Question

With the help of the circuit diagram, explain how a potentiometer is used to compare the emf’s of two primary cells. Obtain the required expression used for comparing the emfs. 

Write two possible causes of one- sided deflection in a potentiometer experiment.

Answer 1

 

Close key K and adjust a suitable constant current in the potentiometer wire with the help of rheostat. When ε₁connected, then we find the position of jockey, where galvanometer shows no deflection. 

Let, 

AJ₁ = l 

Therefore, 

ε₁ is equal to potential difference between the points A and J1 of the potentiometer wire, 

or ε₁ = kl₁ ……..(1) 

Where k = potential gradient across the potentiometer wire. 

Similarly, 

For ε₂, 

If AJ₂ = l₂, then 

ε₂ = kl₂ ………..(2)

From (1) and (2),

\(\frac{ε_1}{ε_2}\) = \(\frac{l_1}{l_2}\)

Two possible causes for one-sided deflection in a potentiometer experiment is :

(i) The emf of the driving cell used in the main circuit of the potentiometer may be less than the potential difference to be measured. 

(ii) If the fall of potential across the potentiometer wire is less than the potential difference to be balanced by the potentiometer wire.