Equivalent resistor of 2011 resistor web

[pincopallino]

1. let's have 2011 point each pair connected by a resitor R. what is the equivalent resistor between each pair of points conneced by R?



2. due to the symmetry the equivalent resistor between two arbitrary point connected by R is always the same. let's take two of this contigouspoins. the voltage at each other point is the same, due to the symmetry (is it right?), but then?

The Attempt at a Solution

 

[berkeman]

1. let's have 2011 point each pair connected by a resitor R. what is the equivalent resistor between each pair of points conneced by R?



2. due to the symmetry the equivalent resistor between two arbitrary point connected by R is always the same. let's take two of this contigouspoins. the voltage at each other point is the same, due to the symmetry (is it right?), but then?

The Attempt at a Solution

Can you please describe the problem in more detail, and maybe show us a diagram?

What do you mean by 2011 points? What is their geometry? 2-dimensional, 3-dimensional? What shape?

 

[pincopallino]

in attachment a diagram for 5 points.

 

[berkeman]

in attachment a diagram for 5 points.

You are required to show your work on schoolwork questions. What is the answer for 3 poinsts? For 5 points? etc.

And where did this question come from?

 

[pincopallino]

it is a question from our teacher.
it is evident that using standard method of resolution (kirchoff or whatever) is absurd, as it would get to a big linear sistem of equations.
for three is easy Req=2/3R. for four, it' s eay as well, as for symmetry reason the free twi point are at hte same voltage, so Req=R/2.
but 5, 6 , 7 ...2011 points? is the generalization that all the other points are equipotential corrct?

 

[Quinzio]

Of course Kirchoff is not an option.
But it looks like you already know the solution, at least partially.

 

[pincopallino]

I would like, but I do not have.
there must be some trick.
isn't there on line in the WEB any free circuit resolutor?

 

[Quinzio]

I would like, but I do not have.
there must be some trick.
isn't there on line in the WEB any free circuit resolutor?

Not in this case.
But your intuition was fine.