JEE Main · 2023 · 01 Feb 2023 · Shift 2 · Current Electricity

The wire of total resistance $R$ makes up the $n$ sides of the polygon, so each side has resistance $\frac{R}{n}$. Between two adjacent corners the current has two paths: the single connecting side of resistance $\frac{R}{n}$, and the remaining $(n-1)$ sides in series, of resistance $\frac{(n-1)R}{n}$. These paths are in parallel, so $R_{eq}=\frac{\frac{R}{n}\cdot \frac{(n-1)R}{n}}{\frac{R}{n}+\frac{(n-1)R}{n}}=\frac{\frac{(n-1)R^2}{n^2}}{R}=\frac{(n-1)R}{n^2}$. The equivalent resistance between adjacent corners is $\frac{(n-1)R}{n^2}$.