the theoretical formula for calculating the Equicalent Resistance of the Complex Resistance Network
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Posted by gestapo21th on October 19, 2008, 10:24 am
Please log in for more thread options I have found the theoretical formula for calculating the Equicalent Resistance of the Complex Resistance Network. If you are interested, please visit this page(http:// cid-8565f2db98f03091.spaces.live.com/), 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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