| ⇦ | 
| ⇨ |
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: - An aeroplane flies 400m due to North and then 300m due to south and then flies
- A particle moving along a straight line OX.At a time (in second ) the distance
- An electric motor runs on DC source of emf 200 V and draws a current of 10 A.
- An ideal gas is compressed to half its initial volume by means of several processes.
- A body travels 200 cm in the first two seconds and 220 cm in the next 4 secs
Topics: Electromagnetic Induction
(76)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- An aeroplane flies 400m due to North and then 300m due to south and then flies
- A particle moving along a straight line OX.At a time (in second ) the distance
- An electric motor runs on DC source of emf 200 V and draws a current of 10 A.
- An ideal gas is compressed to half its initial volume by means of several processes.
- A body travels 200 cm in the first two seconds and 220 cm in the next 4 secs
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