# the theoretical formula for calculating the Equicalent Resistance ofthe Complex Resistance Network

Discussion in 'Electronic Basics' started by gestapo21th, Oct 19, 2008.

1. ### gestapo21thGuest

hi, everyone!
I have found the theoretical formula for calculating the Equicalent
Resistance of the Complex Resistance Network.
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  