Describe briefly, giving the necessary circuit diagram, how potentiometer is used to measure the internal resistance of a cell

The above figure is the circuit diagram for the to measure the internal resistance of a potentiometer. Suppose r represent the internal resistance and R is a load resistance. First we will measure the emf of the cell then we will measure the voltage difference across the resistance and then calculate the current flowing through the circuit. From this we will find out the circuit.
Now suppose the I is the current across the potentiometer circuit. If the the galvanometer is showing null reading at length l₁ then the emf of the cell is $l_1\rho I$, where $\rho$ is the resistance per unit length.
Now suppose i the current flowing through the load if we switch on. Now the if galvanometer is showing null point at l₂ then the potential difference across the resistance is $v=l_2\rho I$. Now  we have
$\frac{\epsilon }{v}=\frac{l_1}{l_2}$
Now $ir=\epsilon -v\Rightarrow r=\frac{\epsilon -v}{i}=\frac{v}{i}\left(\frac{\epsilon }{v}-1\right)=\frac{iR}{i}\left(\frac{l_1}{l_2}-1\right)=R\left(\frac{l_1}{l_2}-1\right)$