When the two known resistances R and S are co | vedclass

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(A) In the first case,the meter bridge balance condition is given by: $\frac{R}{l_1} = \frac{S}{100-l_1} \implies R = S \frac{l_1}{100-l_1}$.In the se

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$\frac{S l_1(100-l_2)}{100(l_2-l_1)}$

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(A) In the first case,the meter bridge balance condition is given by: $\frac{R}{l_1} = \frac{S}{100-l_1} \implies R = S \frac{l_1}{100-l_1}$.In the second case,$X$ is connected in parallel with $S$. The equivalent resistance $S'$ is given by: $S' = \frac{XS}{X+S}$.The new balance condition is: $\frac{R}{l_2} = \frac{S'}{100-l_2} = \frac{XS}{(X+S)(100-l_2)}$.Substituting $R$ from the first equation: $\frac{S l_1}{(100-l_1) l_2} = \frac{XS}{(X+S)(100-l_2)}$.Simplifying: $\frac{l_1}{l_2(100-l_1)} = \frac{X}{(X+S)(100-l_2)} \implies \frac{X+S}{X} = \frac{l_2(100-l_1)}{l_1(100-l_2)}$.$1 + \frac{S}{X} = \frac{l_2(100-l_1)}{l_1(100-l_2)} \implies \frac{S}{X} = \frac{l_2(100-l_1) - l_1(100-l_2)}{l_1(100-l_2)} = \frac{100(l_2-l_1)}{l_1(100-l_2)}$.Therefore,$X = \frac{S l_1(100-l_2)}{100(l_2-l_1)}$.

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