# Thevenin Equivalent Circuit: using open/short circuit method

1. ### november1992

120
1. The problem statement, all variables and given/known data
http://i.imgur.com/fSRqw.png

2. Relevant equations
$V_{th}$ = $V_{oc}$

3. The attempt at a solution

I got the thevenin voltage by using node voltage

$\frac{V2-40}{5}$ - $\frac{80+2V2}{5}$ -8 + V2 = 0

($\frac{1}{5}$ + $\frac{2}{5}$ + 1)V2 = 32

$\frac{8}{5}$V2 = 32
V2 = 20V = Vth

but I don't know how to setup the equation to find the short-circuit current. Since there's a short circuit, I believe that means no current goes through the 1 ohm resistor. I tried using a source transformation, but I got 28A as the current. Since Rth is 0.625, then the current has to be 32A.

http://i.imgur.com/6Twjp.png

### Staff: Mentor

With the short in place you know what the potential at terminal a is and therefore the potential across the resistor, thus handing you ix. Do KCL for the terminal 'a' node.

3. ### november1992

120
I wrote the current of the 1 ohm resistor in terms of v and r, I just forgot that it was in parallel with node 2.