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Thevenin Resistence

  1. Sep 14, 2014 #1
    For the following circuit:
    http://imagizer.imageshack.us/v2/640x480q90/674/aBkkRU.png [Broken]

    I'm solving for the Rth at node 2. In the solution guide, it shows that the Rth = (10k || 10k). I don't understand how this is so, by following the resistence equivalent rules, these two resistance should be in series.

    I'd appreciation for the clarification.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Sep 14, 2014 #2

    Baluncore

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    Science Advisor

    The fixed 10V ideal source has zero dv/di = zero resistance, so the two resistors are actually in parallel.
    If it was a fixed current source, it would be treated as an infinite resistance element.
     
  4. Sep 14, 2014 #3
    When computing the Thevenin resistance, voltage sources become short circuits and current sources become open circuits. In your example, when the voltage source is shorted, the 10k resistors are in parallel
     
  5. Sep 15, 2014 #4

    NascentOxygen

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    Staff: Mentor

    For Thévenin analysis, we first replace the voltage source with a short circuit. Then examine what we are left with.

    To get from node 2 to ground, an electron can go via a path through one of the 10k resistors to ground, OR through the other 10k resistor to ground, but no path to ground takes it through both resistors. That's the characteristic of parallel paths: you can take one path OR the other, and don't go through both in sequence.

    If one path leads inevitably through the other, then and only then, are they are in series.
     
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