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Linear Algebra Problem concerning a circuit

  1. Jul 7, 2007 #1
    1. The problem statement, all variables and given/known data

    [​IMG]

    By applying Kirchoff's laws to the circuit above, we obtain the following equations:

    [tex]i_1\,-\,i_2\,-\,i_3\,=\,0[/tex]

    [tex]i_1\,-\,i_2\,-\,i_3\,=\,0[/tex]

    [tex]R_2\,i_2\,-\,R_3\,i_3\,=\,0[/tex]

    [tex]R_1\,i_1\,-\,R_2\,i_2\,=\,E[/tex]

    [tex]R_1\,i_1\,-\,R_3\,i_3\,=\,E[/tex]

    Obtain the solution set of equations by Gauss elimination. If there is no solution, or if there is a non-unique solution, explain that result in physical terms.

    [tex]R_1\,=\,R_2\,=\,R_3\,\equiv\,R[/tex]



    2. Relevant equations

    Linear algebra, matrices, etc.



    3. The attempt at a solution

    First, I put the four non-identical equations into a matrix.

    [tex]\left(\begin{array}{cccc}1&-1&-1&0\\0&R&-R&0\\R&R&0&E\\R&0&R&E\end{array}\right)[/tex]

    Now I reduce it down to R.R.E.F. using elementary row operations. (Note that one of the equations is redundant)

    [tex]\left(\begin{array}{cccc}1&0&0&\frac{2E}{3R}\\0&1&0&\frac{E}{3R}\\0&0&1&\frac{E}{3R}\end{array}\right)[/tex]

    So then, [itex]i_1\,=\,\frac{2E}{3R}[/itex] and [itex]i_2\,=\,i_3\,=\,\frac{E}{3R}[/itex]?
     
    Last edited: Jul 7, 2007
  2. jcsd
  3. Jul 7, 2007 #2

    Dick

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    Look fine to me.
     
  4. Jul 8, 2007 #3
    yeah, that's right but in the initial equations, it should be:
    I2R2 + I1R1=E
    I3R3 + I1R1 = E
     
  5. Jul 8, 2007 #4

    Dick

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    True. But the signs were correct in the matrix. Took it as a typo.
     
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