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Superposition and Thevenin theorem

  1. Jul 13, 2006 #1
    In exercise i) I had to use the superposition to find the current through R2:removing the voltage source and sticking with the current source,this was my node equation:

    Eq: v1/1000 + v1/1000 + 0.001=0

    Solving for I2 the equation was:

    V1-0/R2=I2 where I2=-5x10E-4

    Removing the current source,the result for I2=2.5x10E-3
    Adding those values in order to get the current passing through the original circuit I got:


    The problem is,I'm not totally sure if this solution is right,because this's a practice exam whose solution isn't provided,so If someone could tell me if my equations are right I'd be glad!:)

    http://i75.photobucket.com/albums/i281/esmeco/supeprositionethevenin.jpg [Broken]

    My other problem is in ii) where I need to find the Thevenin's voltage by opening the circuit on the load resistor (which is R2) and i'm not sure what is the voltage on the load resistor terminals(maybe 5 because it is connected to the voltage source?)...
    The equation I got for the current on the short circuit was:

    Eq: (Vo-5)/1000 + 0.001=0

    Once again I'm not sure if this's right...Just one more question:DO we have to add the resistors on the short circuited circuit in order to calculate Isc?
    Thanks for your time and for all the replies!
    Last edited by a moderator: May 2, 2017
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  3. Jul 13, 2006 #2

    Tom Mattson

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    There's a problem with your image. The right side of the circuit got cut off.
  4. Jul 13, 2006 #3
    Here's a new link...I hope it looks okay now...

    http://i75.photobucket.com/albums/i281/esmeco/superpositionandthevenin.jpg [Broken]
    Last edited by a moderator: May 2, 2017
  5. Jul 13, 2006 #4

    Tom Mattson

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    I agree with your solution. In both cases the current runs from right to left, so you add (as opposed to subtract) the currents.
  6. Jul 13, 2006 #5
    But regarding to the Thévenin problem,is the Thévenin voltage 5v like I said?And is equation for the current on the short circuit right?
  7. Jul 13, 2006 #6
    Well after searching a bit on the net about thevenin, I've redone my calculations and found that vth was:


    I got this equation by finding the voltage across one terminal to the other of the opened circuit.
    I obtained thevenin's resistance by short circuiting the voltage source and open circuit the current source, so Rth is:


    Could I be right on this?
  8. Jul 13, 2006 #7
    Tom's right, I2 = 3mA from right to left. As for the Thevenin problem, you are correct; VTh = 6V (with respect from right to left) and RTh = 1000ohms.
  9. Jul 13, 2006 #8
    So, I2 isn't 2x10E-3?How did you get the 3mA?My equation for the circuit with the voltage source short circuited is:

    Eq.: v1/1000 + v1/1000 +0.001=0

    Thanks anyway for the input!
    Last edited: Jul 13, 2006
  10. Jul 13, 2006 #9
    No mistake in that, and thus V1 = -0.5V. Consequently positive current is right to left. Likewise, positive current is right to left if you consider only the current source, hence the two numbers add.
  11. Jul 14, 2006 #10
    Now that I look at my equation I think I have something worng...Since I've put the equations based on the current leaving the node it should be -0.001 and not +0.001...And that way
    V1=1/2v and not -1/2....DOes this make sense?
    Last edited: Jul 14, 2006
  12. Jul 14, 2006 #11
    Given that the voltage source is deactivated (or shorted), then surely at the mid-left node, 1mA current _is_ leaving the node by the constraint of the current source. Why should you say current leaving the node is -1mA?
  13. Jul 14, 2006 #12
    Well I said that base on the current source polarity(positive on top and negative on the bottom)...
  14. Jul 14, 2006 #13
    On iii) we want to know the value of I1 so that the current through I2=0...So,I added all the currents leaving the node like this:

    I1 + I2 + I3=0 <=> I1 + 0 + I3=0 <=> I1=-I3

    Being I3 the current the passes through R1.
    The equation for the node would be:

    (v1+5)/1000 + v1/1000 + 0.001=0 <=> V1=-3

    I3=(v1+5)/1000 <=> I3=2x10E-3

    Does this look right?
  15. Jul 14, 2006 #14
    You shouldn't be too bothered about the polarity of the current source, only the arrowhead matters and that tells you also exactly where the current goes.

    Assuming that V1 refers to the voltage at the mid-left node, then you got it right in your last post with I1 + I2 + I3 = 0 where I1, I2 and I3 are the currents (left-to-right) in the three resistors R1, R2 and R3, respectively. But I don't see how you can write (v1+5)/1000 + v1/1000 + 0.001 = 0 since we have all agreed that I2 = 0 and I2 is after all = V1/1000, right?

    From I1 = -I3, you should be able to work out I3. Think about it.
  16. Jul 15, 2006 #15
    SO,since the current thta goes through I3 is 5/1000,I1 sould be -5/1000...
  17. Jul 15, 2006 #16
    It is not possible to satisfy the condition I2 = 0 unless one of the two sources is replaced- at present, we have I3 = 1mA (across R3) and I1 = (0+5)/1000 = 5mA (across R1).

    Since I do not understand the language, I can only say that the question is ill-posed.
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