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Thevenin equivalent for an AC circuit

  1. May 30, 2014 #1
    1. The problem statement, all variables and given/known data
    I have the network attached below. I need to find the open circuit voltage and short circuit current between A and B and then use these to find the Thevenin equivalent for the circuit.

    The Thevenin equivalent voltage is of the form VTsinwt and the equivalent impedance is ZT.

    2. Relevant equations
    I know the Thevenin voltage is just the open circuit voltage between A and B, and the Thevenin impedance is the Thevenin voltage over the short circuit current. The only issue is of course finding these.

    3. The attempt at a solution
    My idea is to reduce the series RC combination into a single impedance, and then find the total current. Then the open circuit voltage is that of the voltage source minus the voltage drop across the capacitor. The voltage drop across the capacitor is the total current times the capacitor impedance.

    However this turns out to be very messy, so I'm not sure if it's right or if there is a better way.

    Also for the short circuit current, would that just be the current the flows in a circuit that contained the voltage source in series with C, as no current goes down the resistor path or of the other capacitor.
     

    Attached Files:

  2. jcsd
  3. May 30, 2014 #2

    ehild

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    Your method is correct, and it will not be that messy. Use complex impedances. Show your work.

    ehild
     
  4. Jun 1, 2014 #3
    Shall post it some time soon - sorry for the late reply.
     
  5. Jun 1, 2014 #4
    Here is what I have:

    Open circuit current. Total Impedance here is 1/jwC, then the open circuit current is jwCV0ejwt.

    Open circuit voltage. Circuit is reduced down to a source in series with an impedance 1/jwC and an impedance jwRC/R+jwC (from the parallel resistor/capacitor combination). Let Z=1/jwC+jwRC/R+jwC, then calling the source V, the current flowing is I=V/Z. The voltage drop I'm interested in is the one across the impedance of jwRC/R+jwC (from the parallel resistor/capacitor combination). Call this X. So the open circuit voltage O=IX=VX/Z. This gives O=[V/(1/jwC+jwRC/R+jwC)](jwRC/R+jwC), and things get very ugly from here. I can get this into the form Vw2RC2(w2RC2-R+jwC)/w4R2C4-2w2R2C2+w2C2+R2, but that doesn't seem to be going anywhere.

    My main worry about all of this is that for a DC circuit, my Thevenin voltage is just the open circuit voltage, and thevenin resistance the open circuit voltage over the short circuit current. Here however there's all sorts of things going on with phases and my Thevenin voltage has to contain sin(wt) i.e no phase change. I'm guessing my coefficients can be complex?
     
  6. Jun 1, 2014 #5

    ehild

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    You need the short-circuit current, and then the internal impedance of the Thevenin voltage source is Z=U(open)/I(short).
    But you get that internal impedance also as the impedance between the points A and B, by replacing the ideal source by a single wire.
    Think it over. What is the resultant impedance of the parallel connected R and C?

    You need to work with complex quantities. You have complex voltage of the source, and complex impedances in the circuit. The voltage of the Thevenin equivalent source and also the internal impedance will be complex quantities. Don't worry about phases, let the expressions so as they are. If you got the complex expressions you can bring them to the proper form et the end.
     
  7. Jun 1, 2014 #6
    I know I could use that method too, but the question specifically asks for me to do it via finding the short circuit current.

    Ahhhhhhh silly me, 1/(1/jwC) does not give 1/jwC! That should help.

    Edit:

    Now I have the open circuit voltage as VwRC(2wRC+j)/1+4w2R2C2, as well as the open circuit current as jwCV with V the sinusoidal input voltage.

    So doing the division of the voltage over current gives the equivalent impedance as R(1-2jwRC)/1+4w2R2C2.

    So I have the equivalent impedance ZT=R(1-2jwRC)/1+4w2R2C2.

    I now need to express my voltage in the form VTsinwt which I'm unsure about. I don't know if VT is supposed to be complex or not. I would imagine it would be impossible for it not to be?
     
    Last edited: Jun 1, 2014
  8. Jun 1, 2014 #7

    ehild

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    Express the voltage of the Thevenin equivalent as Asin(wt+θ) where θ is the phase of the open-circuit voltage and A is its absolute value.

    ehild
     
  9. Jun 1, 2014 #8
    The questions asks for an equivalent voltage VTsinwt, and for me to find VT only though. Do you think I should just give VT as a complex number?
     
  10. Jun 1, 2014 #9

    ehild

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    When you write the voltage as VTsin(wt) it is real representation of the AC voltage, you can not use complex VT. Write VT= (magnitude of open circuit voltage)sin(wt+θ) where θ is the phase constant of the complex open-circuit voltage. Or omit the phase. I do not know how your professor want it.


    ehild
     
    Last edited: Jun 1, 2014
  11. Jun 2, 2014 #10
    Yes I will omit the phase.
     
  12. Jun 2, 2014 #11

    donpacino

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    in the future use 's' instead of 'jw'
    it can make life simpler. If you need to get the equation in the form of 'jw' then you can do that at the end after simlification
     
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