Two resistances are connected in the two gaps | vedclass

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(A) Let the two resistances be $R_1$ and $R_2$. In a meter bridge,the balancing condition is $\frac{R_1}{R_2} = \frac{L}{100-L}$.Given $L = 20 \ cm$,s

Vedclass · Accepted Answer

$9$

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(A) Let the two resistances be $R_1$ and $R_2$. In a meter bridge,the balancing condition is $\frac{R_1}{R_2} = \frac{L}{100-L}$.Given $L = 20 \ cm$,so $\frac{R_1}{R_2} = \frac{20}{100-20} = \frac{20}{80} = \frac{1}{4}$.This implies $R_2 = 4R_1$. Since $R_2 = 4R_1$,$R_1$ is the smaller resistance.When $15 \ \Omega$ is connected in series with $R_1$,the new resistance is $R_1' = R_1 + 15$.The new balancing point is $L' = 40 \ cm$.Using the balancing condition again: $\frac{R_1 + 15}{R_2} = \frac{40}{100-40} = \frac{40}{60} = \frac{2}{3}$.Substitute $R_2 = 4R_1$ into the equation: $\frac{R_1 + 15}{4R_1} = \frac{2}{3}$.Cross-multiplying gives: $3(R_1 + 15) = 2(4R_1)$.$3R_1 + 45 = 8R_1$.$5R_1 = 45$,so $R_1 = 9 \ \Omega$.

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