In a meter bridge experiment to determine the value of unknown resistance, first the resistances $2

Question

In a meter bridge experiment to determine the value of unknown resistance, first the resistances $2 \Omega$ and $3 \Omega$ are connected in the left and right gaps of the bridge and the null point is obtained at a distance $l \mathrm{~cm}$ from the left. Now when an unknown resistance $x \Omega$ is connected in parallel to $3 \Omega$ resistance, the null point is shifted by 10 cm to the right of wire. The value of unknown resistance $x$ is $\_\_\_\_$ $\Omega$.

Professor PC Thomas Classes · Answer

Initially, the meter bridge operates at balance with $2\Omega$ and $3\Omega$. The condition dictates $\frac{2}{3} = \frac{l}{100 - l} \implies 200 - 2l = 3l \implies 5l = 200 \implies l = 40 	ext{ cm}$. Once $x$ is connected in parallel with the $3\Omega$ resistor, the new equivalent resistance in the right gap is $R' = \frac{3x}{3+x}$. The null point is specified to shift 10 cm rightward, creating a new balance length $l' = 40 + 10 = 50 	ext{ cm}$. The new balance condition is $\frac{2}{R'} = \frac{50}{100-50} = \frac{50}{50} = 1$. This implies $R' = 2\Omega$. Equating this to the parallel combination gives $\frac{3x}{3+x} = 2 \implies 3x = 6 + 2x \implies x = 6\Omega$.