| ⇦ |
| ⇨ |
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 particles P and Q start from origin and execute simple harmonic motion along X-axis
- Identify the paramagnetic substance
- At 10⁰C the value of the density of a fixed mass of an ideal gas divided by its pressure
- A wheel has angular acceleration of 3.0 rad/sec² and an initial angular speed of 2.00
- If force(F),velocity(v) and time(T) are taken as units, then the dimensions of mass are
Topics: Electromagnetic Induction
(76)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Two particles P and Q start from origin and execute simple harmonic motion along X-axis
- Identify the paramagnetic substance
- At 10⁰C the value of the density of a fixed mass of an ideal gas divided by its pressure
- A wheel has angular acceleration of 3.0 rad/sec² and an initial angular speed of 2.00
- If force(F),velocity(v) and time(T) are taken as units, then the dimensions of mass are
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