Two resistors $2 \Omega$ and $3 \Omega$ are connected in the gaps of bridge as shown in figure. The

Question

Two resistors $2 \Omega$ and $3 \Omega$ are connected in the gaps of bridge as shown in figure. The null point is obtained with the contact of jockey at some point on wire $X Y$. When an unknown resistor is connected in parallel with $3 \Omega$ resistor, the null point is shifted by 22.5 cm toward $Y$. The resistance of unknown resistor is $\_\_\_\_$ $\Omega$.

Professor PC Thomas Classes · Answer

For the initial balance point $l_1$, the bridge condition gives $\frac{2}{3} = \frac{l_1}{100 - l_1} \implies 200 - 2l_1 = 3l_1 \implies l_1 = 40 	ext{ cm}$. When an unknown resistor $R$ is connected in parallel with the $3 \Omega$ resistor, the new resistance in that gap becomes $R' = \frac{3R}{3+R}$. The null point shifts by 22.5 cm towards Y, meaning the balancing length from X increases: $l_2 = 40 + 22.5 = 62.5 	ext{ cm}$. The new balance condition is $\frac{2}{R'} = \frac{62.5}{100 - 62.5} = \frac{62.5}{37.5} = \frac{5}{3}$. Solving for $R'$ gives $R' = \frac{6}{5} = 1.2 \Omega$. Equating this to the parallel combination: $\frac{3R}{3+R} = 1.2 \implies 3R = 3.6 + 1.2R \implies 1.8R = 3.6 \implies R = 2 \Omega$.