How the emf of two cells are compared using potentiometer ?

Explanation

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 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. The cells whose emf `xi_1` 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_1` and `xi_2` should be connected to the same end C.

Procedure: 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 `I_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. Letr 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" ...(1)"`
` xi_2 = Irl_2" ...(2)"`
By dividing equation (1) by (2)
`xi_1/xi_2 = l_1/l_2" ....(3)"`
By including a rheostat (Rh) in the primary circuit, the experiment can be repeated several times by changing the current flowing through it.