Equivalent resistance between the adjacent corners of a regular n...

Given the total resistance of the wire is $R$. If the polygon has $n$ numbers of side, then the resistance \left(r\right)$, as shown in the following figure, of each side can be written as</p><p>$r = \dfrac{R}{n} . . . . . . . . . . . . . . . . . ( 1 )$</p><p style="text-align: center;"><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/6e595b63-e39b-423f-97d2-f24d7beb398a-9896616.png" style="width: 300px; height: 342px;" /></p><p>Among all the resistances,$\left(n - 1\right)$resistances are connected in series combination, whereas, their equivalent resistance is connected in parallel with the${n}^{th}$resistance. The equivalent circuit diagram for the above set of resistances can be depicted as follows:</p><p style="text-align: center;"><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/2d590221-a22b-416e-bf69-c761ee2bb75e-9896616_2.png" style="width: 500px; height: 169px;" /></p><p>From the above equivalent circuit diagram, the net equivalent resistance$\left({R}_{eq}\right)$for the entire combination can be calculated as follows-</p><p>$\begin{cases}\dfrac{1}{{R}_{eq}} & = & \dfrac{1}{\left(n - 1\right)r} + \dfrac{1}{r} \\ & = & \dfrac{1}{\left(n - 1\right)\dfrac{R}{n}} + \dfrac{1}{\dfrac{R}{n}} \\ & = & \dfrac{n}{\left(n - 1\right)R} + \dfrac{n}{R} \\ & = & \dfrac{{n}^{2}}{\left(n - 1\right)R} \\ {R}_{eq} & = & \dfrac{\left(n - 1\right)R}{{n}^{2}}\end{cases}$