Question 115 of 380

Equivalent resistance between the adjacent corners of a regular $n$-sided polygon of uniform wire of resistance $R$ would be :

A$\dfrac{\left(n-1\right)R}{{n}^{2}}$B$\dfrac{\left(n-1\right)R}{\left(2n-1\right)}$C$\dfrac{{n}^{2}R}{n-1}$D$\dfrac{\left(n-1\right)R}{n}$

Show answer and solution▾

Answer: (A) $\dfrac{\left(n-1\right)R}{{n}^{2}}$

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
$r=\dfrac{R}{n}.................(1)$
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:
From the above equivalent circuit diagram, the net equivalent resistance $\left({R}_{eq}\right)$ for the entire combination can be calculated as follows-
$\begin{matrix}\dfrac{1}{{R}_{eq}} & = & \dfrac{1}{\left(n-1\right)r}+\dfrac{1}{r} \\ & = & \dfrac{1}{\left(n-1\right)\dfrac{R}{n}}+\dfrac{1}{\dfrac{R}{n}} \\ & = & \dfrac{n}{\left(n-1\right)R}+\dfrac{n}{R} \\ & = & \dfrac{{n}^{2}}{\left(n-1\right)R} \\ {R}_{eq} & = & \dfrac{\left(n-1\right)R}{{n}^{2}}\end{matrix}$

JEE Main Physics practice question on Current Electricity (Combination of Resistances), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.