State the working principle of potentiometer. Explain with the help of circuit diagram. How the emf of two primary cells are compared by using the potentiometer.

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.”$V = I \rho l$ ,$v \propto l$
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 $\mathrm{R}=\mathrm{\rho }\frac{l}{\mathrm{A}}$
Where ‘$\mathrm{\rho }$ ’ = resistivity of the wire
$\mathrm{V}=\mathrm{I}\cdot \mathrm{\rho }\cdot \frac{l}{\mathrm{A}}=\left(\frac{\mathrm{I\rho }}{\mathrm{A}}\right)l\text{ but }\frac{\mathrm{I\rho }}{\mathrm{A}}=\mathrm{K}$
$\mathrm{V}=Kl\text{ (or) }\mathrm{V\alpha }l$ provided A, $\mathrm{\rho }$ & I are constants.
This is called ‘Principle of potentiometer’. comparisen of emf’s of two cells using potentiometer.

i) The primary circuit consists of a cell of emf ‘E’ internal resistance ‘r’ connected inseries to a key ‘K’ and rheostat ($R_h$)
ii) The secondary circuit consists of two cells emf ‘$E_1$’ and ‘$E_2$’ two plug keys $K_1$ and $K_2$, a galvanometer (G) and jockey ‘J’ are connected as show in figure.
iii) First, the plug keys $K$ and $K_1$ are closed. The position of jockey is adjusted on the potentiometer wire and balance point is obtained at $l_1$.
Then $E_1=I\rho l_1 \to \left(1\right)$
iv) Now plug key $K_1$ is opened and plug keys $K$ and $K_2$ are closed. The position of jockey is adjusted on the potentiometer wire and balance point is obtained at $l_2$.
Then $E_2=I\rho l_2 \to \left(1\right)$
Dividing the above two equations $\frac{E_1}{E_2}=\frac{l_1}{l_2}$
Using the above formula, the emf’s of two given cells is compared.