Question

A potentiometer wire of length 1m has a resistance of 10 ohms. It is connected to a 6V battery in series with a resistance of 5 ohms. Determine the emf of a primary cell 'e' which gives a balance point of 40 cm.

Answer 1

Given ,
length of wire , l = 100cm
Resistance, R = 10Ω
emf of battery , E1 = 6V
Resistance , R1 = 5Ω
& x = 40 cm

Now , Current , I = E1/R+R1

⇒ I = 6/10+5 = 2/5 A

Also , V = IR

⇒V = 2/5 × 10 = 4V

Now , emf of primary cell = Vx/l
= 4×40/100 = 1.6V

Answer 2

I hope this helps you!!!!