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Solve I(x) in the current using supermesh concept?

  1. Feb 19, 2013 #1
    1. The problem statement, all variables and given/known data
    30k6kqt.png

    2. Relevant equations
    1111lqw.png
    4A = I3 - I2 (supplementary eq.)

    Mesh 1:
    6(I1)-6(I2)=2A

    Mesh 2,3 (supermesh):
    6(I2-I1)+6(I2-I4)+12(I3)=0
    -6(I1)+12(I2)+12(I3)-6(I4)=0
    substitute supp. eq.:
    -6(I1)+12(I2)+12(I2+4A)-6(I4)=0
    -6(I1)+24(I2)-6(I4)=-48

    Mesh 4:
    6(I4-I2)=-3A
    -6(I2)+6(I4)=-3A


    3. The attempt at a solution
    |6 -6 0| |I1| = | 2 |
    |-6 24 -6| |I2| = |-48|
    |0 -6 6| |I4| = | -3|

    used cramer's rule (not sure if there was an easier way) and used online matrix calculator http://ncalculators.com/matrix/matrix-determinant-calculator.htm cause i was lazy :(

    I2 was all i needed to calculate:
    I2 =-1764/432= -4.08A

    supp. eq.:
    4A+I2=I3
    I3= -0.08A

    since I(x) is opposite to I3, i got I(x)=0.08A as my answer.
     
  2. jcsd
  3. Feb 20, 2013 #2

    The Electrician

    User Avatar
    Gold Member

    Right away I see a problem in the two equations I've marked in red. I inserted some units and I don't think it's working out very well, is it?

    The equation for mesh 1 should be I1_Amps = 2_Amps and similarly for mesh 4.
     
  4. Feb 20, 2013 #3
    i see my mistake now. i was trying to treat the current source as a voltage source but that's defnitely not right. so the equations are:

    so now this makes everything much simpler. all i need now is the supp. eq. and the eq. for the supermesh:

    -6(2A)+12(I2)+12(I3)-6(-3A)=0
    12(I2)+12(I3)=-6V
    12(I3-4A)+12(I3)=-6V
    24(I3)=42
    I3=1.75A
    I(x)=-1.75A
     
    Last edited: Feb 20, 2013
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