A potentiometer wire of length L and a resistance r are connected in series with a battery of e.m.f.  $E_0$ and a resistance $r_1$. An unknown e.m.f. E is balanced at a length $\mathcal{l}$ of the potentiometer wire. The e.m.f. E will be given by

Given, length of potentiometer wire = L
Resistance of the wire = r
Resistance per unit length $\left(\rho \right)=\frac{r}{L}$
Potentiometer battery emf = E₀
Internal resistance = r₁
 $\therefore$Current length of the potentiometer wire i is  $\mathrm{i}=\frac{{\mathrm{E}}_0}{\left(\mathrm{r}+{\mathrm{r}}_1\right)}$ [From Ohm's Law]
When the unknown emf E is balanced, there will be no flow of current through it.
The balancing length is given as l .
The emf E is equal to the potential drop across length l .
$E=i\times R_{\ell },$
where i is the current and R is the resistance of the length l .
$\begin{array}{l}\Rightarrow R_{\ell }=\ell \times \rho \\ =\ell \times \left(\frac{r}{L}\right)\\ \Rightarrow E=\frac{E_0}{\left(r+r_1\right)}\times \left(\frac{r}{L}\right)\times \ell \\ \frac{E_{0}r}{\left(r+r_1\right)}\times \frac{\ell }{L}\end{array}$