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Did i solve my matrix correctly? (source sub. on 3 window circuit)

  1. Jan 14, 2008 #1
    I've got two equations that i want to put into a matrix and solve for Va and Vb:

    1.7ma = (1/5k + 1/20k) -1/20k Va
    -2ma = -1/20k (1/4k + 1/20k) Vb

    The answers I got were Va= 5.655v and Vb= -5.724

    then, using I= (Vb-Va)/R ----> (-11.379/20k) = 0.65ma




    ..-----Va----20k----Vb-----
    |.......|..................|.......|
    ^.....5k.................4k......V
    1.7ma.|..................|.....2ma
    .| ___|___________|_____|

    (ignore the periods, i needed them for spacing)
     
    Last edited: Jan 14, 2008
  2. jcsd
  3. Jan 14, 2008 #2

    Tom Mattson

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    Your equations look wrong, but I agree with your answers.

    Look again. You've got a negative divided by a positive. That should be negative. However, you've got Vb and Va reversed. Resistors are passive, so to compute the current you subtract the lower potential from the higher one, not the other way around.
     
  4. Jan 14, 2008 #3
    **I set up my equations based on another example from class, what is the correct matrix set up for an example like this?

    ..-------Va----R2--I>---Vb------
    |..........|.....................|.........|
    ^........R1....................R3.......V
    Is1........|.....................|........Is2
    .| _____|_____________|______|



    **so after switching Va and Vb and correcting the neg/pos mistake, the current should be the same, right?
     
  5. Jan 14, 2008 #4

    Tom Mattson

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    Regarding your equations: Let me show you how I interpreted them, using LaTeX.

    I read this as:

    [tex]1.7mA=\left(\frac{1}{5k\Omega}+\frac{1}{20k\Omega}\right)-\frac{1}{20k\Omega}V_a[/tex]

    The equation, as I read it, is dimensionally wrong. The LHS has units of current. The first term on the RHS has units of reciprocal resistance. Also, Vb doesn't appear at all! Here is how the equation should read.

    [tex]1.7mA=\frac{1}{5k\Omega}V_a+\frac{1}{20k\Omega}\left(V_a-V_b\right)[/tex]

    On to the second equation.

    This equation should read as follows.

    [tex]-2mA=\frac{1}{4k\Omega}V_b-\frac{1}{20k\Omega}\left(V_a-V_b\right)[/tex]

    But I checked your answers using KCL at nodes a and b, and they worked.
     
  6. Jan 14, 2008 #5
    oh i see. i had set up a matrix as below where brackets above one another are actually one, but i just cant type them that way here.

    [1.7ma] = [ (1/5k + 1/20k), -1/20k ] [ Va ]
    [-2ma ] = [ -1/20k, (1/4k + 1/20k) ] [ Vb ]

    thanks you for your help!
     
    Last edited: Jan 14, 2008
  7. Jan 14, 2008 #6

    Tom Mattson

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    No problemo. And if you're interested in learning how to use LaTeX here, you should consult the following thread:

    https://www.physicsforums.com/showthread.php?t=8997

    You can also see the LaTeX code for any equation that you see by clicking on it (you will need to allow pop-ups for this).
     
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