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CIRCUIT ANALYSIS: 4 resistor, Current src, Voltage src, V.C.V.S. - Find node Voltages

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

    Determine voltages [itex]V_1[/itex] through [itex]V_3[/itex] in the circuit below.

    [​IMG]


    2. Relevant equations

    KCL, V = iR


    3. The attempt at a solution

    So I added some variables to represent currents and a super-node. The variables are in red and the super-node in light-blue.

    [​IMG]

    [tex]V_A\,=\,V_2[/tex]

    [tex]V_3\,=\,13\,V[/tex]

    [tex]I_1\,=\,\frac{V_1\,-\,V_3}{\frac{1}{2}\Omega}\,=\,2\,V_1\,-\,2\,V_3[/tex]

    [tex]I_2\,=\,\frac{V_1\,-\,0}{1\Omega}\,=\,V_1[/tex]

    [tex]I_3\,=\,\frac{V_2\,-\,0}{\frac{1}{4}\Omega}\,=\,4\,V_2[/tex]

    [tex]I_4\,=\,\frac{V_2\,-\,V_3}{\frac{1}{8}\Omega}\,=\,8\,V_2\,-\,8\,V_3[/tex]

    Now I use KCL at the super-node:

    [tex]I_1\,+\,I_2\,+\,I_3\,+I_4\,=\,2\,A[/tex]

    [tex](2\,V_1\,-\,2\,V_3)\,+\,(V_1)\,+\,(4\,V_2)\,+\,(8\,V_2\,-\,8\,V_3)\,=\,2\,A[/tex]

    [tex]3\,V_1\,+\,12\,V_2\,-\,10\,V_3\,=\,2\,A[/tex]

    [tex]3\,V_1\,+\,12\,V_2\,-\,10(13\,V)\,=\,2\,A[/tex]

    [tex]3\,V_1\,+\,12\,V_2\,=\,132[/tex]

    And get the voltage equation from inside the super-node:

    [tex]V_1\,-\,V_2\,=\,2\,V_A[/tex]

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

    [tex]V_1\,-\,3\,V_2\,=\,0[/tex]

    Now put into a matrix and rref to get [itex]V_1[/itex] and [itex]V_2[/itex]:

    [tex]\left[\begin{array}{ccc}3&12&132\\1&-3&0\end{array}\right]\,\,\longrightarrow\,\,\left[\begin{array}{ccc}1&0&\frac{132}{7}\\0&1&\frac{44}{7}\end{array}\right][/tex]

    So I get these for [itex]V_1[/itex] through [itex]V_3[/itex]:

    [tex]V_1\,=\,\frac{132}{7}\,V\,\approx\,18.86\,V[/tex]

    [tex]V_2\,=\,\frac{44}{7}\,V\,\approx\,6.286\,V[/tex]

    [tex]V_3\,=\,13\,V[/tex]

    Does this look right?
     
    Last edited: Jan 22, 2007
  2. jcsd
  3. Jan 22, 2007 #2
    Yup, looks right to me.
     
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