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Thevenin equivalent phasor domain dependent voltage source

  1. Apr 28, 2013 #1
    1. The problem statement, all variables and given/known data
    determine thevenin equivalent of circuit at terminals (a,b):
    cqeer.png

    2. Relevant equations
    Z(L)=jwL
    Z(C)=-j/wC
    V(th)=V(oc)
    R(th)=V(ex)/I(ex)

    3. The attempt at a solution
    actually 1st i need help with finding V(th). i do know that V(th) would be the voltage across the 2ohm resistor, so 2*I. i suppose i just need to find I, but i have no idea how to do that, since there's no independent voltage/current sources. i have tried mesh analysis for instance: 0.2I+ 0.2I + 0.2I = 0. maybe i am trying the wrong method. for R(th) i know i would have to try the V(ex) method, but 1st i need to find V(th).
     
    Last edited: Apr 28, 2013
  2. jcsd
  3. Apr 28, 2013 #2

    gneill

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    Staff: Mentor


    I don't see a 2 Ohm resistor in the diagram.

    If there are no independent sources to stimulate the dependent source, you've got no choice but to add one...
     
  4. Apr 28, 2013 #3
    sorry meant 0.2ohm resistor. and does that mean i need to add an external voltage source V(ex) in order to find v(th)?
     
  5. Apr 28, 2013 #4

    gneill

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    Staff: Mentor

    Well, if you pin the output with a fixed voltage then you lose the Vth as a variable. But you can find the Rth that way by dividing the applied potential by the current driven into the port.

    On the other hand, it looks as though the Vth should be zero since the open-circuit output voltage is zero.
     
  6. Apr 28, 2013 #5
    OK, so I add an external voltage source at terminals (a,b) on the right side so there is no open circuit anymore. at the top node the KCL is:
    V(ex)/0.2 + (V(ex)-0.2I)/0.2 - I(ex) = 0

    also, since I= -V(ex)/0.2:
    V(ex)/0.2 + (V(ex)+V(ex))/0.2 = I(ex)
    I(ex) = (3*V(ex))/0.2 = 15*V(ex)

    R(th)= V(ex)/I(ex) = 1/15= 0.066ohm. is that right?
     
    Last edited: Apr 28, 2013
  7. Apr 29, 2013 #6

    gneill

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    Staff: Mentor

    If you put a voltage source Vex at a-b the node voltage won't also be Vex; there are components between the node and Vex.
     
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