A potentiometer wire of length 1 m has a resistance of `100Omega`. It is connected to a 6V battery in series with a resistance of `5Omega`. Determine the emf of the primary cell which gives a balance point at 40cm.

Explanation

Here , l=1m `R_1=10 Omega , V=6V , R_2=5Omega`
Current flowing in potentiometer wire,
`I=V/(R_1+R_2)=6/(10+5)=6/15`=0.4 A
Potential drop across the potentiometer wire
V = IR = 0.4 × 10 = 4V
Potential gradient, `K=(V')/1=4/1`=4V/m
Emf of the primary cell = KI
= 4 × 0.4 = 1.6 V