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Thevenin Equivalent Circuit: using open/short circuit method

  1. Oct 6, 2012 #1
    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
     
  2. jcsd
  3. Oct 7, 2012 #2

    gneill

    User Avatar

    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.
     
  4. Oct 7, 2012 #3
    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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