Thevenin Equivalent Circuit: using open/short circuit method

[november1992]

Homework Statement

http://i.imgur.com/fSRqw.png

Homework Equations

V_{th} = V_{oc}

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

 

[gneill]

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

 

[november1992]

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.