Verified Solution • 2025–26 Session moderate

QJEEM23Ph1a9d: JEE Main 2023 Physics - Class 12 JEEM23Ph1a9d Solution: Equivalent resistance between the adjacent corners of a regular …

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

Multiple Choice Options

  • A (n−1)Rn2\frac{(\mathrm{n}-1) \mathrm{R}}{\mathrm{n}^{2}}
  • B n2Rn−1\frac{n^{2} R}{n-1}
  • C (n−1)R(2n−1)\frac{(n-1) R}{(2 n-1)}
  • D (n−1)Rn\frac{(n-1) R}{n}

Step-by-Step Verified Solution

When, a uniform wire of resistance R\mathrm{R} is shaped into a regular n\mathrm{n} -sided polygon, the resistance of each side will be,

Rn=R1\frac{\mathrm{R}}{\mathrm{n}}=\mathrm{R}_1

Let R1R_1 and R2R_2 be the resistance between adjacent corners of a regular polygon

∴\therefore The resistance of (n−1)(\mathrm{n}-1) sides, R2=(n−1)Rn\mathrm{R}_2=\frac{(\mathrm{n}-1) \mathrm{R}}{\mathrm{n}}

Since two parts are parallel, therefore,

amp;Req=R1R2R1+R2=(Rn)(n−1n)R(Rn)+(n−1n)Ramp;⇒Req=(n−1)R2n2×nR+nR−Ramp;⇒Req=(n−1)Rn2\begin{aligned}& \mathrm{R}_{\mathrm{eq}}=\frac{\mathrm{R}_1 \mathrm{R}_2}{\mathrm{R}_1+\mathrm{R}_2}=\frac{\left(\frac{\mathrm{R}}{\mathrm{n}}\right)\left(\frac{\mathrm{n}-1}{\mathrm{n}}\right) \mathrm{R}}{\left(\frac{\mathrm{R}}{\mathrm{n}}\right)+\left(\frac{\mathrm{n}-1}{\mathrm{n}}\right) \mathrm{R}} \\\\& \Rightarrow \mathrm{R}_{\mathrm{eq}}=\frac{(\mathrm{n}-1) \mathrm{R}^2}{\mathrm{n}^2} \times \frac{\mathrm{n}}{\mathrm{R}+\mathrm{nR}-\mathrm{R}} \\\\&\Rightarrow \mathrm{R}_{\mathrm{eq}}=\frac{(\mathrm{n}-1) \mathrm{R}}{\mathrm{n}^2}\end{aligned}

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