A potentiometer wire of length 10 m and resistance $20 \Omega$ is connected in series with a 25 V battery and an external resistance $30 \Omega$. A cell of emf E in secondary circuit is balanced by 250 cm long potentiometer wire. The value of E (in volt) is $\frac{x}{10}$. The value of $x$ is ______.

JEE Main · 2022 · 24 Jun 2022 · Shift 2 · Current Electricity

A potentiometer wire of length 10 m and resistance

20Ω

is connected in series with a 25 V battery and an external resistance

30Ω

. A cell of emf E in secondary circuit is balanced by 250 cm long potentiometer wire. The value of E (in volt) is

\frac{x}{10}

. The value of

x

is ______.

Answer: 25

Total resistance in primary circuit:

R_{t o t a l} = 20 + 30 = 50Ω

Current through potentiometer wire:

I = \frac{25}{50} = 0.5

A

Potential drop across the 10 m wire:

Δ V = I R_{w i r e} = 0.5 \times 20 = 10

V

Potential gradient:

k = \frac{10 V}{10 m} = 1

V/m

For 250 cm = 2.5 m:

E = k \times 2.5 = 1 \times 2.5 = 2.5

V

Given

E = \frac{x}{10}

2.5 = \frac{x}{10}

, so

x = 25

Answer: 25

JEE Main · 2022 · 24 Jun 2022 · Shift 2 · Integer answer