Thevenin Equivalent

Discussion in 'Homework Help' started by tjgear17, Dec 9, 2010.

  1. tjgear17

    Thread Starter New Member

    Dec 9, 2010
    1
    0
    Ok so I am trying to do the Thevenin equivalent to this circuit. I have been starring at this circuit for a while now a while I have the concept of Thevenin down pretty good. My problem is that since a and b are across a corners of the circuit I am not sure how to start my analysis. My instincts tell me that E_th is the voltage across the capacitor and resistor; and since the inductor is parallel to the resistor it is equivalent to it. But if this is correct I am not sure how to analyze for Z_th
     
  2. blah2222

    Well-Known Member

    May 3, 2010
    554
    33
    Well it's already pretty much setup in a thevenin/norton configuration.

    Set the current source to zero (open circuit) and find the equivalent impedance: Zeq = -j5 + j10||8 = 4.878 - j1.098 ohms = 5<-12.68 ohms

    Set the circuit back up in terms of the current source and the equivalent impedance. The current source is in parallel with Zeq and Vab is the voltage in parallel with both of them.

    This is the Norton equivalent of the circuit. To get the Thevenin circuit, find Vth by the relation Vth = In*Zeq = (2<-45)*(5<-12.68) = 10<-57.68 V

    The Thevenin circuit is just Vth in series with Zeq, and Vab is the voltage between the open end of Zeq and the negative terminal of Vth.
     
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