Question

Explain the principle of a potentiometer. How can emf of two cells be compared using potentiometer?

Answer 1

Potentiometer is the device which is to be used in the comparison of 2 cell for measuring their cell internal resistance & Potential difference towards the resistor. It consist large uniform cross sectional area wire which is 10 m long . also the wire having highly Resistivity. 

Principle of Potentiometer -: In any constant current flowing wire, Potential drop towards length of the wire is directly prop. to that length.∴ V α I .

Comparison of e.m.f of Two Cells :- Refer to the Diagrammatic attachment . See in the figure ,

In close key k :-
For finding the emf across e , 

Connect cell 1→3 .
Slide it until galvanometer stops deflections 
Now Length = l₁
potential diff.  v = kl₁ 
so,
e₁ = kl₁  ----→ (i)

When jockey is connected , Connect 2→ 3 
length =l₂
then similarly 
e₂ = kl₂ ----→(ii)

Dividing eq i & ii
e₁/e₂ = kl₁/kl₂
or  
___________
e₁ /e₂ = l₁/l₂ .
___________

Hence proved.


Hope it Helps. :-)

Answer 2

Potentiometer is an instrument used for measuring the emf of a cell or potential difference between two points in an electrical circuit .

Principle of Working :

    Fall in potential across any portion of wire is directly proportional to length of that portion provided the wire is of uniform area of cross section and a constant current is flowing through it.

Comparisons of emfs of two cells using potentiometer:

  A standard cell is connected between the terminals A and B of potentiometer wire with a rheostant Rh and a key K in series.
Two cells of emfs ε1 and ε2 to be compared are connected to A where the positive terminal of standard cell was connected.
The negative terminals of these two cells are connected to two terminals 1 and 2 of a two way key.
The common terminal 3 of the two way key is connected to jockey J through galvanometer.

The key K is closed and the rheostat Rh is adjusted so that a suitable constant current flows in the potentiometer wire.
The terminals 1 and 3 of the two way key are connected and the position of jockey on the wire is adjusted to obtain the position J1 for which there is no deflection in the galvanometerε.
The length AJ1=l1 is measured.
Now, potential difference across first cell is equal to potential difference across the length l1.
ε1=Φ l1where Φ1 is the potential drop per unit length of wire.

Next, the terminals 2 and 3 of the two keys are connected and the position J2 of the jockey on the wire is obtained for which there is no deflection in the galvanometer.
the length AJ2=l2 is measured.
Now, potential difference across the second cell is equal to potential difference across the wire of length l2.
ε2=Φl2
Thus, ε1/ε2= Φl1/Φl2=l1/l2

By this relation, the ratio of emfs of two cells is calculated. If emf of one cell is known, emf of other cell an be found out.