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The Formula for the Equivalent Resistance of Complex Resistance Network(Circuits)

Discussion in 'General Electronics Discussion' started by Deli Zhang, Apr 10, 2015.

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  1. Deli Zhang

    Deli Zhang

    3
    0
    Apr 8, 2015
    Hi All,
    I am back!
    In 2008 I posted the thread https://www.electronicspoint.com/thr...ance-ofthe-complex-resistance-network.121263/ and just provided a example. Now I have time to complete the English paper The Formula for the Equivalent Resistance of Complex Resistance Network.pdf as below attachment.
    In this paper I given the formula of equivalent resistance for arbitrary circuits and proof. To understand the paper you may have to learn Graph Theory and Metrix Theory, the wording statement of the paper is professional to a mathematics student, even so I will help you understand it.

    The example that year took is still the best.
    --------------------------------------------------------------------------------
    Assuming a four vertices graph (means a circuit with 4 nodes), the conductance between two vertices(it is
    symmetric. ):
    g(1,2)=1; g(1,3)=2; g(1,4)=3; g(2,3)=4; g(2,4)=5; g(3,4)=6;

    The corresponding metrix for the graph:
    upload_2015-4-10_23-25-8.png
    then got the matrix of matrix-tree theory,
    M4 =
    upload_2015-4-10_23-25-29.png

    Attention on the main diagonal elements, which are the sum of all
    conductance on the same row and then multiplied by -1

    Then, remove the last row and the last column, we get determinant T3.
    T3 =
    upload_2015-4-10_23-25-45.png

    Next, let the conductance g(1,2) between vertex 1 and vertex 2
    replaced by number '1', we got T3((g1,2)=1). which is same to T3, because g(1,2)=1 originally.
    And let the conductance g(1,2) replaced by number '0', we got
    T3(g(1,2)=0) =
    upload_2015-4-10_23-25-59.png

    Finally, we get the Equivalent Resistance G(1,2) between vertex 1 and
    vertex 2
    G(1,2)
    = |T3| / (|T3(g(1,2)=1)| - |T3((g(1,2)=0)|)
    = (-556) / ((-556)-(-424))
    = 139/33
    = 4.2121
    Equivalent Resistance R(1,2) = 1 / G(1,2) = 0.2374

    -------------------------------
    Same method, you can get any others,
    e.g. G(2,4)

    the raw value of g(2,4) is 5, now replaced by 1, so
    T3(g(2,4)=1) =
    upload_2015-4-10_23-26-16.png
    and replaced by 0, got
    T3(g(2,4)=0) =
    upload_2015-4-10_23-26-30.png

    So
    G(2,4) = |T3| / (|T3(g(2,4)=1)| - |T3((g(2,4)=0)|)
    = (-556) / ((-284)-(-216))
    = 139/17
    = 8.1765
    Equivalent Resistance R(2,4) = 1 / G(2,4) = 0.1223
    ... ...

    Any question, please raise.

    Thanks,
    Deli
     
  2. Deli Zhang

    Deli Zhang

    3
    0
    Apr 8, 2015
    Another example. You will feel how magical the formula is!


    upload_2015-4-11_20-40-20.png

    What's the equivalent resistance R(A,B) between A and B?

    Solution:
    Use conductance to describe the circuit.
    upload_2015-4-11_20-40-20.png


    So the equivalent resistance R(A,B)=5/6
     
  3. Laplace

    Laplace

    1,252
    184
    Apr 4, 2010
    Here is an alternate procedure of solving for the lattice resistance using KCL with node equations. Not sure which method should be regarded as the most magical.
     

    Attached Files:

  4. pebe

    pebe

    83
    11
    Sep 3, 2013
    I make it 1ohm.

    Edit: Oops! My mistake.
     
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