A meter bridge with two resistances R_{1} and | vedclass

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(C) For a meter bridge, the balancing condition is given by $\frac{R_{1}}{R_{2}} = \frac{l}{100-l}$.Initially, the null point is at $40 	ext{ cm}$, s

Vedclass · Accepted Answer

$R_{2}=8 \ \Omega, R_{1}=\frac{16}{3} \ \Omega$

Vedclass · Answer

(C) For a meter bridge, the balancing condition is given by $\frac{R_{1}}{R_{2}} = \frac{l}{100-l}$.Initially, the null point is at $40 	ext{ cm}$, so $l = 40 	ext{ cm}$:$\frac{R_{1}}{R_{2}} = \frac{40}{60} = \frac{2}{3} \implies R_{1} = \frac{2}{3}R_{2} \quad ... (1)$When a $16 \ \Omega$ resistance is connected in parallel to $R_{2}$, the new equivalent resistance $R_{2}'$ is $\frac{R_{2} 	imes 16}{R_{2} + 16}$.The new null point is at $50 	ext{ cm}$, so $l = 50 	ext{ cm}$:$\frac{R_{1}}{R_{2}'} = \frac{50}{50} = 1 \implies R_{1} = R_{2}' = \frac{16R_{2}}{R_{2} + 16} \quad ... (2)$Equating $(1)$ and $(2)$:$\frac{2}{3}R_{2} = \frac{16R_{2}}{R_{2} + 16}$$\frac{1}{3} = \frac{8}{R_{2} + 16} \implies R_{2} + 16 = 24 \implies R_{2} = 8 \ \Omega$.Substituting $R_{2} = 8 \ \Omega$ into equation $(1)$:$R_{1} = \frac{2}{3} 	imes 8 = \frac{16}{3} \ \Omega$.

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