State the working principle of potentiometer. Explain with the help of circuit diagram. How the potentiometer is used, to determine the internal resistance of the primary cell?

Principle of potentiometer : “It works on the principle that the potential difference across any part of a uniform wire is directly proportional to the length, when a constant current flows through the wire.” $V=I\rho l, V \alpha l$
Let ‘V’ be the potential difference across the wire of length
‘R’ resistance
‘A’ be uniform area of cross-section
‘I’ be the current passing through the wire then according to ohms law V = IR

But $R=\rho \frac{l}{A}\text{ Where ' }\rho \text{ ' = }$resistivity of the wire
$V=I\cdot \rho \cdot \frac{l}{A}=\left(\frac{I\rho }{A}\right)l\text{ but }\frac{I\rho }{A}=K$
$V=Kl$ (or)
$V \alpha l\text{ provided A, }\mathrm{\rho } \mathrm{\&} l$ are constants.
This is called ‘Principle of potentiometer’. Determination of internal resistance of a cell:
i) The primary circuit consists of a cell of emf ‘E’ internal resistance ‘r’ connected in series to a key ‘K’ and rheostat ($R_h$)
ii) The secondary circuit consisting a cell of emf & and internal resistance ‘r’ connected in series with a galvanometer(G) and Jockey ‘J’. A resistance box and a plug key ‘$K_1$’ are connected across the cell as shown in figure.
iii) The plug key ‘K’ is closed, and ‘$K_1$’ is opened. The position of jockey is adjusted on potentiometer wire and balance point is obtained ‘$l_1$’.
Then $E_1=I\rho l_1\to \left(1\right)$
iv)Now, both plug keys ‘K’ and ‘$K_1$’ are closed. Let ‘R’ be resistance chosen in resistance box. If ‘V’ is terminal potential difference of the cell and the position of jockey is adjusted on potentiometer wire and balance point is obtained $l_2$.
$V=I\rho l_2\to \left(2\right)$
Dividing equation $\frac{\left(1\right)}{\left(2\right)},\frac{E_1}{V}=\frac{l_1}{l_2}\to \left(3\right)$
But $V=IR,E_1=I(r+R)$
$\frac{E_1}{V}=\frac{I(r+R)}{IR}\to \left(4\right)$
from (3) and (4)
$\begin{array}{l}\frac{r+R}{R}=\frac{l_1}{l_2},\frac{r}{R}+1=\frac{l_1}{l_2}\\ \frac{r}{R}=\frac{l_1}{l_2}-1,\frac{r}{R}=\frac{l_1-l_2}{l_2}\\ r=R\left[\frac{l_1-l_2}{l_2}\right]\end{array}$
Thus internal resistance of cell is determined.