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Homework Help: CIRCUITS: Current Controlled Voltage source, Indep. Voltage source, Four resistors

  1. Oct 18, 2006 #1
    Find [itex]v_1,\,v_2,\,and\,v_3[/itex] in the circuit below using nodal analysis:

    PracticeProb3-4.jpg

    My work so far:

    [tex]I\,=\,\frac{v_1}{2\Omega},\,\,I_1\,=\,\frac{v_2}{4\Omega},\,\,I_2\,=\,\frac{v_3}{3\Omega},\,\,I_3\,=\,\frac{v_1\,-\,v_3}{6\Omega}[/tex]

    KVL @ loop1 => [tex]-I\,(2\Omega)\,+\,10\,V\,+\,I_1(4\Omega)\,=\,0[/tex]

    Which equals:
    [tex]-\left(\frac{v_1}{2\Omega}\right)(2\Omega)\,+\,10\,V\,+\,\left(\frac{v_2}{4\Omega}\right)(4\Omega)\,=\,0[/tex]

    Which equals:
    [tex]-v_1\,+\,10\,V\,+\,v_2\,=\,0[/tex]

    KVL @ loop2 => [tex]-v_2\,-\,5\,I\,+\,v_3\,=\,0[/tex]

    KVL @ loop3 => [tex]-10\,V\,+\,v_1\,-\,v_3\,+\,5\,I\,=\,0[/tex]

    KCL @ v1 => [tex]I\,+\,I_3\,+\,I_4\,=\,0[/tex]

    KCL @ v2 => [tex]I_4\,=\,I_1\,+\,I_5[/tex]

    KCL @ v3 => [tex]I_2\,=\,I_5\,+\,I_3[/tex]

    KCL @ Super Node 1 => [tex]I_4\,+\,I_3\,=\,I_1\,+\,I_2[/tex]

    When I combine these equations to get 4 equations with 4 variables, I get the following matrix:

    [tex]\left[\begin{array}{cccc|c}
    -1 & 1 & 0 & 0 & -10 \\
    0 & -1 & 1 & -5 & 0 \\
    1 & 0 & -1 & 5 & 10 \\
    \frac{1}{2} & \frac{1}{4} & \frac{1}{3} & 0 & 0
    \end{array}\right][/tex]

    The columns go like this: v1, v2, v3, I, constant

    But this matrix has infinte solutions! How do I solve?
     
    Last edited: Oct 18, 2006
  2. jcsd
  3. Oct 18, 2006 #2
    I figured it out!

    The last variable column of the matrix can be eliminated because [tex]I\,=\,\frac{v_1}{2\Omega}[/tex]

    This gives the matrix:

    [tex]\left[\begin{array}{ccc|c}-1 & 1 & 0 & -10 \\-\frac{5}{2} & -1 & 1 & 0 \\ \frac{7}{2} & 0 & -1 & 10 \\ \frac{1}{2} & \frac{1}{4} & \frac{1}{3} & 0 \end{array}\right][/tex]

    This RREF's out to:

    [tex]v_1\,=\,\frac{70}{23}\,=\,3.043\,V[/tex]

    [tex]v_2\,=\,-\frac{160}{23}\,=\,-6.956\,V[/tex]

    [tex]v_3\,=\,\frac{15}{23}\,=\,0.6522\,V\,V[/tex]

    NOTE: The third row of the matrix is not required!
     
    Last edited: Oct 18, 2006
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