Search any question & find its solution
Question:
Answered & Verified by Expert
A varying current in a coil changes from $10 \mathrm{~A}$ to zero in $1.5 \mathrm{~s}$. If the average emf induced in the coil is $200 \mathrm{~V}$, the self-inductance of the coil is
Options:
Solution:
2559 Upvotes
Verified Answer
The correct answer is:
$30 \mathrm{H}$
Self-induced emf in coil is given by
$E=L\left(\frac{d I}{d t}\right)$
Here, $E=200 \mathrm{~V}, \frac{d I}{d t}=\left(\frac{10-0}{1.5}\right) \mathrm{A} / \mathrm{s}$
So, self-inductance of coil, $L=\frac{E}{d I / d t}=\frac{200}{\left(\frac{10-0}{1.5}\right)}$
$\Rightarrow L=\frac{200 \times 1.5}{10}=30 \mathrm{H}$
$E=L\left(\frac{d I}{d t}\right)$
Here, $E=200 \mathrm{~V}, \frac{d I}{d t}=\left(\frac{10-0}{1.5}\right) \mathrm{A} / \mathrm{s}$
So, self-inductance of coil, $L=\frac{E}{d I / d t}=\frac{200}{\left(\frac{10-0}{1.5}\right)}$
$\Rightarrow L=\frac{200 \times 1.5}{10}=30 \mathrm{H}$
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.