Physics - Current Electricity Question with Solution | TestHub

Let the initial balance point be $l_1$ cm from the left end.

Using the principle of the metre bridge:

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

$200-2l_1=3l_1$

$5l_1=200\Rightarrow l_1=40$ cm

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

Since $R^{\prime }<3\Omega$, the balance point shifts to the right. The new balance point $l_2$ is:

$l_2=40+22.5=62.5$ cm

Applying the metre bridge principle again:

$\frac{2}{R^{\prime }}=\frac{62.5}{100-62.5}$

$\frac{2}{R^{\prime }}=\frac{62.5}{37.5}=\frac{5}{3}$

$R^{\prime }=\frac{6}{5}=1.2\Omega$

Substituting the expression for $R^{\prime }$:

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

$3x=1.2(3+x)$

$3x=3.6+1.2x$

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

Answer: 2