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Equivalent resistance, using node voltage analysis

  1. Mar 29, 2014 #1
    1. The problem statement, all variables and given/known data

    3aWU4B6.jpg
    (I added in the R values for reference)

    2. Relevant equations
    GV=I
    KCL


    3. The attempt at a solution

    I am not concerned with the MATLAB portion of the problem, only the setup.

    I think what the problem wants me to do is:
    1) connect a 1A test current from the ground source to v1 (like shown in the picture), find v1 using node voltage analysis, then find the equivalent resistance by dividing the test voltage (v1) by the test current, which is 1A.

    2) connect a 1A test current from the ground source to v2. Repeat steps, where now v2 is the test voltage, to get equivalent resistance.

    3) same as above, except with v4.

    When I set up my system of equations I get too many unknowns.
    For example, trying KCL on case 1 (where 1A is connected to v1), I get this system of equations:

    v1/R1 + (v1-v2)/R2 + (v1-v3)/R6 = itest

    (v2-v1)/R2 + (v2-v4)/R7 + (v2-v7)/R3 = 0

    (v4-v2)/R7 + (v4-v3)/R11 + (v4-v6)/R12 = 0

    Plugging in 1 for i and for all R values still gives me 6 different voltages to solve for, and only 3 equations.



    Using the same method for case 2 and 3 (where vtest is v2 and v4 respectively), I get nearly identical systems of equations, the only difference being which equation is set equal to i instead of 0.
     
  2. jcsd
  3. Mar 29, 2014 #2

    gneill

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    Staff: Mentor

    There are seven essential nodes, plus the reference node. So you'll need to write seven separate node equations. That's a lot of equations to bash away at by hand :smile: which is why the problem suggests using a computer and the matrix method; The node equations can be written by inspection into matrix form.
     
  4. Mar 29, 2014 #3
    Thanks, this helped a bit but I am still stuck.
    I wrote GV=I in matrix form, where G is a 3x7 matrix with all values being 1, and then V must be a 7x3 matrix. This means that I must be a 3x3 matrix. However, when I did the node voltage equations it ended up being a 1x7 matrix: [1; 0; 0; 0; 0; 0; 0]

    Even when I make I a 3x3 matrix with 1 for the top left value and all others 0, MATLAB tells me I have the wrong dimensions when I perform the calculation.

    Also, the problem says to use this in MATLAB: V=G\I

    since the formula is GV=I, shouldn't the command then be V=I\G ?
     
  5. Mar 29, 2014 #4

    gneill

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    Staff: Mentor

    You should have 7 equations in 7 unknowns, the unknowns being the node voltages. The G matrix will be a square 7 x 7, the V and I vectors 1 x 7.

    I must admit that I am not familiar with the particulars of MATLAB equation syntax.
     
  6. Mar 29, 2014 #5

    gneill

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    Staff: Mentor

    I should have mentioned... also note that not all the entries of the G matrix will be 1. The entries on the diagonal will be the sum of the conductances connected to the given node, while the off-diagonal entries will have negative values (or zeros).
     
    Last edited: Mar 29, 2014
  7. Mar 29, 2014 #6
    Yeah I was just figuring this out. I ended up with a 7x7 G matrix, with 3's across the diagonal and either -1 or 0 values for the off-diagonals. I set I=[1;0;0;0;0;0;0], and using MATLAB obtained these voltage values:

    v1 = 0.5V
    v2 = v3 = 0.25V
    v4 = v5 = v6 = v7 = 0.125V

    Do these values make sense?

    EDIT: They don't seem to make sense to me. Since Req = vtest/itest, I get Req = (0.5)*(1) = 0.5 Ω
    That can't be true because the equivalent resistance is R1 which is 1 Ω just like all the other resistors.
     
    Last edited: Mar 29, 2014
  8. Mar 29, 2014 #7

    gneill

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    Staff: Mentor

    They aren't the node voltages that I'm seeing.

    The equivalent resistance between node 1 and the reference node won't be R1 since there are many more paths between those nodes than just the one through R1.

    Recheck your G matrix. If you can, post an image of it.
     
  9. Mar 29, 2014 #8
    I tried to check the values by part C) 3) of this problem. I ended up getting 0.111Ω for the total equivalent resistance, and 0.333Ω for the equivalent resistance at node 1. (Not sure which one I am supposed to be looking for)

    This is what I got for my G matrix:

    f4dKvfI.jpg

    EDIT: Just realized I made a mistake in row 4. Here is the new G matrix:

    aOjhHXU.jpg

    (The only difference is a -1 value on the 4th row, 2nd column)

    These are my newly obtained values:

    v1 = 0.5833V
    v2 = v3 = 0.3750V
    v4 = 0.3333V
    v5 = v7 = 0.2083V
    v6 = 0.2500V
     
    Last edited: Mar 29, 2014
  10. Mar 29, 2014 #9

    gneill

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    Staff: Mentor

    Yup. Matches what I'm seeing now. That's for the current source driving node 1.

    You'll have to move the current source to other nodes to find their resistances to ground.
     
  11. Mar 29, 2014 #10
    So then would this mean that for the current source driving node 1, Req = v1/itest = 0.5833/1 = 0.5833Ω ?

    These are my results for the other 2 nodes:

    node 2:
    v1 = v3 = v6 = v7 = 0.3750V
    v2 = 0.7500V
    v4 = 0.5000V
    v5 = 0.2500V

    Req = 0.7500/1 = 0.75Ω

    node 4:
    v1 = v5 = v7 = 0.3333V
    v2 = v3 = v6 = 0.5000V
    v4 = 0.8333V

    Req = 0.8333/1 = 0.8333Ω
     
  12. Mar 30, 2014 #11

    gneill

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    Staff: Mentor

    Yes that looks good for nodes 2 and 4 :smile:
     
  13. Mar 30, 2014 #12
    Great, much appreciated!
     
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