Search any question & find its solution
Question:
Answered & Verified by Expert
An average induced emf of $0.5 \mathrm{~V}$ appears in a coil when the current in it is changed from $10 \mathrm{~A}$ in one direction to $10 \mathrm{~A}$ in opposite direction in 0.8 second. The self-inductance of the coil is
Options:
Solution:
2875 Upvotes
Verified Answer
The correct answer is:
50 mH
$e=L \frac{d I}{d t}=L\left(\frac{20}{0.5}\right)=40 L$ and $e=0.5 \mathrm{~V}$
$\therefore L=\frac{2}{40} \mathrm{H}=0.05 \mathrm{H}=50 \mathrm{mH}$
$\therefore L=\frac{2}{40} \mathrm{H}=0.05 \mathrm{H}=50 \mathrm{mH}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.