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A varying current in a coil changes from \(10 \mathrm{~A}\) to zero in \(0.5 \mathrm{sec}\). If the average e.m. \(f\) induced in the coil is \(220 \mathrm{~V}\), the self-inductance of the coil is
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The correct answer is:
\(11 \mathrm{H}\)
Initial current \(\left(\mathrm{I}_1\right)=10 \mathrm{~A}\); Final current \(\left(I_2\right)=0\); Time \((t)=0.5 \mathrm{sec}\) and induced e.m.f. \((\varepsilon)=\) \(220 \mathrm{~V}\).
Induced e.m.f. (\(\varepsilon)\)
\(=-\mathrm{L} \frac{\mathrm{dI}}{\mathrm{dt}}=-\mathrm{L} \frac{\left(\mathrm{I}_2-\mathrm{I}_1\right)}{\mathrm{t}}=-\mathrm{L} \frac{(0-10)}{0.5}=20 \mathrm{~L}\)
or, \(\mathrm{L}=\frac{220}{20}=11 \mathrm{H}\)
[where \(\mathrm{L}=\) Self inductance of coil]
Induced e.m.f. (\(\varepsilon)\)
\(=-\mathrm{L} \frac{\mathrm{dI}}{\mathrm{dt}}=-\mathrm{L} \frac{\left(\mathrm{I}_2-\mathrm{I}_1\right)}{\mathrm{t}}=-\mathrm{L} \frac{(0-10)}{0.5}=20 \mathrm{~L}\)
or, \(\mathrm{L}=\frac{220}{20}=11 \mathrm{H}\)
[where \(\mathrm{L}=\) Self inductance of coil]
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