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

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

Vedclass · Accepted Answer

$30, 20$

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(C) For a meter bridge,the balancing 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 $R_2 = P + Q$. Given $l = 37.5 	ext{ cm}$.$\frac{30}{P+Q} = \frac{37.5}{100-37.5} = \frac{37.5}{62.5} = \frac{3}{5}$$P+Q = 30 	imes \frac{5}{3} = 50 \Omega$ ... $(i)$Case $II$: $P$ and $Q$ are in parallel,so $R_2 = \frac{PQ}{P+Q}$. Given $l = 71.4 	ext{ cm}$.$\frac{30}{\frac{PQ}{P+Q}} = \frac{71.4}{100-71.4} = \frac{71.4}{28.6} \approx 2.5$$\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),$P$ and $Q$ are roots of the quadratic equation $x^2 - (P+Q)x + PQ = 0$,which is $x^2 - 50x + 600 = 0$.$(x-30)(x-20) = 0$Thus,the values are $30 \Omega$ and $20 \Omega$.

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