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CIRCUIT ANALYSIS: FInd the Thevenin equivalent of the circuit. 2 res - 1 IVS - 1 VCVS

  1. Jan 29, 2007 #1
    1. The problem statement, all variables and given/known data

    Find the Thevenin equivalent at terminals a-b of the circuit below.

    [​IMG]


    2. Relevant equations

    v = i R, KCL, KVL, [itex]R_{TH}\,=\,\frac{V_{OC}}{I_{OC}}[/itex].


    3. The attempt at a solution

    To get [itex]R_{TH}[/itex], I removed the independent voltage source and I added a test voltage of 1V, an unknown test current [itex]I_{OC}[/itex] and two KVL loops.

    [​IMG]

    [tex]V_0\,=\,-1\,V,\,\,4\,V_0\,=\,-4\,V[/tex] <----- Right?

    For KVL Loop 1)

    [tex]10000\,I_1\,+\,(1\,V)\,=\,0[/tex]

    [tex]I_1\,=\,-\frac{1}{10000}\,A\,=\,-0.0001\,A[/tex]


    For KVL Loop 2)

    [tex](-1\,V)\,+\,20000\,I_2\,+\,4\,V_0\,=\,0[/tex]

    [tex]I_2\,=\,\frac{5}{20000}\,A\,=\,0.00025\,A[/tex]

    KCL at node between two resistors)

    [tex]I_1\,+\,I_{OC}\,=\,I_2\,\,\longrightarrow\,\,I_{OC}\,=\,I_2\,-\,I_1[/tex]

    [tex]I_{OC}\,=\,(0.00025\,A)\,-\,(-0.0001\,A)\,=\,0.00035\,A[/tex]

    Now I use the equation mentioned above for [itex]R_{TH}[/itex] to get the Thevenin equivalent resistance)

    [tex]R_{TH}\,=\,\frac{(1\,V)}{0.00035\,A}\,\approx\,2857\Omega[/tex]

    Now I need to get the Thevenin equivalent voltage at the terminals a-b. I redrew the circuit to include the 70V independent voltage source and added two node voltages.

    [​IMG]

    [tex]V_0\,=\,70\,-\,V_1,\,\,V_2\,=\,4\,V_0[/tex] <----- Right?

    [tex]V_2\,=\,4\,V_0\,=\,4\,(70\,-\,V_1)\,=\,280\,-\,4\,V_1[/tex]

    KVL Loop 1)

    [tex](-70\,V)\,+\,V_0\,+\,V_{TH}\,=\,0[/tex]

    [tex]V_1\,-\,V_{TH}\,=\,70[/tex]

    KVL Loop 2)

    [tex]-V_{TH}\,+\,V_1\,+\,V_2\,=\,0[/tex]

    [tex]3\,V_1\,+\,V_{TH}\,=\,280[/tex]

    Now I have two equations in two variables and I can solve. I get [itex]V_1\,=\,87.5\,V[/itex] and [itex]V_{TH}\,=\,17.5\,V[/itex]. But wouldn't the Thevenin equivalent voltage include the contributions from both sources? (i.e. - [itex]V_{TH}[/itex] is actually [itex]V_1[/itex], [itex]V_{TH}\,=\,87.5\,V[/itex] and not [itex]V_{TH}\,=\,17.5\,V[/itex])
     
    Last edited: Jan 29, 2007
  2. jcsd
  3. Jan 29, 2007 #2
    Does the [itex]R_{TH}[/itex] look correct? Shouldn't [itex]V_1[/itex] and [itex]V_{TH}[/itex] be equal?
     
  4. Sep 13, 2008 #3

    CEL

    User Avatar

    Re: CIRCUIT ANALYSIS: Find the Thevenin equivalent of the circuit. 2 res - 1 IVS - 1

    In the equation for loop 1 you have made a wrong substitution. You should have
    [tex]V_1\,-\,V_{TH}\,=\,0[/tex]
    instead of
    [tex]V_1\,-\,V_{TH}\,=\,70[/tex]
    Of course this leads to [itex]V_{TH}=V_1[/itex]
     
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