Question

To compare the emf of two cells using potentiometer

Answer 1

Theory.....

When we keep key (K1) closed and (K2) open , let the null point found be lAJ1

E1=KlAj1 (1)

When we keep K1 open and K2 closed , let null point obtained by lAJ2 .

E2=KlAj2 (2)

(1)/ (2)

E1 / E2 = KlAJ1 / KlAj2

E1/E2 =l1/l2

Where E1 and E2 are the emf of two given cells .

Procedure.......

Arrange the apparatus as shown in the circuit diagram .
Connect the positive poles of the cells to the terminal and the negative poles to the terminal a and b of the two way key .
Insert the plug in the key K and also in between the terminals a and c of the two way key .
Slide the jockey gently over the potentiometer wires until you obtain a point of no deflection .
Note the length l1 at the point.
Repeat this with E2 by disconnecting E1 and inserting plug into gap a and c of two way key .
Record l2 at null point .
Repeat this different resistance .