Thevenin Equivalent Circuit: using open/short circuit method


by november1992
Tags: circuit, equivalent, method, open or short, thevenin
november1992
november1992 is offline
#1
Oct6-12, 10:17 PM
P: 120
1. The problem statement, all variables and given/known data
http://i.imgur.com/fSRqw.png


2. Relevant equations
[itex]V_{th}[/itex] = [itex]V_{oc}[/itex]


3. The attempt at a solution

I got the thevenin voltage by using node voltage

[itex]\frac{V2-40}{5}[/itex] - [itex]\frac{80+2V2}{5}[/itex] -8 + V2 = 0

([itex]\frac{1}{5}[/itex] + [itex]\frac{2}{5}[/itex] + 1)V2 = 32

[itex]\frac{8}{5}[/itex]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
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gneill
gneill is offline
#2
Oct7-12, 08:02 AM
Mentor
P: 11,409
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.
november1992
november1992 is offline
#3
Oct7-12, 10:55 AM
P: 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.


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