| ⇦ | 
| ⇨ |
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:
- Two identical circular coils A and B are kept on a horizontal tube
- Heat is not being exchanged in a body. If its internal energy is increased, then
- In a series resonant LCR circuit, the voltage across R is 100 volts and R =1kΩ
- Which of the following phenomena does not show the wavenature of light?
- A thermocouple of negligible resistance produces an e.m.f. of 40 µV/⁰C in the linear range
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