# Open Circuit Voltage and Short Circuit Voltage

1. Sep 5, 2016

### Marcin H

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

2. Relevant equations
V=IR
Current Loop method, KCL, Node Voltage

3. The attempt at a solution
I am struggling with this problem because I haven't seen a circuit like this before. I think I found the Voc correctly (4V) but I am stuck with finding the short circuit current. I tried using the current loop rule with 2 loops but it's confusing and doesn't seem to work.

I also tried this way (above) with one big loop and one small but then I get stuck because I dont know how to find the voltage across the 2i source in my problem. I dont have a resistance so I can't really make an equation for the small loop. I ended up using KCL and the big loop which gave me something, but I don't know if what I did is valid or correct. The current source in the middle is really throwing me off and I am not too sure how to deal with it.

Last edited: Sep 5, 2016
2. Sep 5, 2016

### cnh1995

It all looks correct to me.

3. Sep 5, 2016

### Marcin H

How do you do the loop method in this problem like in the colored picture? That pic was from khan Academy and they said that is a special case and that's the way to do it. How can that method be applied to my circuit? We don't know the voltage across the current source and I can't think of a way to find it.

4. Sep 5, 2016

### cnh1995

The loop method doesn't work in such cases. Since both the loops have a common current source and you directly don't know the voltage across it, you have to take help of KCL here. Look up the concepts 'supermesh' and 'supernode'.

5. Sep 5, 2016

### The Electrician

Connect a resistor R across the a/b terminals. Solve the circuit leaving the resistor R as a symbolic variable. You will get a value for ix involving the variable R. Take the limit as R→0 and you will get the value for ix you already have. Allow R→∞ and get ix when the a/b terminals are open circuited. Knowing ix (and thus i), you can calculate the voltage drop across the 1Ω resistor in the left loop and then the voltage across a/b.