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 5 Ω. Determine the emf of the primary cell which gives a balance point at 40 cm.
Solution
Given,
Total length of the potentiometer wire, L = 1m
Resistance of the wire, R = 10 Ω
Voltage of the battery = 6 V
Resistance of the battery = 5 Ω
Therefore, total resistance of the circuit, R = (RAB + 5) Ω = 15 Ω
Using the figure given above, we have
Current in the circuit, I = V R = 6 x 15 A
Therefore,
Voltage across AB, VAB = i.RAB = 4 V
Emf of the cell, e = 
... (1)
Here,
Balance point is obtained at, l = 40 cm
Total length, AB = L = 1 m = 100 cm
Putting the values in equation (1), we have
