7.Alternating Current
medium

एक ज्यावक्रीय वोल्टता $V ( t )=100 \sin (500 t )$ एक विशुद्ध प्रेरकत्व $L =0.02\; H$ पर लगाई जाती है। कुण्डली से प्रवाहित धारा है

A

$10\,cos\,(500t)$

B

$-10\,cos\,(500t)$

C

$10\,sin\,(500t)$

D

$-10\,sin\,(500t)$

(JEE MAIN-2014)

Solution

In a pure inductive circuit current always

lags behind the emf by $\frac{\pi}{2}$

If $v(t)=v_{0} \sin \omega t$

then $\mathrm{I}=\mathrm{I}_{0} \sin \left(\omega \mathrm{t}-\frac{\pi}{2}\right)$

Now, given $v(t)=100 \sin (500 \,t)$

and $\mathrm{I}_{0}=\frac{\mathrm{E}_{0}}{\omega \mathrm{L}}=\frac{100}{500 \times 0.02}[\because \mathrm{L}=0.02 \,\mathrm{H}]$

$\mathrm{I}_{0}=10 \sin \left(500 \mathrm{t}-\frac{\pi}{2}\right)$

$\mathrm{I}_{0}=-10 \cos (500 \mathrm{t})$

Standard 12
Physics

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