How the emf of two cells are compared using potentiometer ?

Explanation

Comparison of emf of two cells with a potentiometer: To compare the emf of two cells, the circuit connections are made as shown in figure. Potentiometer wire CD is connected to a battery Bt and a key K in series. This is the primary circuit. The end C of the wire is connected to the terminal M of a CDPDT (Double Pole Double Throw) switch and the other terminal N is connected to a jockey through a galvanometer G and a high resistance HR. The cells whose emfs,`xi_(1)` and `xi_(2)`to be compared are connected toMi EN the terminals `M_(1) N_(1)` and `M_(2)N_(2)` of the DPDT switch. The positive terminals of Bt, `xi_(1)` and `xi_(2)`should be connected to the same end C.The DPDT switch is pressed towards `M_(1) N_(1)` so that cell `xi_(1)` is included in the secondary circuit and the balancing length `l_(1)`, is found by adjusting the jockey for zero deflection. Then the second cells,`xi_(2)` is included in the circuit and the balancing length `l_(2)` is determined. Let r be the resistance per unit length of the potentiometer wire and I be the current flowing through the wire
we have `xi_(1)=Irl_(1)ldots(1)`
`xi_(2)=Irl_(2)ldots(2)`
By dividing equation (1) by (2)`(xi_(1))/(xi_(2))=(l_(1))/(l_(2))`ldots(3)
By including a rheostat (Rh) in the primary circuit, the experiment can be repeated several times by changing the current flowing through it