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.


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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – David Z
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

Not the answer you're looking for? Browse other questions tagged or ask your own question.