the theoretical formula for calculating the Equicalent Resistance of the Complex Resistance Network

Oct 19, 2008 0 Replies

hi, everyone! I have found the theoretical formula for calculating the Equicalent Resistance of the Complex Resistance Network. If you are interested, please visit this page

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which is a Chinese webpage. I am sorry that I have not translated into corresponding English papers.



Let's see an example, then you will know how to calculate it.



Assuming a four vertex graph, the resistances between two vertex(it is symmetric): r(1,2)=1; r(1,3)=1/2; r(1,4)=1/3; .... r(3,4)=1/6;



The corresponding table:



  • 1 1/2 1/3

1 * 1/4 1/5
1/2 1/4 * 1/6
1/3 1/5 1/6 *

then we have this form into a matrix, which use conductance to replace resistance, T4 =



-(1+2+3) 1 2 3



1 -(1+4+5) 4 5
2 4 -(2+4+6) 6
3 5 6 -(3+5+6)

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 T3. T3 =



-(1+2+3) 1 2



1 -(1+4+5) 4
2 4 -(2+4+6)
3 5 6

Next, let the conductance (g1,2) between vertex 1 and vertex 2 replaced by number '1', we get T3(g1,2=1). And let the conductance (g1,2) replaced by number '0', we get T3(g1,2=0).



Finally, we get the Equicalent Resistance (R1,2) between vertex 1 and vertex 2 R1,2 = (|T3(g1,2=1)| - |T3(g1,2=0)|) / |T3| = (556-424)/556 = 0.2374



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