| ⇦ |
| ⇨ |
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: - When a radioactive substance emits an ∝ particle, its position in the periodic table
- The respective speeds of the molecules are 1,2,3,4 and 5 km/s.
- All components of the electromagnetic spectrum in vacuum have the same
- Moment of inertia, along the diameter of a ring is
- A proton is moving in a uniform magnetic field B in a circular path of radius a
Topics: Electromagnetic Induction
(76)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- When a radioactive substance emits an ∝ particle, its position in the periodic table
- The respective speeds of the molecules are 1,2,3,4 and 5 km/s.
- All components of the electromagnetic spectrum in vacuum have the same
- Moment of inertia, along the diameter of a ring is
- A proton is moving in a uniform magnetic field B in a circular path of radius a
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