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.

Answer 1

From the figure above:

Total resistance of the circuit, R = (R_AB + 5) Ω = 15 Ω

Current in the circuit, `i=V/R=6/15A`

∴ Voltage across AB, V_AB = i.R_AB = 4 V

emf of the cell, `e=l/LV_0`

Here:
l = 40 cm (balance point)

AB = L = 1 m = 100 cm (total length of the wire)

`:.e=40/100(4)=1.6V`