In the given figure, there is a circuit of potentiometer of length $AB=10 \mathrm{~m}$. The resistance per unit length is $0.1 \Omega$ per cm. Across AB, a battery of emf E and internal resistance 'r' is connected. The maximum value of emf measured by this potentiometer is:

JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Current Electricity

In the given figure, there is a circuit of potentiometer of length

A B = 10 m

. The resistance per unit length is

0.1Ω

per cm.
Across AB, a battery of emf E and internal resistance 'r' is connected. The maximum value of emf measured by this potentiometer is:

Answer: (C)

The wire carries

0.1 Ω

per cm and

A B = 10

= 1000

cm, so

R_{A B} = 100 Ω

. The driving loop is the

6

V cell with the

20 Ω

resistor in series with the wire, so the current along the wire is

I = \frac{6}{20 + 100} = 0.05 A

A potentiometer balances the unknown emf against the fall of potential from

A

to the balance point, and the largest fall available is the one across the whole wire:

V_{A B} = I R_{A B} = 0.05 \times 100 = 5 V

The largest emf this arrangement can measure is

5

JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Single correct option