When two resistances R_1 and R_2 are connecte | vedclass

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(B) For a meter bridge,the balancing condition is $\frac{P}{Q} = \frac{l}{100-l}$,where $P$ is the resistance in the left gap and $Q$ is the resistanc

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$30$

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(B) For a meter bridge,the balancing condition is $\frac{P}{Q} = \frac{l}{100-l}$,where $P$ is the resistance in the left gap and $Q$ is the resistance in the right gap.Case $1$: $R_1$ and $R_2$ are in series. $P = R_1 + R_2$,$Q = 10 \ \Omega$,$l = 60 \ cm$.$\frac{R_1 + R_2}{10} = \frac{60}{100-60} = \frac{60}{40} = \frac{3}{2}$.$R_1 + R_2 = 10 	imes \frac{3}{2} = 15 \ \Omega$.Case $2$: $R_1$ and $R_2$ are in parallel. $P = \frac{R_1 R_2}{R_1 + R_2}$,$Q = 3 \ \Omega$,$l = 40 \ cm$.$\frac{R_1 R_2 / (R_1 + R_2)}{3} = \frac{40}{100-40} = \frac{40}{60} = \frac{2}{3}$.$\frac{R_1 R_2}{3(R_1 + R_2)} = \frac{2}{3} \Rightarrow R_1 R_2 = 2(R_1 + R_2)$.Substituting $R_1 + R_2 = 15 \ \Omega$:$R_1 R_2 = 2 	imes 15 = 30 \ \Omega^2$.

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