In a meter bridge,a 30 Omega resistance is co | vedclass

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(C) For a meter bridge,the balance condition is given by $\frac{R_1}{R_2} = \frac{l}{100-l}$,where $R_1$ is the resistance in the left gap and $R_2$ i

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$30$,$20$

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(C) For a meter bridge,the balance condition is given by $\frac{R_1}{R_2} = \frac{l}{100-l}$,where $R_1$ is the resistance in the left gap and $R_2$ is the resistance in the right gap.Case $I$: $P$ and $Q$ are in series,so $R_2 = P + Q$.$\frac{30}{P+Q} = \frac{37.5}{100-37.5} = \frac{37.5}{62.5} = 0.6$$P+Q = \frac{30}{0.6} = 50 \Omega$ ... $(i)$Case $II$: $P$ and $Q$ are in parallel,so $R_2 = \frac{PQ}{P+Q}$.$\frac{30}{\frac{PQ}{P+Q}} = \frac{71.4}{100-71.4} = \frac{71.4}{28.6} \approx 2.5$$\frac{30(P+Q)}{PQ} = 2.5$Since $P+Q = 50$,we have $\frac{30 	imes 50}{PQ} = 2.5$$PQ = \frac{1500}{2.5} = 600 \Omega^2$ ... $(ii)$From $(i)$ and $(ii)$,we have a quadratic equation $x^2 - 50x + 600 = 0$.$(x-30)(x-20) = 0$.Thus,the values are $30 \Omega$ and $20 \Omega$.

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