$n$ identical resistance are taken in which $\frac{n}{2}$ resistors are joined in series in the left gap and the remaining $\frac{n}{2}$ resistances are joined in parallel in the right gap of a metre bridge. Balancing length in $\mathrm{cm}$ is

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Aicharya · Accepted Answer

Meter bridge is shown in the figure below,

When $\frac{n}{2}$ resistances are joined in series in left gap each of resistance $R_1$, then equivalent resistance in left gap. $R=\frac{R_1}{2}+\frac{R_1}{2}+\frac{R_1}{2}+\ldots \frac{n}{2} \cdot \text { times }=\frac{R_1 n}{2}$ When $\frac{n}{2}$ resistors are joined in parallel in right gap, then the equivalent resistance in right gap. $\begin{aligned} \frac{1}{S} & =\frac{1}{R_1}+\frac{1}{R_1}+\frac{1}{R_1}+\ldots. \frac{n}{2} \text { times }=\frac{n}{2 R_1} \\ \Rightarrow \quad S & =\frac{2 R_1}{n} \end{aligned}$ If $l$ be the balancing length in the meter bridge wire, then $\begin{aligned} & \frac{R}{S}=\frac{l}{100-l} \Rightarrow \frac{\frac{R_1 n}{2}}{\frac{2 R_1}{n}}=\frac{l}{100-l} \\ & \frac{n^2}{4}=\frac{l}{100-l} \Rightarrow l=\frac{100 n^2}{n^2+4} \end{aligned}$

\(n\) identical resistance are taken in which \(\frac{n}{2}\) resistors are joined in series in the left gap…

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