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Thevenin Theorem - Open Voltage and Short Circuit Current

  1. Aug 22, 2011 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations
    Thevenin Theorem, Kirchoff's Voltage Law, mesh analysis

    3. The attempt at a solution

    I'm kind of confused as to how to approach this question as a whole because I'm getting put off from the two voltage sources.

    I was thinking that maybe I could use mesh current analysis so then I'd get:

    -10 + 1200([itex]i_{1}[/itex]) + 3300([itex]i_{1}[/itex] - [itex]i_{2}[/itex]) + 12 = 0
    where [itex]i_{2}[/itex] = 0 since it's within an open circuit, so then
    1200([itex]i_{1}[/itex]) + 3300([itex]i_{1}[/itex]) = -2
    4500([itex]i_{1}[/itex]) = -2
    [itex]i_{1}[/itex] = [itex]\frac{-2}{4500}[/itex] A

    [itex]V_{oc}[/itex] = [itex]\frac{-2}{4500}[/itex] * 3300
    [itex]V_{oc}[/itex] = 1.5 V

    but that doesn't sound right, so I'm not sure how else to approach it. Thanks for help in advance!
    Last edited: Aug 22, 2011
  2. jcsd
  3. Aug 22, 2011 #2


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    Staff: Mentor

    There's a single closed loop, as you discovered when you realized that i2 = 0.

    The equation that you've written for loop 1 does not appear to be self consistent in terms of voltage polarities. Voltage rises and drops should always correspond to what occurs while traversing a given component in the direction of the assumed current.

    So, suppose that you have the assumed current direction for i1 as you've shown in the diagram and begin "walking around" the loop at the - terminal of the 10V supply. Can you write the KVL equation?
  4. Aug 22, 2011 #3
    Is it meant to be:
    -10 + 1200(i1) + 3300(i1) - 12 = 0 ?
  5. Aug 22, 2011 #4


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    Staff: Mentor

    That will do :smile: I would have given the terms the opposite signs when writing out the equation, but as long as you're rigidly consistent in your methodology everything will work out.

    When I write an equation for a loop from a diagram, after choosing the current direction for the mesh I imagine "walking around the loop" in the direction of the current, and record the rises and falls of the potential as I pass through components.

    Consider: start at the "-" terminal of the 10V supply and moving through that supply in the direction of the current the potential *rises* by 10V. That is, +10V. Proceeding on to move through R1 the potential will *drop* by i1*R1, and so on. So:

    10V - i1*R1 - i1*R2 + 12V = 0.

    That's how I do it, but your equation will yield the same result for the current.
  6. Aug 23, 2011 #5


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    Staff: Mentor

    You could analyze it using superposition. Having recognized that i2 is zero, the analysis is much simpler.

    First, pretend source S2 is replaced by a short circuit (0 volts and 0 ohms), and the circuit can be see to be a potential divider fed by S1. Determine Vab.

    Next, pretend S1 is replaced by a short circuit (a solid wire). You again have a potential divider, but this time it is fed by S2. Determine Vab in this arrangement.

    Now, because the circuit is composed of linear elements, the actual value of Vab will be the sum of its two components determined above.
  7. Aug 23, 2011 #6
    Thanks so much for the help :) but I just wanted to make sure that my answer was correct so:

    -10 + 1200*i1 + 3300*(i1 - i2) - 12 = 0
    since i2 = 0 A,
    -22 + 4500*i1 = 0
    i1 = [itex]\frac{22}{4500}[/itex] A = 0.0049 A

    Then, for Voc:
    3300*(i2 - i1) + Voc + 12 = 0
    since i2 = 0 A,
    3300*([itex]\frac{-22}{4500}[/itex]) + Voc + 12 = 0
    Voc = [itex]\frac{62}{15}[/itex] V = 4.13 V
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