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

Question

A potentiometer wire of length 10 m and resistance 20 Ω is connected in series with a 25 V battery and an external resistance 30 Ω. A cell of emf E in secondary circuit is balanced by 250 cm long potentiometer wire. The value of E (in volt) is (x/10). The value of x is . (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/28b898e876fb33d6ac4d9bbf46174688-.png" /> I =(25/50)=(1/2) A ∴ Δ V =10 V 10 m arrow 10 V 2.5 m arrow 2.5 V

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

I=5025​=21​A ∴ΔV=10V 10m→10V 2.5m→2.5V

$I = \frac{25}{50} = \frac{1}{2} A$
$∴ Δ V = 10 V$
$10 m \to 10 V$
$2.5 m \to 2.5 V$