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7.Alternating Current
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A sinusoidal voltage $V(t) = 100\,sin\,(500t)$ is applied across a pure inductance of $L = 0.02\,H.$ The current through the coil is
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
