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

Explanation

Current `I=V/(R_1+R_2)`
`=5/(15+5)`=0.05 A
Potential drop across the potentiometer wire
V = IR= 0.25 × 15 = 3.75 V
Potential gradient,
`k=V/t=3.75/1.0 =3.75` V/m
`therefore` Unknown emf of the cell = KI

= 3.75 × 0.6 = 2.25 V.