Question

A potentiometer circuit is shown in Figure 3 below. AB is a uniform metallic wire having length of 2 m and resistance of 8 Ω. The batteries E₁ and E₂ have emfs of 4 V and 1.5 V and their internal resistances are 1 Ω and 2 Ω respectively.

1. When the jockey J does not touch the wire AB, calculate:
   1. the current flowing through the potentiometer wire AB.
   2. the potential gradient across the wire AB.
2. Now the jockey J is made to touch the wire AB at a point C such that the galvanometer (G) shows no deflection. Calculate the length AC.

Answer 1

i. When jockey J does not touch the wire AB:

a. E₁ = 4 V

R = (7 + 8) = 15 Ω

r = 1 Ω

Current through AB,

I₁ = `E_1/(R + r)`

= `4/(15 + 1)`

= `4/16`

= 0.25 A

b. Potential difference across wire AB,

V_AB = I₁ × 8

= 0.25 × 8

= 2 V

Potential gradient in AB,

k = `V_(AB)/L`

= `2/2`

= 1 Vm⁻¹

ii. Let, AC = l metre

∴ E₂ = kl

⇒ 1.5 = 1 × l

⇒ l = 1.5 m