Question

A uniform wire of resistance $R$ is shaped into a regular $n-$sided polygon $(n$ is even $)$. The equivalent resistance between any two corners can have

• A. the maximum value $\frac{R}{4}$
• B. the maximum value $\frac{R}{n}$
• C. the minimum value $R\left(\frac{n-1}{n^2}\right)$
• D. the minimum value $\frac{R}{n}$

Answer 1

Answer: (A), (C)

The correct options are
A the maximum value $\frac{R}{4}$
C the minimum value $R\left(\frac{n-1}{n^2}\right)$

When, a uniform wire of resistance $R$ is shaped into a regular n-sided polygon, the resistance of each side of polygon is $\frac{R}{n}$
Let, $R_1$ and $R_2$ be the resistances between two parts of polygon between opposite corners. Polygon will get divided into $\frac{n}{2}$ sides on both sides. Therefore,
$R_1=R_2=\left(\frac{n}{2}\right)\left(\frac{R}{n}\right)$
$R_1=R_2=\frac{R}{2}$
The two parts are parallel to each other hence, equivalent resistance between two opposite corners is
$R^{\prime }=\frac{R_1{R}_2}{R_1+R_2}$
$R^{\prime }=\frac{\left(\frac{R}{2}\right)\left(\frac{R}{2}\right)}{\frac{R}{2}+\frac{R}{2}}$
$R^{\prime }=\frac{R^2}{4R}=\frac{R}{4}$
Also, the polygon is equivalent to the combination of two resistances.
The resistance of one side, $R_1=\frac{R}{n}$
The resistance of $(n-1)$ side, $R_2=\frac{(n-1)R}{n}$
Since, the two parts are parallel, hence equivalent resistance between two adjacent corners is
$R^{\prime \prime }=\frac{R_1{R}_2}{R_1+R_2}$
$R^{\prime \prime }=\frac{\left(\frac{R}{n}\right)\left(\frac{(n-1)R}{n}\right)}{\frac{R}{n}+\frac{(n-1)R}{n}}$
$R^{\prime \prime }=\frac{(n-1)R^2}{n^2}\times \frac{n}{R+nR-R}$
$R^{\prime \prime }=\frac{(n-1)R}{n^2}$