Question

Describe the experiment to compare the e.m.f. of two cells using potentiometer under the following points :
(i) Labelled electric circuit diagram
(ii) Derivation of formula
(iii) Observation table
(iv) Two precautions.

Answer 1

AB = Potentiometer wire
C = Lead cell, $E_1,E_2-$ Experimetntal cells
$K_1$ = Plug key, $K_2-$ Two way key
$Rh$ = Rheostat, G-Galvanometer
J - Jockey
(2) Derivation of Formula
Let the first cell is having e.m.f. ${}_1$ and the balancing point is obtained at distance $l_1$, then by the principle of potentiometer:
$E_1=\rho l_1$ ...... (1)
Let $E_2$ is the e.m.f. of second cell whose balancing point is at $l_2$ then
$E_2=\rho l_2$ ....... (2)
Divi. $1÷2$
$\frac{E_1}{E_2}=\frac{\rho l_1}{\rho l_2}=\frac{E_1}{E_2}$
(3) Observation table

Sr. No.

Length of null point for $E_1=l_1$ cm

Lenght of null point for $E_2=l_2$ cm

($\frac{E_1}{E_2}=\frac{l_1}{l_2}$

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