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Thevenin Equivalent Circuit: using open/short circuit method 
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#1
Oct612, 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{V240}{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 shortcircuit 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 


#2
Oct712, 08:02 AM

Mentor
P: 11,846

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.



#3
Oct712, 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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