Explain the comparison of the electromotive force (emf) of two cells using a potentiometer with a necessary diagram.

(A) As shown in the figure,a battery having emf $\varepsilon$ and internal resistance $r$,a variable resistance $R$,and a switch $K_{1}$ are connected between the two ends $A$ and $B$ of the potentiometer wire.
To compare the emf of two cells $\varepsilon_{1}$ and $\varepsilon_{2}$,the positive terminals of both cells are connected to point $A$. The negative terminals of the cells are connected to points $1$ and $2$ of a two-way switch. Terminal $3$ of the switch is connected to a galvanometer $(G)$ and a jockey in series. The jockey can be moved along the potentiometer wire.
First,points $1$ and $3$ of the switch are connected,placing cell $\varepsilon_{1}$ in the circuit. By sliding the jockey along the wire,a balance point $N_{1}$ is obtained such that the galvanometer shows zero deflection. Let $AN_{1} = l_{1}$.
By applying Kirchhoff's second law for the loop $AN_{1}G31A$:
$\phi l_{1} - \varepsilon_{1} = 0$
$\therefore \varepsilon_{1} = \phi l_{1} \quad .....(1)$
where $\phi$ is the potential gradient of the wire.
Next,points $2$ and $3$ of the switch are connected,placing cell $\varepsilon_{2}$ in the circuit. By sliding the jockey,a balance point $N_{2}$ is obtained such that the galvanometer shows zero deflection. Let $AN_{2} = l_{2}$.
By applying Kirchhoff's second law for the loop $AN_{2}G32A$:
$\phi l_{2} - \varepsilon_{2} = 0$
$\therefore \varepsilon_{2} = \phi l_{2} \quad .....(2)$
Dividing equation $(1)$ by $(2)$:
$\frac{\varepsilon_{1}}{\varepsilon_{2}} = \frac{\phi l_{1}}{\phi l_{2}} = \frac{l_{1}}{l_{2}}$
Thus,the ratio of the emfs of the two cells is equal to the ratio of their respective balancing lengths.