymore. Analyse the new circuit from scratch. The node equation approach is convenient, but don't assume a value for V1. Instead, solve for it then obtain the short circuit current as the current through the 9 Ω resistor branch.

Hint: Identify the variable $i_x$ with the auxiliary equation $i_x = \frac{8 - V1}{2}$ in order to eliminate $i_x$ from the node equation.

22. Sep 10, 2016

### Marcin H

Is my way from the picture wrong? Are you saying I should take the equation for ix ( the one you just wrote) and plug it into 3ix and then solve for V1? How will V1 help me find Isc?

23. Sep 10, 2016

### Staff: Mentor

When you write the node equation you have two unknowns: $V1$ and $i_x$. The auxiliary equation allows you to eliminate $i_x$ and solve for $V1$.

$I_{sc}$ is the current through the 9 Ω resistor. What's the potential across it?

Last edited: Sep 10, 2016
24. Sep 10, 2016

### Staff: Mentor

Alternatively you could use mesh analysis to solve for the current in the second loop. But you'll need to use a supermesh, and another auxiliary equation tying the mesh currents together with the controlled source.

25. Sep 10, 2016

### Marcin H

8V is the potential across so would the In just be I=V/R = 8V/9ohm = 8/9A?