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[Circuits] Finding the Thevenin and Norton Equivalents #2

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

    g05KQuK.png

    2. Relevant equations



    3. The attempt at a solution

    Attached to this post. I'm not sure how to proceed.
     

    Attached Files:

  2. jcsd
  3. Feb 3, 2014 #2
    First try to write one nodal equation for only one unknown nodal voltage Vx = Va = Vth.
    Next, short point A and B and find this short circuit current Isc. And from there
    Vth = Vx = Va and Rth = Vth/Isc
     

    Attached Files:

  4. Feb 3, 2014 #3

    gneill

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    Staff: Mentor

    Here's a hint that might help.

    If you consider a general Thevenin model with a resistive load attached, you've really got a simple voltage divider:

    attachment.php?attachmentid=66262&stc=1&d=1391463799.gif

    If you were to tack a load resistor onto the output of your circuit and use node voltage analysis to solve for the output voltage, then you should be able to rearrange the resulting expression into the form of the voltage divider expression and just pick out Vth and Rth by inspection, thus solving for both at the same time.

    This technique can be very handy; as long as the circuit in question has at least one power source (so any controlled sources can be "stimulated"), you don't have to bother with sticking external sources onto the output to find resistances or voltages.
     

    Attached Files:

  5. Feb 10, 2014 #4
    Sorry, I wasn't clear enough in my original question. I'm not sure how I forgot put these details but here they are now. I've calculated the Thevenin equivalent and I'm currently trying to get the Norton equivalent. I realize that I can just use source transformation to convert from the Thevenin equivalent to the Norton equivalent but I am trying to get the Norton equivalent directly from the circuit.

    Since I already did the Thevenin equivalent, I already know that Vx=10/21 V and Rth=10/21 Ω. Now, I'm trying to calculate IN. I've connected nodes a and b but I am having trouble applying mesh analysis. I've already applied mesh analysis to mesh 1 but after looking at it, it seems to be incorrect. Here is the correct mesh analysis:

    $$-50+12{ i }_{ 1 }+60({ i }_{ 1 }-{ i }_{ 2 })=0\\ -50+12{ i }_{ 1 }+\frac { 10 }{ 21 } =0$$

    Edit: Actually, I don't even know what Vx is. That was incorrect.

    I've tried applying nodal analysis at node X:

    $$\frac { { v }_{ x }-50 }{ 12 } +\frac { { V }_{ x } }{ 60 } +2{ V }_{ x }+{ I }_{ N }=0$$

    but I'm still stuck.
     
    Last edited: Feb 10, 2014
  6. Feb 10, 2014 #5

    gneill

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    Staff: Mentor

    Instead of nodal analysis, first redraw your circuit with the short circuit in place. A whole bunch of things get "squashed" by this short circuit. For example, what would Vx be?
     
  7. Feb 10, 2014 #6
    Hmmm, Vx would be 0V, right? And that means I can remove the 60 ohm resistor and the dependent current source, right?
     
  8. Feb 10, 2014 #7

    gneill

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    Staff: Mentor

    Try it and see :smile:

    Your results should match your Thevenin values via the simple transformation.
     
  9. Feb 10, 2014 #8
    Haha, I got it. Wow, I was stressing out over such a simple problem.
     
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