Question

How the emf of two cells are compared using potentiometer?

Answer 1

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 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 ξ₁ and ξ₂ to be compared are connected to the terminals M_1, N₁ and M₂, N₂ of the DPDT switch.

Potentiometer

The positive terminals of Bt, ξ₁ and ξ₂ should be connected to the same end C. The DPDT switch is pressed towards M₁, N₁ so that cell ξ₁ is included in the secondary circuit and the balancing length l₁ is found by adjusting the jockey for zero deflection, Then the second cells ξ₂ is included in the circuit and the balancing length l₂ 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.

ξ₁ = Irl₁ …… (1)

ξ₂ = Irl₂ ……. (2)

By dividing (1) by (2)

`ξ^1/ξ^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.