| ⇦ | 
| ⇨ |
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: - In nuclear fission, the percentage of mass converted into energy is about
- The Davisson- Germer experiment is the direct evidence of
- In an LCR circuit, capacitance is changed from C to 2C. For the resonant frequency
- In a parallel plate capacitor of capacitance C, a metal sheet is inserted
- A circular road of radius 1000m has banking angle 45°.The maximum safe speed of a car
Topics: Electromagnetic Induction
(76)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- In nuclear fission, the percentage of mass converted into energy is about
- The Davisson- Germer experiment is the direct evidence of
- In an LCR circuit, capacitance is changed from C to 2C. For the resonant frequency
- In a parallel plate capacitor of capacitance C, a metal sheet is inserted
- A circular road of radius 1000m has banking angle 45°.The maximum safe speed of a car
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