Using superposition to find Thevenin equivalent circuit

Discussion in 'Homework Help' started by ltj_nufan, Jan 7, 2009.

  1. ltj_nufan

    Thread Starter New Member

    Nov 23, 2008
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    Hi,

    I'm trying to study for an exam I have coming up and one of the questions in a past paper asks me to use superposition to determine the Thevenin equivalent at terminals AB of the circuit included below.

    I know that the Thevenin resistance is 10 + j10 but I can't quite figure out how to get the Thevenin voltage.
    I know that the voltage source needs to be short circuited in order to determine the effect of the current supply, and that the current supply must be made open to determine the effect of the voltage supply. But from there I can't get my head around it.

    Any help will be much appreciated.

    Thanks.
     
  2. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    If you remove ZL and open circuit the current source, do you know how to determine the voltage that appears at the A-B terminals?

    Then you short circuit the voltage source and determine the voltage that the current source produces at the A-B terminals.

    Finally, you add those two voltages.

    If you are still having problems, show your work and you will get some more help.
     
  3. ltj_nufan

    Thread Starter New Member

    Nov 23, 2008
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    By removing the current source and ZL you are left with a series circuit comprising of a 20V supply and a (10 + j10)ohm impedance, so will the voltage there just be 20V across the terminals?

    For determining the voltage due to the current supply I'm stuck here.

    Thanks for the response.
     
  4. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    For the 20v due to the voltage source, you are correct.

    For the voltage due to the current, short the voltage source. Then the current passes through the two inductors, and the voltage at the output is just the voltage across the j10 inductor.
     
  5. ltj_nufan

    Thread Starter New Member

    Nov 23, 2008
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    OK thanks very much for clearing that up.

    Can I just ask why is it just the current through the j10 inductor?
     
  6. hgmjr

    Moderator

    Jan 28, 2005
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    Having replaced the voltage source with a short circuit, you are left with the current source which has infinite impedance. The voltage developed across the inductive reactance j20 does not influence the voltage drop across the j10 inductive reactance. The load resistor is also removed so the entire 20A @ 90 degrees is flowing through the inductive reactance j10. This is why the current through the inductive reactance j10 is all you need to consider for that component of voltage associated with the 20A @ 90 degrees.

    Once you have successfully calculated the voltage at the junction of inductive reactance j10 and j20, you can sum it with the voltage at that the same junction that is due to the 20V @ 0 degrees to arrive at the final value for the Thevenin voltage.

    hgmjr
     
  7. hgmjr

    Moderator

    Jan 28, 2005
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    Have you arrived at a solution for the Thevenin equivalent voltage?

    hgmjr
     
  8. ltj_nufan

    Thread Starter New Member

    Nov 23, 2008
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    Yeah sorry for not replying sooner.

    I think I'm doing something wrong because I keep getting Vth = 0V.
     
  9. hgmjr

    Moderator

    Jan 28, 2005
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    If you are making a mistake then so am I. You have obtained the same answer that I did.

    hgmjr
     
  10. ltj_nufan

    Thread Starter New Member

    Nov 23, 2008
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    Ah maybe not then. Thanks very much for all your help.
     
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