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Question
How the emf of two cells are compared using potentiometer?
Solution
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 M1, 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.
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