Potentiometer and determination of internal resistance of a cell

Potentiometer and determination 

of internal resistance of a cell

 

1) A potentiometer is arranged as like previously discussed.

2) But instead of adding two cells we are going to add only one cell of emf e₁ connecting parallelly to the potentiometer wire.

3) A resistance box R is also connected parallelly to the cell e₁  with a key K₁ in series. Through a galvanometer G.

4) When key K₁ is open, the Jockey is slided over the potentiometer wire and found no deflection of the galvanometer at the position J₁

5) Now f is the potential drop per unit length of the wire. And from A to J₁ length of the wire is  l₁

6) Now applying Kirchhoff’s rule

e1  -  f l1  =  0

e₁  =   f l₁       ............    3.59

7) Now a known resistance R is added to the resistance box, then key K₁  is closed, again the Jockey is slided over the wire and found no deflection of the galvanometer at J₂  position of the wire.

8) Then if V is the potential difference across the cell e₁ then,

  V -  f l2  =  0

 

  V   =   f l2       

But

V = I R

So,

  IR   =   f l2       ............    3.86

9) But we know that

e₁  = I (r + R)

Where r is the internal resistance of the cell, now putting the value of  e₁  on 3.59 we get

 I (r + R)  =   f l₁       ............  3.91

10) now

 this is how internal resistance of a cell can be determined by using a potentiometer.