How does this circuit technique work?

[simpleton]

Hi all,

Recently I came across a circuit problem:

Consider a cube circuit, where there is a resistor on each of the edges (Thus there are 12 resistors in total). All resistors have resistance R. Find the effective resistance between any two adjacent points.

I used the usual method of equipotential points and wire-cutting and got the answer (7R/12 in case anyone is interested). However, I came across a method that I have never seen before, and I am amazed by how it works.

In this method, suppose the vertices A and B are adjacent, and you want to find the effective resistance between them.

Firstly, you input current I into vertex A, and output I/7 current from the remaining 7 vertices. The current from vertex A will split itself evenly and flow equally into the 3 adjacent vertices (I/3 to each vertx)

Then, you output current I from vertex B, and input I/7 current from the remaining 7 vertices. The current from the 3 vertices adjacent to vertex B will equally flow to vertex B (I/3 from each vertex).

Then you superpose these two scenarios together. Other than vertex A and B, all other vertices will not have any current inputted or outputted from them. A will have 8I/7 current inputted to it and B will have 8I/7 current outputted from it. The edge AB will have current 2I/3 due to superposition.

Let K be the effective resistance of the cube circuit. Then:

8I/7 * K = 2I/3 * R => K = 7K/12

I am amazed by this method, because it is so simple and elegant! However, I am not sure when, why and how it works. Why is it that current will split itself equally into 7 parts and flow in/out of the 7 remaining vertices? Why is it that the current I flowing into A will split itself equally to the 3 adjacent vertices? I don't see why the adjacent vertices are equipotential.

Also, how do you choose where to input, output the current? Do you input/output the current at every juncture (which is why you input/output current from all of the 7 remaining vertices?)

And when does this work? After testing this method on some random circuits, I came up with a conjecture: this method only works if the effective resistance between any two adjacent vertices is the same. Or in other words, all the vertices have the same degree. I am not sure whether this works, and if it works, why it works.

Anyway, I find this technique really fascinating and hope you people can help shed some light on it. Thanks in advance! :D

 

[Dale]

Why is it that current will split itself equally into 7 parts and flow in/out of the 7 remaining vertices?

That is not a natural feature, that is a boundary condition. So imagine it as though you hook up each corner to a current source and force the current into that configuration.

Why is it that the current I flowing into A will split itself equally to the 3 adjacent vertices? I don't see why the adjacent vertices are equipotential.

By symmetry. Would the circuit look any different if you rotated it 1/3 of the way? No. So since there is no way to distinguish the circuit with the 3 different paths swapped then the 3 different paths must have the same current and voltage for each.

Also, how do you choose where to input, output the current? Do you input/output the current at every juncture (which is why you input/output current from all of the 7 remaining vertices?)

You could set up any boundary conditions that you like, but the idea is to have two boundary conditions that, by the principle of superposition, result in the final configuration that you want (all current entering one vertex and leaving an adjacent vertex). This is the simplest configuration that would do that, but not the only configuration, particularly if you allow more than 2. Also, the symmetry helps solve it easier.

And when does this work? After testing this method on some random circuits, I came up with a conjecture: this method only works if the effective resistance between any two adjacent vertices is the same. Or in other words, all the vertices have the same degree. I am not sure whether this works, and if it works, why it works.

Yes, if the vertices have different resistance then the symmetry is broken and you cannot assume that the current is equally distributed.

 

[simpleton]

Hi,

I am still not really sure why all 3 vertices are equilpotential. If I do not connect the to the cube at points A and B, then I agree with you that by symmetry, the potential of all 3 vertices is the same. However, now you are connecting the cube resistor to a circuit at points A and B. In this case, I believe that the symmetry is broken. I agree that the other two vertices are equipotential, but I am not sure why vertex B is equipotential with the other two vertices. Can I know how you determined them to by symmetric?

Also, the currents are boundary conditions that I can force, as long as they superpose to get to the final condition I want, doesn't that mean I can get multiple answers? Say instead of allowing current to flow out from 7 other vertices, I allow it to flow out from 6 other vertices, I/6 current from each vertex. Then I will get a different answer (I think 4R/7). This seems to imply that there are some stricter conditions that I need to follow when I set boundary conditions.

Sorry, as you can see, I don't really understand this technique :(

 

[Dale]

I am still not really sure why all 3 vertices are equilpotential. If I do not connect the to the cube at points A and B, then I agree with you that by symmetry, the potential of all 3 vertices is the same. However, now you are connecting the cube resistor to a circuit at points A and B. In this case, I believe that the symmetry is broken. I agree that the other two vertices are equipotential, but I am not sure why vertex B is equipotential with the other two vertices. Can I know how you determined them to by symmetric?

You are analyzing two different symmetric situations, then by superposition you add them together to get the asymmetric situation. That is one of the important features of superposition. It allows you to break a complicated problem up into several simpler problems and add the results.

Let's call Condition A the situation where you inject 1 A at A and remove 1/7 A at all of the other vertices. Condition A is highly symmetric and can easily be analyzed. Let's call Condition B the situation where you remove 1 A at B and inject 1/7 A at all of the other vertices. Condition B is also highly symmetric and can easily be analyzed. Then Condition A+B is the situation where you inject 1 A at A and remove 1 A at B. This condition is not highly symmetric and is difficult to analyze e.g. by wire-cutting, but by superposition you can simply add the results of the symmetric analyses above to easily get the analysis of Condition A+B.

Also, the currents are boundary conditions that I can force, as long as they superpose to get to the final condition I want, doesn't that mean I can get multiple answers? Say instead of allowing current to flow out from 7 other vertices, I allow it to flow out from 6 other vertices, I/6 current from each vertex. Then I will get a different answer (I think 4R/7). This seems to imply that there are some stricter conditions that I need to follow when I set boundary conditions.

Suppose the vertex C is opposite A. Let's call Condition AC the condition where 1 A is injected into A, 0 A go through C, and 1/6 A is removed at all of the other vertices. This condition is highly symmetric and easy to analyze. Let's call Condition BC the condition where 1 A is removed at B, 0 A go through C, and 1/6 A is injected at all of the other vertices. This condition is not highly symmetric and must be analyzed e.g. by wire-cutting. Once you have done that then Condition AC+BC is the same as Condition A+B and the analysis is obtained simply by adding the two and it will agree with the previous result.

 

[simpleton]

Hmm ok. I think I understand it better now. The desired configuration is not symmetric. However, what we are doing is that we are superposing two symmetric situations (and therefore we can split the currents equally) to create the resultant asymmetric situation. Then we can easily get the solution.

Thanks for your help! :D

 

[Dale]

Exactly! And you are welcome.