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Homework Help: 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.

    ch4prob14.jpg


    My work so far:

    ch4prob14_Part1.jpg

    [tex]v_1\,=\,2\,i_2\,+\,v_{0\,1}[/tex]

    [tex]v_1\,=\,4\,i_1[/tex]

    [tex]v_{0\,1}\,=\,3\,i_3[/tex]

    [tex]v_{0\,1}\,=\,6\,i_4[/tex]

    KCL @ v1:
    [tex]1\,A\,=\,i_1\,+\,i_2[/tex]

    [tex]i_2\,=\,i_3\,+\,i_4[/tex]


    Using these six equations, with 6 variables:

    [tex]v_{0\,1}\,=\,\frac{12}{7}\,V[/tex]

    Is that correct?

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

    Thanks:tongue:

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

    [tex]i_1\,=\,\frac{1}{4}\,v_1[/tex]

    [tex]i_2\,=\,\frac{1}{2}\,v_1\,-\,\frac{1}{2}\,v_{01}[/tex]

    [tex]i_3\,=\,\frac{1}{3}\,v_{01}[/tex]

    [tex]i_4\,=\,\frac{1}{6}\,v_{01}[/tex]

    [tex]i_1\,+\,i_2\,=\,1\,A\,\,\,\Rightarrow\,\,\,\frac{3}{4}\,v_1\,-\,\frac{1}{2}\,v_{01}\,=\,1[/tex]

    [tex]i_3\,+\,i_4\,=\,i_2\,\,\,\Rightarrow\,\,\,\frac{1}{3}\,v_{01}\,+\,\frac{1}{6}\,v_{01}\,=\,\frac{1}{2}\,v_1\,-\,\frac{1}{2}\,v_{01}[/tex]

    [tex]v_{01}\,=\,1\,V[/tex]
     
    Last edited: Oct 4, 2006
  2. jcsd
  3. Oct 4, 2006 #2
    Apply superposition two:

    ch4prob14_Part2.jpg

    [tex]i_3\,=\,i_2\,-\,2\,A[/tex]

    [tex]i_1\,=\,i_0\,+\,i_2[/tex]

    [tex]v_{0\,2}\,=\,6\,i_1\,=\,6\,i_3\,=\,3\,i_o[/tex]

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

    [tex]v_{0\,2}\,=\,-6\,V[/tex]


    Next, I do the last superposition:

    ch4prob14_Part3.jpg

    The circuit above can be simplified:

    ch4prob14_Part4.jpg

    [tex]v_{0\,3}\,=\,3\,I[/tex]

    [tex]v_{0\,3}\,=\,3\,\left(\frac{20\,V}{3\,\ohm}\right)\,=\,20\,V[/tex]

    Total them up:

    [tex]v_0\,=\,v_{0\,1}\,+\,v_{0\,2}\,+\,v_{0\,3}[/tex]

    [tex]v_0\,=\,\left(\frac{12}{7}\,V\right)\,+\,(-6\,V)\,+\,(20\,V)[/tex]

    [tex]v_0\,=\,15.7\,V[/tex]

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

    Päällikkö

    User Avatar
    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):

    [tex]
    \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]
    [/tex]




    * 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?

    [tex]v_{03}\,=\,3\,I[/tex]

    [tex]v_{03}\,=\,3\,\left(\frac{20\,V}{6\,\ohm}\right)\,=\,10\,V[/tex]

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

    Päällikkö

    User Avatar
    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].

    [tex]v_0\,=\,v_{01}\,+\,v_{02}\,+\,v_{03}[/tex]

    [tex]v_0\,=\,1V\,+\,0V\,+\,10V\,=\,11V[/tex]

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

    Päällikkö

    User Avatar
    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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