Curious problem with resistors

In summary: I assume you mean that the currents in the other three vertices are the same as the current in the vertex.
  • #1
Tolya
23
0
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)?
 
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  • #2
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?
 
  • #3
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 ;)
 
  • #4
Tolya said:
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?
 
  • #5
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" [Broken]
 
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  • #6
Hmm. The problem is also pretty easy for the vertex between adjacent and opposite. Then the symmetry breaks down...
 
  • #7
If it so easy for you, please, write me your answer :) We will check.
 
  • #8
Tolya said:
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.
 
  • #9
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.
 
  • #10
Dick said:
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?
 
  • #12
Avodyne said:

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.
 
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  • #13
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'.
 

1. What is a resistor?

A resistor is an electrical component that is used to restrict the flow of current in a circuit. It is typically made of a material with high resistance, such as carbon or metal, and is designed to dissipate heat generated by the flow of electricity.

2. How do resistors affect the flow of current in a circuit?

Resistors restrict the flow of current by creating resistance in a circuit. This resistance slows down the flow of electrons and can help control the amount of current in a circuit.

3. What causes resistors to heat up?

When current flows through a resistor, it encounters resistance, which causes some of the electrical energy to be converted into heat. The more resistance a resistor has, the more heat it will generate.

4. How do I calculate the resistance of a resistor?

The resistance of a resistor can be calculated using Ohm's Law, which states that resistance (R) is equal to the voltage (V) divided by the current (I). This can be expressed as R = V/I. The unit of resistance is ohms (Ω).

5. What are some common problems with resistors?

Some common problems with resistors include overheating, which can lead to malfunction or failure, and incorrect resistance values, which can cause issues with the functioning of a circuit. It is important to choose the right type and value of resistor for a particular circuit to avoid these problems.

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