Connect with us

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

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

Scroll to continue with content
  1. gestapo21th

    gestapo21th Guest

    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(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
     
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day

-