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CIRCUIT ANALYSIS: Use superposition - 2 Current source, 1 Voltage source, 4 resistors

  1. Oct 4, 2006 #1
    Using superposition, find [itex]v_0[/itex] in the following circuit.


    My work so far:






    KCL @ v1:


    Using these six equations, with 6 variables:


    Is that correct?

    My real question is how to put the six equations above into matrix form to enable solving using RREF.


    EDIT: I have fixed this part with Paallikko's guidance (thanks!)







    Last edited: Oct 4, 2006
  2. jcsd
  3. Oct 4, 2006 #2
    Apply superposition two:





    From these five equations, I put into a matrix and solve.


    Next, I do the last superposition:


    The circuit above can be simplified:




    Total them up:




    Does this look right? If not, please explain the errors and how to correct them. Thanks:tongue2:
  4. Oct 4, 2006 #3


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    Homework Helper

    Well, you've understood how to use superposition (setting the sources to 0), you even seem to have most of the equations right*. However, all your calculations are wrong.

    The first one in matrix form would be (solve for ix from the first equations and substitute into the KCL equations):

    \left[ {\begin{array}{*{20}c}
    {3/4} & { - 1/2} \\
    { - 1/2} & 1 \\

    \end{array} } \right]\left[ {\begin{array}{*{20}c}
    {V_1} \\
    {V_2} \\

    \end{array} } \right] = \left[ {\begin{array}{*{20}c}
    1 \\
    0 \\

    \end{array} } \right]

    * the two equations below are wrong
    [tex]i_1=i_0 + i_2[/tex]
    [tex]v_{0\,3}\,=\,3\,\left(\frac{20\,V}{3\,\ohm}\right) \,=\,20\,V[/tex]

    PS. If you're not familiar with PSpice, there's a free version available for download eg. here: http://www.electronics-lab.com/downloads/schematic/013/ you can quickly check if your answers are correct with it.
    PPS. the topic's more engineering or introductory physics than advanced physics.
    Last edited: Oct 4, 2006
  5. Oct 4, 2006 #4
    For the first incorrect equation (in the 2nd part of problem), I see that all the currents are going out of the node. Does this mean that the actual equation is [tex]i_0\,+\,i_1\,+\,i_2\,=\,0[/tex]?

    In the third part of the problem, is the error becuase I didn't include both resistors in the I = equation?



    Is that right now?
  6. Oct 4, 2006 #5


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    Homework Helper

    Yep, they're right now.
  7. Oct 4, 2006 #6
    How do I solve the second part? I keep getting 0V for [tex]v_{02}[/tex].

    I also used Pspice and got an 8V differential at the resistor for [tex]v_0[/tex].



    This is 3V higher than Pspice reports though.
  8. Oct 5, 2006 #7


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    Homework Helper

    You've written:
    v02 = 3i0 = 6i1 = 6i3 (1)
    i2 - i3 = 2 (2)
    i0 + i1 + i2 = 0 (3)

    Substituting (2) into (3) (Eliminating i2):
    i0 + i1 + i3 = -2

    Substituting (1) into the above:
    v02(1/3 + 1/6 + 1/6) = -2

    => v02 = -3 (V)
    Last edited: Oct 5, 2006
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