In the given figure, there is a circuit of potentiometer of length AB = 10 m. The resistance per unit length is 0.1 Ω 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: (a) 5 V (b) 2.25 V (c) 6 V (d) 2.75 V

Answer: (a). Wire resistance: 1000 cm×0.1 Ω/cm = 100 Ω. From the figure, driver is 6 V with 20 Ω: I = (6)/(100+20) = 0.05 A. Maximum measurable emf = VAB = 0.05×100 = 5 V So the answer is (a) 5 V.

In the given figure, there is a circuit of potentiometer of length AB… | JEE Main 2021 (25 Jul, Shift 1)

Q61JEE Main 2021 · 25 JUL · SHIFT 1Medium

Potentiometer

In the given figure, there is a circuit of potentiometer of length AB = 10 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:

Show answer and solution

Answer: (a)

Wire resistance: $1000\text{ cm}\times0.1\ \Omega/\text{cm} = 100\ \Omega$ .
From the figure, driver is 6 V with $20\ \Omega$ : $I = \dfrac{6}{100+20} = 0.05$ A.
Maximum measurable emf $= V_{AB} = 0.05\times100 = 5$ V
So the answer is (a) 5 V.

Report a mistake

Potentiometer question from JEE Main 2021 (25 Jul, Shift 1)

More questions on Potentiometer