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Thevenin equivalent

  1. Mar 29, 2006 #1
    i have an attempted solution to a HW problem, and I want to know if it is right or wrong. Could you guys please help me out. I have attached the image as a pdf

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    Last edited: Mar 29, 2006
  2. jcsd
  3. Mar 30, 2006 #2
    Before you can actually start solving the problem, you will need to identify firstly the two points A and B for which the Thevenin equivalent circuit is defined. So where are these points upon which you want to find the Thevenin equivalent circuit? Across the 10k resistor?
  4. Mar 30, 2006 #3
    if you look at the second circuit i have labeled the two nodes V_A and V_AB, those are the two nodes in which i am going to do nodal analysis in.
  5. Mar 30, 2006 #4
    I didn't calculate the Vth, but your work nodal analysis looks correct to me. On a side note, you didn't really need to convert the mA to A or the kohms to ohms. The prefixes will cancel giving you V during the calculations. Now find Rth to complete the problem.
  6. Mar 31, 2006 #5
    GREAT!!!!! thanks teknodude!!
  7. Apr 1, 2006 #6
    I have a few concerns.

    1. Why remove the 10k resistor? Shouldn't the load resistor (if there should be one) be lying between the two points where the Thevenin equivalent circuit is defined?

    2. I don't think [tex]V_{AB} = V_{Th}[/tex] as you have written in the solution. If the Thevenin equivalent circuit is to be determined across nodes AB and A, then [tex]V_{Th} = V_{AB} - V_A[/tex].
  8. Apr 1, 2006 #7
    you may be right about the load resistor lying betwen the two points. But even if i dont remove the load resistor, and use the same technique to find V_AB, then id still have the same equations because there would still be no current through the 10k resistor.
  9. Apr 2, 2006 #8
    That's true. But you should let the 10k resistor be and [tex]V_{Th}[/tex] should be as I stated above.
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