Question

Describe experiment to compare the e.m.f.'s of two cells by potentiometer under the following headings: (i) Circuit diagram (ii) Formula (iii) Observation table (iv) Two precautions.

Answer 1

(i) Circuit diagram
AB=AB= Potentiometer wire
E1E1, E2=E2= Experimental cells
C=C= Lead cell,
K2=K2= Two way key , K_1=K1= Plug key,
G=G= Galvanometer
Rh=Rh= Rheostat,
J=J= Jockey
(ii) Formula: Let the first cell is having e.m.f. E1E1 and the balancing point is obtained at distance l1l1, then by the principle of potentiometer:
E1=ρl1E1=ρl1 .......(1)
Let E2E2 is the e.m.f. of second cell whose balancing point is at l2l2 then:
E2=ρl2E2=ρl2 .......(2)
divided by equation (1) ÷÷ (2)
E1E2=ρl1ρl2=l1l2=E1E2E1E2=ρl1ρl2=l1l2=E1E2
where E1E1 and E2E2 are two experimental cells and l1l1, l2l2 length respectively.
(iii) Observation table:

S.No.

Length of null point for E1=l1cmE1=l1cm

Length of null point for E2=l2cmE2=l2cm

E1E2=l1l2E1E2=l1l2

(iv) Precautions:
(1) The e.m.f. of lead cell should be greater than experimental cell.
(2) All positive terminals should be connected to one point.