Physics - Current Electricity Question with Solution | TestHub

Let the initial balancing length be $l_1$. Using the principle of the metre bridge:

$\frac{2}{3}=\frac{l_1}{100-l_1}$

$3l_1=200-2l_1\Rightarrow 5l_1=200\Rightarrow l_1=40$ cm

When a shunt of $X$ $\Omega$ is connected across the $3$ $\Omega$ resistor, the equivalent resistance in the right gap becomes $R_p=\frac{3X}{3+X}$.

Since the resistance in the right gap decreases, the balancing point shifts to the right. The new balancing length is:

$l_2=40+22.5=62.5$ cm

Applying the metre bridge principle again:

$\frac{2}{R_p}=\frac{62.5}{100-62.5}$

$\frac{2}{R_p}=\frac{62.5}{37.5}=\frac{5}{3}$

$R_p=\frac{6}{5}=1.2$ $\Omega$

Equating the two expressions for $R_p$:

$\frac{3X}{3+X}=1.2$

$3X=3.6+1.2X$

$1.8X=3.6\Rightarrow X=2$ $\Omega$

Answer: $2$