Question

In a meter bridge circuit, resistance in the left hand gap is $2\Omega$ and an unknown resistance X is in the right hand gap as shown in Figure. The null point is found to be $40$cm from the left end of the wire. What resistance should be connected to X so that the new null point is $50$cm from the left end of the wire?

Answer 1

Case 1 :

Null point is found at $l_1=40$ cm

Using balanced Wheatstone bridge condition, $\frac{2}{X}=\frac{l_1}{100-l_1}$

$\therefore$ $\frac{2}{X}=\frac{40}{100-40}$

$⟹$ $X=3\Omega$

Case 2 :

New null point is found at $l_2=50$ cm

Using balanced Wheatstone bridge condition, $\frac{2}{X^{\prime }}=\frac{l_2}{100-l_2}$

$\therefore$ $\frac{2}{X^{\prime }}=\frac{50}{100-50}$

$⟹$ $X^{\prime }=2\Omega$

Since the unknown resistance gets decreased, thus we have to connect a resistance $R$ in parallel to $X$ so that $X^{\prime }$ comes out to be $2\Omega$.

Thus resistance of parallel combination $X^{\prime }=\frac{XR}{X+R}$

$\therefore$ $2=\frac{3R}{3+R}$

$⟹$ $R=6\Omega$

Hence $6\Omega$ must be connected in parallel to unknown resistance $X$.