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[Circuits] Finding the Thevenin and Norton Equivalents #3

  1. Feb 3, 2014 #1
    1. The problem statement, all variables and given/known data

    vtG9YnW.png

    2. Relevant equations



    3. The attempt at a solution

    Attached to this post. I had to take the absolute value of Rth. What did I do wrong to get a negative value for Rth?
     

    Attached Files:

  2. jcsd
  3. Feb 3, 2014 #2

    gneill

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    Staff: Mentor

    No doubt it has to do with the direction of the current that you're "injecting" into the circuit versus the polarity of the resulting voltage across the terminals. The resistance of a component (or circuit for that matter) is defined as the current into the port divided by the resulting potential across that port. The polarity of the voltage "measurement" is such that the terminal where the current is injected is taken to be positive:

    attachment.php?attachmentid=66261&stc=1&d=1391463145.gif
     

    Attached Files:

  4. Feb 10, 2014 #3
    So my mistake was that the 1A current source was pointing down when it should've been pointing up?
     
  5. Feb 10, 2014 #4

    gneill

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    Staff: Mentor

    Not a mistake per se, not if you take it into account when you assign the polarity of the voltage. You are looking for the potential at terminal a with respect to terminal b, so automatically that implies that the a terminal is the where you'd put the + lead of your meter (if you were measuring it). You are driving the current into the b terminal, so it becomes the "+" terminal for the resistance determination. That's contrary to the desired arrangement, and will result in a reversed sign for the voltage Vab.
     
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