$I=\frac{E_{0}}{\left(r+r_{1}\right)}$

and the potential difference across the wire is

$V=I r=\frac{E_{0} r}{\left(r+r_{1}\right)}$

The potential gradient along the potentiometer wire is

$k=\frac{V}{L}=\frac{E_{0} r}{\left(r+r_{1}\right) L}$

As the unknown e.m.f. $E$ is balanced against length $l$ of the potentiometer wire,

$\therefore \quad E=k l=\frac{E_{0} r}{\left(r+r_{1}\right)} \frac{l}{L}$