⇦ | ⇨ |
In an inductor when current changes from 2 A to 18 A in 0.05 sec, the e.m.f. induced is 20 V. The inductance L is
Options
(a) 62.5 mH
(b) 625 mH
(c) 6.25 mH
(d) 0.625 mH
Correct Answer:
62.5 mH
Explanation:
Induced e.m.f. |e| = L (dI / dt) = L(18 – 2) / 0.05 or,
L = (20 × 0.05) / 16 = 0.0625 H = 62.5 mH
Related Questions: - Two tuning forks are sounded together and produce 2 beats per second.
- A projectile is thrown in the upward direction making an angle of 60°with the horizontal
- A particle having a mass of 10⁻² kg carries a charge of 5 x 10⁻⁸C
- The physical quantity which does not have the same dimensions as the other three is
- A body of weight W newton is at the surface of the earth.
Topics: Electromagnetic Induction
(76)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Two tuning forks are sounded together and produce 2 beats per second.
- A projectile is thrown in the upward direction making an angle of 60°with the horizontal
- A particle having a mass of 10⁻² kg carries a charge of 5 x 10⁻⁸C
- The physical quantity which does not have the same dimensions as the other three is
- A body of weight W newton is at the surface of the earth.
Topics: Electromagnetic Induction (76)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends
Leave a Reply