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

Deli Zhang

Apr 8, 2015
3
Joined
Apr 8, 2015
Messages
3
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
 

Attachments

  • The Formula for the Equivalent Resistance of Complex Resistance Network.pdf
    302.5 KB · Views: 127

Deli Zhang

Apr 8, 2015
3
Joined
Apr 8, 2015
Messages
3
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
 

Laplace

Apr 4, 2010
1,252
Joined
Apr 4, 2010
Messages
1,252
You will feel how magical the formula is!
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.
 

Attachments

  • Lattice-Resistor.pdf
    52.4 KB · Views: 150
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