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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
- Ionization potential of hydrogen atom is 13.6 eV. Hydrogen atoms in the ground state
- If the dipole moment of a short bar magnet is 1.25 A-m², the magnetic field on its
- A particle is executing the motion x=Acos (ωt-θ). The maximum velocity of the particle is
- A ball is dropped from a high rise platform at t = 0 starting from rest
- The electric field in a certain region is acting radially outward and is given by E=Ar.
Topics: Electromagnetic Induction (76)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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