Question

A potentiometer wire of length 1 m has a resistance of 10 Ω. 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 = 1m, R₁ = 10, V = 6V, R₂ = 5Ω

Current flowing in potentiometer wire,

\(I=\frac{V}{R_1\,+\,R_2}=\frac{6}{10+5}=\frac{6}{15}=0.4A\)

Potential drop across the potentiometer wire

V′= IR = 0.4 × 10 = 4V

Potential gradient,

\(K=\frac{V'}{l}=\frac{4}{1}=4V/m\)

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