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Resistor Cube Problem

  1. Sep 25, 2011 #1
    1. The problem statement, all variables and given/known data
    So I need to find the total voltage across the current source.

    Looking at the attachment, I took

    front square: B_______C
    ...................|..............|
    ...................|..............|
    ...................A_______D

    back square: E_______H
    ...................|..............|
    ...................|..............|
    ...................F_______G

    2. Relevant equations
    node analysis


    3. The attempt at a solution
    So I used node analysis at each corner to get a set of equations:
    -g6Vc-g1Ve-g3Vg+(g3+g1+g6)Vh=0.001
    -g5Vd-g4Vf+(g5+g4+g3)Vg-g3Vh=0
    -g8Va-g2Ve+(g8+g4+g2)Vf-g4Vg=0
    -g7Vb+(g1+g7+g2)Ve-g2Vf-g1Vh=0
    -g12Va+(g12+g9+g7Vb)-g9Vc-g7Ve=0
    -g9Vb+(g9+g10+g6)Vc-g10Ve-g6Vh=0
    -g11Va-g10Vc+(g5+g10+g11)Vd-g5Vg=0
    (-g12-g11-g8)Va+g12Vb+g11Vd+g8Vf=0.001

    where g(x) is the conductance=1/R(x)

    So I notice this as 8 equations with 8 unknowns. However, when I plug this into matlab,
    I get (Warning: Matrix is close to singular or badly scaled.)

    Is there a problem with my equations?
    Any help would be appreciated.
     

    Attached Files:

  2. jcsd
  3. Sep 25, 2011 #2

    The Electrician

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    Gold Member

    Node A is the reference node (ground). You don't need to solve for the voltage at that node, so you should eliminate that node from your equations. If you solve for 7 unknowns with 7 equations you shouldn't get that error message.
     
  4. Sep 25, 2011 #3
    alright so I removed the first column in matrix a as well as the last row

    so i did it and got that
    Vb=0.7196
    Vc=1.5696
    Vd=0.7541
    Ve=1.7336
    Vf=1.3792
    Vg=1.4909
    Vh=2.3588

    So my voltage across the current source would be 2.3588V?
    But when I check my answers looking around loops, the total voltage does not equal 0.
     
  5. Sep 25, 2011 #4

    The Electrician

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    Gold Member

    It's easy to make a mistake entering all the coefficients in a problem like this. The first thing you should do is make sure that the transpose of your coefficient matrix is the same as the matrix itself.

    If you could paste the matrix from Matlab into a post here, I could check it out.
     
  6. Sep 26, 2011 #5
    g1=1/1000
    g2=1/3000
    g3=1/4000
    g4=1/1000
    g5=1/7000
    g6=1/5000
    g7=1/2000
    g8=1/6000
    g9=1/4000
    g10=1/3000
    g11=1/2000
    g12=1/1000

    a=[0, -g6, 0, -g1, 0, -g3, g3+g1+g6;
    0, 0, -g5, 0, -g4, g5+g4+g3, -g3;
    0, 0, 0, -g2, g8+g4+g2, -g4, 0;
    -g7, 0, 0, g1+g7+g2, -g2, 0, -g1;
    g12+g9+g7, -g9, 0, -g7, 0, 0, 0;
    -g9, g9+g10+g6, 0, -g10, 0, 0, -g6;
    0, -g10, g5+g10+g11, 0, 0, -g5, 0]

    b=[0.001; 0; 0; 0; 0; 0;0]

    then I do x=a\b
     
  7. Sep 26, 2011 #6

    The Electrician

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    Gold Member

    You must arrange the matrix so that the elements containing the driving point admittances (such as g3+g1+g6) are all on the main diagonal. If you don't do this, then the columns and rows don't apply to your node variables in the same order.

    Right now, the rows of your matrix are associated with the nodes in this order:

    H
    G
    F
    E
    B
    C
    D

    Rearrange the last three rows so the order is:

    H
    G
    F
    E
    D
    C
    B

    and fix them up so that the main diagonal elements are the sums of the 3 admittances connected to the particular node. It will be less confusing and easier for me to help you if you do this.

    Your matrix should be (showing the first four fixed):

    a=[g3+g1+g6, 0, -g3, 0, -g1, 0, -g6;
    -g3, g5+g4+g3, -g4, 0, 0, 0, -g5;
    0, -g4, g8+g4+g2, -g2, 0, 0, 0;
    -g1, 0, -g2, g1+g7+g2, -g7, 0, 0;
    ............
    ............
    ............]

    The solution vector I get when this is done is:

    Vh = 2.0601
    Vg = 1.2625
    Vf = 1.1724
    Ve = 1.4885
    Vd = 0.5560
    Vc = .91499
    Vb = .49719

    Keep in mind that I could have made a mistake with all the numbers that have to be typed in, so I can't give an absolute guarantee that these are correct. But I did recheck several times.
     
  8. Sep 26, 2011 #7
    Alright so I rearranged the matrix to get:

    a=[g3+g1+g6, -g3, 0, -g1, 0, -g6, 0;
    -g3,g5+g4+g3,-g4,0,-g5,0,0;
    0,-g4,g8+g4+g2,-g2,0,0,0;
    -g1,0,-g2,g1+g7+g2,0,0,-g7;
    0,-g5,0,0,g5+g10+g11,-g10,0;
    -g6,0,0,-g10,0,g9+g10+g6,-g9;
    0,0,0,-g7,0,-g9,g12+g9+g7]

    however, this does not look like the one you started?

    And I also get the same answer as before.
     
  9. Sep 26, 2011 #8

    The Electrician

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    Gold Member

    The reason it doesn't look like the one I started is that I mixed up some of your rows.
    You see how easy it is to do! :-(

    I think you've got it correct except for one little problem.

    I think it should be:

    a=[g3+g1+g6, -g3, 0, -g1, 0, -g6, 0;
    -g3,g5+g4+g3,-g4,0,-g5,0,0;
    0,-g4,g8+g4+g2,-g2,0,0,0;
    -g1,0,-g2,g1+g7+g2,0,0,-g7;
    0,-g5,0,0,g5+g10+g11,-g10,0;
    -g6,0,0,0,-g10,g9+g10+g6,-g9;
    0,0,0,-g7,0,-g9,g12+g9+g7]

    This shows that if you arrange your matrix so that the driving point admittances are on the main diagonal, the matrix will be equal to its transpose (for networks of linear, passive components). Checking that would have revealed your last little error.

    Try that and see what you get.

    I think you've got the method ok; you're just having problems with the details (as I did when trying to rearrange your matrix).

    See the attached image for my solution. I multiplied everything by 1000 to make the matrix more compact. If you still don't get what I get, compare yours to the attached image.
     

    Attached Files:

    Last edited: Sep 26, 2011
  10. Sep 26, 2011 #9
    yep I got
    Vh = 2.0601
    Vg = 1.2625
    Vf = 1.1724
    Ve = 1.4885
    Vd = 0.5560
    Vc = .91499
    Vb = .49719

    ya nice catch on that little mistake there. thanks a lot.
     
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