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Mesh Current Analysis

  1. Jan 30, 2007 #1
    Hi I need to find Vab in the following circuit using Mesh Currents...Ive tried this a million times, I understand the method...but im definitely doing something wrong. Whould I Simplify the top loop?...if its possible. Ive been doing it as 3 loops, but i dont know what to use as a voltage for the top.

    the file is attached,

    Here is my matrix:[tex] \left( \begin{array}{ccc} 8 & -6 & -2\\-6 & 19 & -5\\-2 & -5 &7\end{array}\right) \left( \begin{array}{c} I_1\\I_2\\I_3\end{array}\right) = \left( \begin{array}{c} 28\\0\\20\end{array}\right)[/tex]

    [tex]V_{ab}[/tex] is suppose to equal 10V...

    Thanks
     

    Attached Files:

    Last edited: Jan 30, 2007
  2. jcsd
  3. Jan 30, 2007 #2

    berkeman

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    Staff: Mentor

    I assume you have I1 in top loop, I2 in left loop, I3 in right loop, all going clockwise? Please show us your full initial equations so we can check how you got to the final matrix. Like, for the top loop you get:

    8(I1) + 5(I1-I3) + 6(I1-I2) = 0

    Right? What are the other two loop equations, and then show the steps to get to your matrix equation, and then show how you solved it with the determinant.
     
  4. Jan 30, 2007 #3
    The way we were taught to do it was for the i1ji position in the matrix, you sum all the resistances in that loop. i2j2 is the same for that loop and i3j3 is the same for the 3rd loop. the i1j2 position is the resistances that are common to the 1st and second loop...etc..
     
  5. Jan 30, 2007 #4

    berkeman

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    Staff: Mentor

    Well, okay, but if it were me and I was getting the wrong answer, I'd do it the long way that I mentioned just to check my answer.....
     
  6. Apr 2, 2009 #5
    The last equation coefficient on the right side vector should be -20 based upon the equation you wrote for Loop 3. If you then take the difference between the 2 loop currents coupling the 2 Ohm resistor and use Ohm's Law you should get 10 Volts.
    In your case this would be (i1-i3)2=10 Volts. i1 = 4 A; i2 = 1 A; i3 = -1 A
     
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