⇦ | ⇨ |
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: - Light of two different frequencies whose photons have energies 1eV
- Three identical spherical shells, each of mass m and radius r are placed
- An object is projected with a velocity of 20 m/s making an angle of 45⁰ with horizontal.
- In circular coil, when number of turns is doubled and resistance becomes 1/4th
- The time period of a thin bar magnet in Earth’s magnetic field is T
Topics: Electromagnetic Induction
(76)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Light of two different frequencies whose photons have energies 1eV
- Three identical spherical shells, each of mass m and radius r are placed
- An object is projected with a velocity of 20 m/s making an angle of 45⁰ with horizontal.
- In circular coil, when number of turns is doubled and resistance becomes 1/4th
- The time period of a thin bar magnet in Earth’s magnetic field is T
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