⇦ | ⇨ |
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: - The electric field associated with an e.m. wave in vacuum is given by
- A 200 W sodium street lamp emits yellow light of wavelength 0.6 µm
- If two bodies are projected at 30⁰ and 60⁰ respectively with the same velocity, then
- The identical cells connected in series are needed to heat a wire of length one meter
- Two thin lenses of focal lengths f₁ and f₂ are in contact and coaxial.
Topics: Electromagnetic Induction
(76)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The electric field associated with an e.m. wave in vacuum is given by
- A 200 W sodium street lamp emits yellow light of wavelength 0.6 µm
- If two bodies are projected at 30⁰ and 60⁰ respectively with the same velocity, then
- The identical cells connected in series are needed to heat a wire of length one meter
- Two thin lenses of focal lengths f₁ and f₂ are in contact and coaxial.
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