Question

In the figure a long uniform potentiometer wire AB is having a constant potential gradient along its length. The null points for the two primary cells of emfs ε₁ and ε₂ connected in the manner shown are obtained at a distance of 120 cm and 300 cm from the end A. Find (i) ε₁/ ε₂ and (ii) position of null point for the cell ε₁.

How is the sensitivity of a potentiometer increased?

Answer 1

(i) Apply Kirchhoff’s law in loop ACFGA:-

Φ (120) ε₁ − ε₂

Φ = potential drop per unit length.

Or, ε₁ = ε₂ + Φ (120) ... (1)

Loop AEHIA:-

Φ (300) = ε₂ + ε₁

Or, ε₂ +(ε₂ + Φ (120)) = Φ (300) (By substituting value of ε₁ from equation (1))

Or, 2ε₂ = (300 − 120) Φ

Or, ε₂ = 90Φ … (2)

Thus, ε₁ = 90Φ + 120Φ

ε₁ = 210Φ … (3)

 Hence,`epsi_1/epsi_2 = 210/90 = 7/3`

(ii) As we know,

ε = Φl

Thus, from equation (2) and (3)

Null point for cell ε₂ is 90 cm.

And for cell ε₁, it is 210 cm.

Sensitivity of the potentiometer can be increased by :

(a) Increasing the length of the potentiometer wire

(b) Decreasing the resistance in the primary circuit