In potentiometer experiment, the balancing length with a cell $\mathrm{E}_{1}$ of unknown e.m.f. is…
In potentiometer experiment, the balancing length with a cell $\mathrm{E}_{1}$ of unknown e.m.f.
is $\mathcal{\ell}_{1}{ }^{\prime} \mathrm{cm}$. By shunting the cell with resistance $\mathrm{R} \Omega$, the balancing length becomes $\frac{\ell_{1}}{2} \mathrm{~cm}$, the internal resistance (r) of a cell is
$\mathrm{r}=0$
$r=\frac{R}{2}$
$\mathrm{r}=2 \mathrm{R}$
$\mathrm{r}=\mathrm{R}$
Solution
$\mathrm{r}=\mathrm{R}\left(\frac{\ell_{1}}{\ell_{2}}-1\right) \quad \ell_{2}=\frac{\ell_{1}}{2}$
Putting the values and solving we get $r=R$