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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
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Topics: Electromagnetic Induction
(76)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- At the first minimum adjacent to the central maximum of a single-slit diffraction
- The half life of radium is 1620 yr and its atomic weight is 226 kg per kilo mol.
- The temperature at which the vapour pressure of a liquid becomes equal to the external
- A particle falls towards earth from infinity. Its velocity on reaching the earth
- Which component of electromagnetic spectrum have maximum wavelength?
Topics: Electromagnetic Induction (76)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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