Physics - Current Electricity Question with Solution | TestHub

The potential difference across the cell when shunted by a resistance $R$ is given by $V=\frac{ER}{R+r}$, where $E$ is the emf and $r$ is the internal resistance of the cell.

The balancing length $l$ on the potentiometer wire is proportional to $V$, so $l\propto \frac{R}{R+r}$.

For the first case, $R_1=3\Omega$ and $l_1=1$ m:

$1\propto \frac{3}{3+r}$

For the second case, $R_2=6\Omega$ and $l_2=1.5$ m:

$1.5\propto \frac{6}{6+r}$

Dividing the two equations, we get:

$\frac{1.5}{1}=\frac{6}{6+r}\times \frac{3+r}{3}$

$1.5=2\left(\frac{3+r}{6+r}\right)$

$1.5(6+r)=2(3+r)$

$9+1.5r=6+2r$

$0.5r=3$

$r=6\Omega$

Answer: $6\Omega$