JEE MAIN - Physics (2023 - 1st February Evening Shift - No. 4)
Equivalent resistance between the adjacent corners of a regular n-sided polygon of uniform wire of resistance R would be :
$$\frac{(\mathrm{n}-1) \mathrm{R}}{\mathrm{n}^{2}}$$
$$\frac{n^{2} R}{n-1}$$
$$\frac{(n-1) R}{(2 n-1)}$$
$$\frac{(n-1) R}{n}$$
Explanation
When, a uniform wire of resistance $\mathrm{R}$ is shaped into a regular $\mathrm{n}$-sided polygon, the resistance of each side will be,
$$ \frac{\mathrm{R}}{\mathrm{n}}=\mathrm{R}_1 $$
Let $R_1 $ and $ R_2$ be the resistance between adjacent corners of a regular polygon
$\therefore$ The resistance of $(\mathrm{n}-1)$ sides, $\mathrm{R}_2=\frac{(\mathrm{n}-1) \mathrm{R}}{\mathrm{n}}$
Since two parts are parallel, therefore,
$$ \begin{aligned} & \mathrm{R}_{\mathrm{eq}}=\frac{\mathrm{R}_1 \mathrm{R}_2}{\mathrm{R}_1+\mathrm{R}_2}=\frac{\left(\frac{\mathrm{R}}{\mathrm{n}}\right)\left(\frac{\mathrm{n}-1}{\mathrm{n}}\right) \mathrm{R}}{\left(\frac{\mathrm{R}}{\mathrm{n}}\right)+\left(\frac{\mathrm{n}-1}{\mathrm{n}}\right) \mathrm{R}} \\\\ & \Rightarrow \mathrm{R}_{\mathrm{eq}}=\frac{(\mathrm{n}-1) \mathrm{R}^2}{\mathrm{n}^2} \times \frac{\mathrm{n}}{\mathrm{R}+\mathrm{nR}-\mathrm{R}} \\\\ &\Rightarrow \mathrm{R}_{\mathrm{eq}}=\frac{(\mathrm{n}-1) \mathrm{R}}{\mathrm{n}^2} \end{aligned} $$
$$ \frac{\mathrm{R}}{\mathrm{n}}=\mathrm{R}_1 $$
Let $R_1 $ and $ R_2$ be the resistance between adjacent corners of a regular polygon
$\therefore$ The resistance of $(\mathrm{n}-1)$ sides, $\mathrm{R}_2=\frac{(\mathrm{n}-1) \mathrm{R}}{\mathrm{n}}$
Since two parts are parallel, therefore,
$$ \begin{aligned} & \mathrm{R}_{\mathrm{eq}}=\frac{\mathrm{R}_1 \mathrm{R}_2}{\mathrm{R}_1+\mathrm{R}_2}=\frac{\left(\frac{\mathrm{R}}{\mathrm{n}}\right)\left(\frac{\mathrm{n}-1}{\mathrm{n}}\right) \mathrm{R}}{\left(\frac{\mathrm{R}}{\mathrm{n}}\right)+\left(\frac{\mathrm{n}-1}{\mathrm{n}}\right) \mathrm{R}} \\\\ & \Rightarrow \mathrm{R}_{\mathrm{eq}}=\frac{(\mathrm{n}-1) \mathrm{R}^2}{\mathrm{n}^2} \times \frac{\mathrm{n}}{\mathrm{R}+\mathrm{nR}-\mathrm{R}} \\\\ &\Rightarrow \mathrm{R}_{\mathrm{eq}}=\frac{(\mathrm{n}-1) \mathrm{R}}{\mathrm{n}^2} \end{aligned} $$
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