Question

A potentiometer wire of length 100 cm has a resistance of 30 ohms. It is connected in series with a resistance of 20 ohms and accumulator of emf 8V having negligible internal resistance. A source of 1.2V is balanced against a length L of the potentiometer wire. What is the value of L?

(a) 20

(b) 25

(c) 30

(d) 35

This question was posed to me during an interview.

My query is from Current Electricity topic in chapter Current Electricity of Physics – Class 12

Answer 1

Correct option is (b) 25

Explanation: The current passing through the potentiometer wire:

I = \(\frac {8}{(30 + 20)} = \frac {8}{50}\) = 0.16A

The potential difference across the potentiometer wire:

V = current × resistance = 0.16 × 30 = 4.8V

Length of the wire = 100 cm

k = \(\frac {V}{l} = \frac {4.8}{100}\) = 0.048

The emf 1.2V is balanced against the length L of the wire, i.e. 1.2 = kL

Length = \(\frac {1.2}{k} = \frac {1.2}{0.048}\) = 25 cm

Therefore, the length L is 25 cm.