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Question about Thevenin Equivalent Vth

  1. Dec 7, 2012 #1
    Hi guys,

    I have a question that is nagging me quite a lot. Let us say that we have a circuit who has two independent sources (any combination of voltage and current sources). In the determination of Vth, can we choose whether or not to use superposition to find this Vth? Because my understanding was that you could use superposition to simplify the analysis or choose to leave both the sources in and find the equivalent voltage. Am I correct?

    Thanks guys :).
  2. jcsd
  3. Dec 7, 2012 #2
    Yes you can use superposition to find this Vth voltage.
  4. Dec 7, 2012 #3
    Thanks! Appreciate it. I have another question related to the determination of Vth. I have an example I am doing where the A and B terminals are in the middle. What does one do to determine this Vth then? Because when the terminals are on the left or the right, it makes sense how we are taking a look at this circuit's inner workings. But the confusion is there when it is in between.

    Here's a picture of the problem:
  5. Dec 7, 2012 #4
    Well In your diagram A,B terminals are not in the middle.
    Because we always can redraw it to this form


    And Vth = - 1mA*1KΩ = -1V and Rth = 1KΩ

    Attached Files:

  6. Dec 7, 2012 #5
    Oh, okay that makes sense. One last diagram I want to ask about is this:


    I somewhat understand how we can get Vth to be 6V using the left side of the circuit. Why can't we use the 2mA and the 2kΩ resistor to determine the same Vth?
  7. Dec 7, 2012 #6
    The answer is that the second Kirchhoff's law doesn't allow us to do so.
    Why? I hope that this diagram explains everything


    As you can see Vth voltage is equal to

    Vht = V2 + V3 and also Vth = V1 (In a parallel circuit, the voltage across each of the components is the same)


    Vth = 2mA * 3KΩ = 6V

    Attached Files:

    • 1.PNG
      File size:
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  8. Dec 7, 2012 #7
    That makes perfect sense! Thank you so much :). I truly appreciate it
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