Determnation of Internal Resisitance of a cell by Potentiometer Method

A battery of emf ε is connected the end terminal A and B of potentiometer wire with rheostat R_h, and key K in series. This is called an auxiliary circuit. Cell of emf which to be measure is connected as shown in the figure.

Now we find the no deflection position of galvanometer by closing the key K. we get

ε=Kl₁                                                                                                                                                        …..1

Now we close the key K₁, resistance R is also added in circuit and we find the no deflection position of galvanometer at l₂ therefore

 V= Kl₂                                                                                                                                                   …..2

Divide 1by 2 we get

ε/ V= l₁/l₂                                                                                            …..3

We know that the internal resistance r₁ of a cell of emf E, when resistance R connected in its circuit is given by

r₁ = (ε-V) R/ V = (ε/ V-1) R                                                        …..4

Putting the value from 3 to 4, we get

r₁ = (l₁/l₂ -1) R