State the working principle of potentiometer. With the help of the circuit diagram, explain how a potentiometer is used to compare the emf's of two primay cells. Obtain the required expression used for comparing the emfs. Write two possible causes for one sided deflection in a potentiometer experiment.

Explanation

(a) Working Principle of Potentiometer : When a constant current is passed through a wire of uniform area of cross - section, the potential drop across any portion of the wire is directly proportional to the length of that portion.
Applications of Potentlometer for comparing emf's of two cells : The following figure shows an application of the potentiometer to compare the emf of two cells of emf `E_(1) and E_(2). E_(1) , E_(2)` are the emf of the two cells 1, 2, 3 form a two way key.
When 1 and 3 are connected, E, is connected to the galvanometer (G).
Jokey is moved to `N_(1)`, which is at a distance `l_(1)` from A, to find the balancing length.
Applying loop rule to `AN_(1)g31A` ,
`phi l_(1)+0-E_(1)=0 " " ` ...(i)
where, `phi` is the potential drop per unit length.
Similarly, for `E_(2)` balanced against `l_(2)(AN_(2)),`
`phi l_(2) +0-E_(2)=0 " " ` ...(ii)
From equations (i) and (ii)
`(E_(1))/(E_(2))=(l_(1))/(l_(2)) " " ` ...(iii)
Thus we can compare the emf's of any two sources. Generally, one of the cells is chosen as a standard cell whose emf of the other cell is then calculated from equation (iii).
(b) (i) The emf of the cell connected in main circuit may not be more than the emf of the primary cells whose emf's are to be compared.
(ii) The positive ends of all cells are not connected to the same end of the wire.