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CIRCUIT ANALYSIS: 3 resistrs, I src, V src, V.C.V.S. in series w/ resistor - Find V_A

  1. Jan 22, 2007 #1
    1. The problem statement, all variables and given/known data

    Use nodal analysis to find [itex]V_A[/itex] in the circuit below.

    [​IMG]


    2. Relevant equations

    KVL, KCL, v = i R, super-node?


    3. The attempt at a solution

    I made a few currents and showed the obvious KVL loop.

    [​IMG]

    [tex]V_1\,=\,V_A[/tex] <---- Right?

    [tex]I_1\,=\,\frac{V_1\,-\,0}{2\Omega}[/tex]

    [tex]I_2\,=\,\frac{V_3\,-\,0}{16\Omega}[/tex]

    [tex]I_3\,=\,\frac{V_1\,-\,V_2}{4\Omega}[/tex] <------Right?


    I know KCL for [itex]V_1[/itex]:

    [tex]\frac{30\,-\,V_1}{1\Omega}\,=\,\frac{V_1\,-\,0}{2\Omega}\,+\,\frac{V_1\,-\,V_2}{4\Omega}[/tex]

    [tex]7\,V_1\,-\,V_2\,=\,120[/tex]


    I know KCL for [itex]V_3[/itex]:

    [tex]\frac{V_1\,-\,V_2}{4\Omega}\,+\,3\,=\,\frac{V_3\,-\,0}{16\Omega}[/tex]

    [tex]4\,V_1\,-\,4\,V_2\,-\,V_3\,=\,-48[/tex]


    For KVL1:

    [tex]-V_1\,+\,4\,I_3\,+2\,V_A\,+\,V_3\,=\,0[/tex]

    [tex]2\,V_1\,-\,V_2\,+\,V_3\,=\,0[/tex]


    Now, putting those three EQs into a matrix and rref:

    [tex]\left[\begin{array}{cccc}7&-1&0&120\\4&-4&-1&-48\\2&-1&1&0\end{array}\right]\,\,\longrightarrow\,\,\left[\begin{array}{cccc}1&0&0&\frac{648}{29}\\0&1&0&\frac{1056}{29}\\0&0&1&-\frac{240}{29}\end{array}\right][/tex]

    This gives me [itex]V_1\,=\,V_A[/itex] equal to 22.34 V.

    Thanks for the help mjsd!
     
    Last edited: Jan 22, 2007
  2. jcsd
  3. Jan 22, 2007 #2

    mjsd

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    all your equations look sound to me..
     
  4. Jan 22, 2007 #3

    mjsd

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    oh..hang on.... 2nd line for KCL for V3 is wrong where is the 48 gone?
     
  5. Jan 22, 2007 #4
    Corrected, thanks much!
     
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