Explain the principle of a potentiometer. How will you compare the e.m.f. of two primary cells by using potentiometer? Explain with proper circuit diagram.

Explanation

Potentiometer: It is a device which is used to compare emfs of two cells or to measure the internal resistance of the cell. It consists of a number of segments of wire (made of manganin or constantan) of uniform area of cross-section stretched over a wooden board between two thick copper strips. Comparison of emfs of two cells: The circuit diagram for the comparison of emfs of two cells is given below:
The auxilliary circuit consists of a battery, an ammeter a rheostat and key k.
The cells, whose emfs have to be compared, have their positive terminals connected to end X. while their negative terminals are connected to the terminals of a two way key. The common terminal of two way key is connected to a jockey (J) thro. The positive terminal of the auxilliary battery is connected to the end X which is taken to the zero of the scale. The gap 1 of the two way key is closed and the jockey J is slided along the wire to get a balance point, say, at P, i.e., the point where galvanometer gives zero deflection. This will be achieved when e.m.f. `E_(1)` of the cell is equal to the difference of potential across XP,
`E_(1)= V_(XP)`
But `V_(XP) prop l_(1)`
`rArr V_(XP) = K l_(1)`
Using (1) and (2), we get
`E_(1)= K l_(1)` ...(3)

where `l_(1)` is the length of wire from X to P and K is assumed to be the potential difference per unit length of the potentiometer wire. Now, the plug from gap 1 is removed and is put in the gap 2. Again the jockey is slided to get a balance point say at P.. If `l_(2)` is the length of wire giving the balance point with the cell of e.m.f. `E_(2)` then
`E_(2)= V_(XP)` ...(4)
But `V_(XP) prop l_(2)`
But `V_(XP) prop l_(2)` ...(5)
and `V_(XP)= K l_(2)`
Using (4) and (5), we get `E_(2)= K l_(2)`
Dividing equation (5) by (6), we get, `(E_(1))/(E_(2))= (K l_(1))/(K l_(2))`
or `(E_(1))/(E_(2))= (l_(1))/(l_(2))`
Putting the values of `l_(1) and l_(2)`, the value of `E_(1)//E_(2)` can be calculated.