Curious problem with resistors

[Tolya]

We have an infinite net of regular hexagons. Each side of hexagons has a resistance R. What is the resistance between two opposite vertexes of hexagon(s)?

 

[Dick]

That's a variation on an old problem with an infinite net of squares. Picture pushing 1 amp into the circuit at a vertex and taking it out at infinity. 1/3 amp flows through each resistor away from the vertex. Then forget that and picture taking 1 amp out of the circuit and feeding it in at infinity. Now we have 1/3 amp flowing through each resistor into the vertex. Now add the two, putting the 1 amp in at one vertex and taking it out at an adjacent one. Now the total going to infinity is zero, there is a total of 2/3 amp flowing through the connecting resistor and a total of 1 amp flowing between the two vertices through the whole network. What's the voltage across the two vertices? What's the total resistance?

 

[Tolya]

This is a misunderstanding. :) I have already solved this problem. I was only trying to represent it to the people, who are interested in it. But I suppose I wrote this problem in the wrong forum section... Dick, please, if you have the answer, write me a private message. We'll check the result ;)

 

[Dick]

This is a misunderstanding. :) I have already solved this problem. I was only trying to represent it to the people, who are interested in it. But I suppose I wrote this problem in the wrong forum section... Dick, please, if you have the answer, write me a private message. We'll check the result ;)

Actually, I think I misread your question. I was taking the resistance between adjacent vertices, not opposite. Is that correct?

 

[Tolya]

We must find the resistance between opposite vertices, not adjacent. The second problem is too easy rather than the first :)
Here is the picture of the problem discussed:
http://rghost.ru/3908/download/22312f648e26dabce002e347898f5242c521aa58/Hexagons.pdf"

 

[Dick]

Hmm. The problem is also pretty easy for the vertex between adjacent and opposite. Then the symmetry breaks down...

 

[Tolya]

If it so easy for you, please, write me your answer :) We will check.

 

[Dick]

If it so easy for you, please, write me your answer :) We will check.

I wasn't saying that the opposite vertex was easy. It looks hard, I'm still thinking about it. I was saying the other vertex in between is nearly as easy as the adjacent.

 

[Tolya]

Excuse me, I didn't catch what you wrote when I read your post first time...
You are right, the symmetry breakes down! The problem is not easy.

 

[KFC]

That's a variation on an old problem with an infinite net of squares. Picture pushing 1 amp into the circuit at a vertex and taking it out at infinity. 1/3 amp flows through each resistor away from the vertex. Then forget that and picture taking 1 amp out of the circuit and feeding it in at infinity. Now we have 1/3 amp flowing through each resistor into the vertex. Now add the two, putting the 1 amp in at one vertex and taking it out at an adjacent one. Now the total going to infinity is zero, there is a total of 2/3 amp flowing through the connecting resistor and a total of 1 amp flowing between the two vertices through the whole network. What's the voltage across the two vertices? What's the total resistance?

I occasional found this thread. The problem is very interesting ... But I don't quite understand why you get 1/3 amp out if you input 1 amp in? Any new idea and anyone explains this to me?

 

[Avodyne]

http://arxiv.org/abs/cond-mat/9909120

 

[KFC]

http://arxiv.org/abs/cond-mat/9909120

That's too complicate to me.

I find a brief introduction to the solution of related problem
http://f2.org/maths/resnet/". I understand the last two step. But for the first one : "a source current of 1amp injected at a vertex gives a current of ¼A flowing through each adjacent resistor by symmetry", I have now idea how to use symmetry to find out that the current will be 1/4 amp.

 

[Dick]

It's Kirchoff's law. If you put 1 amp into a vertex then 1 amp must flow out. There are four different paths it could take and if that is the ONLY current source then there is no reason for it to prefer one path over another. Hence 1/4 amp in each resistor. By 'symmetry'.