Explain the method to measure the internal resistance of a cell using a potentiometer.

(N/A) To measure the internal resistance $(r)$ of a cell $(\varepsilon)$,a potentiometer circuit is set up as shown in the figure.
The primary circuit consists of a battery $(B)$,a variable resistor $(R)$,and a key $(K_1)$ connected in series with the potentiometer wire $AC$.
The cell $(\varepsilon)$ whose internal resistance $(r)$ is to be measured is connected in parallel with a resistance box $(R_{ext})$ and a key $(K_2)$. The positive terminal of the cell is connected to point $A$,and the negative terminal is connected to a galvanometer $(G)$,which is then connected to a jockey.
$1$. With key $K_2$ open,the cell $(\varepsilon)$ is in an open circuit. The null point $N_1$ is found on the wire $AC$ such that the galvanometer shows zero deflection. Let the balancing length be $AN_1 = l_1$. Since the cell is in an open circuit,the potential difference across it is equal to its $EMF$ $(\varepsilon)$.
$\varepsilon = \phi l_1$ ... $(1)$,where $\phi$ is the potential gradient along the wire.
$2$. Now,close the key $K_2$ so that a current flows through the resistance box $(R_{ext})$ and the cell. The null point $N_2$ is found on the wire $AC$. Let the balancing length be $AN_2 = l_2$. In this condition,the potential difference across the cell is equal to its terminal voltage $(V)$.
$V = \phi l_2$ ... $(2)$.
Dividing equation $(1)$ by $(2)$,we get:
$\frac{\varepsilon}{V} = \frac{l_1}{l_2}$
We know that for a cell,$\varepsilon = I(R_{ext} + r)$ and $V = IR_{ext}$,so $\frac{\varepsilon}{V} = \frac{R_{ext} + r}{R_{ext}} = 1 + \frac{r}{R_{ext}}$.
Equating the two expressions for $\frac{\varepsilon}{V}$:
$1 + \frac{r}{R_{ext}} = \frac{l_1}{l_2}$
$\frac{r}{R_{ext}} = \frac{l_1}{l_2} - 1 = \frac{l_1 - l_2}{l_2}$
$r = R_{ext} \left( \frac{l_1 - l_2}{l_2} \right)$