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Question:

A potentiometer wire of length 1.0 m has a resistance of 15Ω. It is connected to a 5V battery in series with a resistance of 5Ω. Determine the emf of the primary cell which gives a balance point at 60 cm.

Solution:

current flowing I = V/R = 5/(5+15) = 1/4 A
Resistance of 60 cm wire is (15/100) × 60Ω = 9Ω
voltage drop on 60 cm wire is V = IR = (1/4) × 9 = 2.25V