Question

A potentiometer wire of length 1 m has a resistance of 10Ω . It is connected to a 6 V battery in series with a resistance of 5Ω. Determine the emf of the primary cell which gives a balance point at 40 cm.

Answer 1

 Here, l = 1 m, R₁ = 10Ω  , V = 6 V, R₂ = 5Ω  

Current flowing in potentiometer wire,

Potential drop across the potentiometer wire 

V’ = IR = 0.4 x 10 = 4V

Potential gradient, K = V'/l = 4/1  = 4 V/m

Emf of the primary cell = KI = 4 × 0.4 = 1.6 V