Comparison of emfs of two given Cells Using Potentiometer - QS Study

The potentiometer wire AB is connected in series with a battery (Bt), Key (K), rheostat (Rh) as shown in Figure. This forms the primary circuit. The end A of potentiometer is connected to the terminal C of a DPDT switch (six way key−double pole double throw). The terminal D is connected to the jockey (J) through a galvanometer (G) and high resistance (HR). The cell of emf E1 is connected between terminals C1 and D1 and the cell of emf E2 is connected between C2 and D2 of the DPDT switch.

Fig: comparison of emf of two cells

Let I be the current flowing through the primary circuit and r be the resistance of the potentiometer wire per metre length.

The DPDT switch is pressed towards C1, D1 so that cell E1 is included in the secondary circuit. The jockey is moved on the wire and adjusted for zero deflection in galvanometer. The balancing length is l1.

The potential difference across the balancing length l1 = Ir l1. Then, by the principle of potentiometer,

E1 = Ir l1 … (1)

The DPDT switch is pressed towards E2. The balancing length l2 for zero deflection in galvanometer is determined. The potential difference across the balancing length is l2 = Ir l2, then

E2 = Ir l2 … (2)

Dividing (1) and (2) we get

E1/E2 = l1/ l2

If emf of one cell (E1) is known, the emf of the other cell (E2) can be calculated using the relation.

E2 = E1 (l2/ l1)