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Yesterday while doing some questions, I came across a formula that if a wire with uniform resistance R is made into n sided regular polygon then the net resistance between any two corners with $x-1$ vertices in between is

$$R_{net} =\frac{Rx(n-x)} {n^2} $$

It's just written in my book with no proof. I have tried proving it but could not. I want the physics behind it rather than just the formula. I would appreciate any kind of help.

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closed as off-topic by David Z May 10 at 12:35

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  • $\begingroup$ Possible duplicate: physics.stackexchange.com/q/478964/2451 $\endgroup$ – Qmechanic May 9 at 15:40
  • $\begingroup$ @Satwik Hello Satwik I recommend you to post the proof on this website. $\endgroup$ – Unique May 9 at 15:44
  • $\begingroup$ Hi Satwik. Welcome to Phys.SE. If you haven't already done so, please take a minute to read the definition of when to use the homework-and-exercises tag, and the Phys.SE policy for homework-like problems. $\endgroup$ – Qmechanic May 9 at 16:55
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So you've made an $n$ sided polygon out of a wire of resistance $R$. Each segment of the polygon therefore has resistance $R/n$. If there are $x - 1 $ vertices between two points, we have basically formed a parallel circuit where one branch has $x$ segments and one has $n - x$. We then use the formula for a parallel resistance: $$ \frac{1}{R_\mathrm{total}} = \frac{1}{R_1} + \frac{1}{R_2} $$ $R_1 = x R/n$ and $R_2 = (n - x) R/n$. As a result: $$ \frac{1}{R_1} + \frac{1}{R_2} = \frac{n}{x R} + \frac{n}{ (n - x) R} = \frac{n(n -x)R + nx R}{(n -x)x R^2} = \frac{n^2}{(n -x )x R} $$

Finally, inverting this yields the following formula: $$ R_\mathrm{total} = \frac{ R x (n-x)}{n^2} $$

I don't see what an even number of sides has to do with it, but maybe I'm missing some obvious geometry.

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  • $\begingroup$ Agreed, the formula appears to be general, as far as I can tell. $\endgroup$ – probably_someone May 9 at 15:36
  • $\begingroup$ Yeah i guess nothing to do with even sides $\endgroup$ – Satwik May 9 at 15:42

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