How the emf of two cells are compared using potentiometer ?

Explanation

(i) To compare the emf of two cell, the circuit connections are made as shown in Figure Potentiometer wire CD is connected to a battery Bt and a key K in series.
(ii) This is the primary circuit. The end C of the wire is connected to the terminal M of a DPDT (Double Pole Double Throw) switch and the other terminal N is connected to a jockey through a galvanometer G and a high resistance HR.
(ii) The cells whose emf `xi and xi_(2)` to be compared are connected to the terminals `M_(1),N_(1) and M_(2), N_(2)` of the DPDT switch. The positive terminals of `Bt, xi and xi_(2)` should be connected to the same end C.

(iii) The DPDT switch is pressed towards `M_(1),N_(1)` so the cell `xi` is included in the secondary circuit and the balancing length `l_(1)` is found by adjusting the jockey for zero deflection. Then the second cell `xi_(2)` is included in the circuit and the balancing length `l_(2)` is determined. Let r be resistance per unit length of the potentiometer wire and I be the current flowing through the wire.
we have `xi_(1)=Irl_(1) " "...(1)`
`xi_(2)=Irl_(2) " "...(2)`
By dividing equation (1) by (2)
`(xi_(1))/(x_(2))=(l_(1))/(l_(2))" "....(3)`
(iv) By including a rheostat (Rh) in the primary circuit, the experiment can be repeated several times by changing the current flowing through it.