0A potentiometer wire of length 1 m has a resistance 10 ohm It is connected to 6V battery in series with a resistance of 5 ohm. Determine the emf of the primary cell which gives a balance point at 40 cm.
Current flowing in potentiometer wire,
I = V/R +R’
I = 6/10+5 A = 0.4 A
Potential drop across the potentiometer wire
V =IR
= 0.4 x 10 V = 4.0 V
Potential gradient k = V/l = 4.0 V/m
.’. Unknown emf of the cell (E) = kl’
= 4.0 x 4V=1.6V
