Question

A piece of conducting wire of resistance $R$ is cut into $2n$ equal parts. Half the parts are connected in series to form a bundle and remaining half in parallel to form another bundle. These bundles are then connected to give the maximum resistance. The resistance of the combination is:

• A. $\frac{R}{2}(1+\frac{1}{n^2})$
• B. $\frac{R}{2}(1+n^2)$
• C. $\frac{R}{2(1+n^2)}$
• D. $R(n+\frac{1}{n})$

Answer 1

The correct option is A $\frac{R}{2}(1+\frac{1}{n^2})$

Resistance of each part $=\frac{R}{2n}$

For 'n' such parts connected in series, equivalent resistances, say $R_1=n\left[\frac{R}{2n}\right]=\frac{R}{2}$

Similarly, equivalent resistance say $R_2$ for another set of n identical respectively in parallel would be $\frac{1}{n}\left(\frac{R}{2n}\right)=\frac{R}{2n^2}$

If they are connected in parallel, equivalent resistance is : $R_{eq}=\frac{\frac{R}{2}\times \frac{R}{2n^2}}{\frac{R}{2}+\frac{R}{2n^2}}=\frac{R}{2(n^2+1)}$

If $R_1$ and $R_2$ are connected in series, to get the maximum equivalent resistance, we get:

$R_{eq}^{\prime }=R_1+R_2=\frac{R}{2}(1+\frac{1}{n^2})$

Clearly, $R_{eq}^{\prime }>R_{eq}$.

Hence, they need to be connected in series.