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(C) In an $LC$ circuit,the charge on the capacitor varies as $q(t) = q_0 \cos(\omega t)$,where $\omega = \frac{1}{\sqrt{LC}}$.At $t=0$,the energy is m

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$2 \pi 	imes 10^{-4} \ s$

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(C) In an $LC$ circuit,the charge on the capacitor varies as $q(t) = q_0 \cos(\omega t)$,where $\omega = \frac{1}{\sqrt{LC}}$.At $t=0$,the energy is maximum,meaning the charge is maximum $(q = q_0)$.The current in the circuit is given by $i(t) = -\frac{dq}{dt} = q_0 \omega \sin(\omega t)$.The current is maximum when $\sin(\omega t) = 1$,which occurs at $\omega t = \frac{\pi}{2}$,or $t = \frac{\pi}{2\omega}$.Given $L = 16 	ext{ mH} = 16 	imes 10^{-3} 	ext{ H}$ and $C = 10 \mu	ext{F} = 10 	imes 10^{-6} 	ext{ F} = 10^{-5} 	ext{ F}$.Calculate $\omega = \frac{1}{\sqrt{16 	imes 10^{-3} 	imes 10^{-5}}} = \frac{1}{\sqrt{16 	imes 10^{-8}}} = \frac{1}{4 	imes 10^{-4}} = 0.25 	imes 10^4 = 2500 	ext{ rad/s}$.Now,$t = \frac{\pi}{2 	imes 2500} = \frac{\pi}{5000} = \pi 	imes 2 	imes 10^{-4} 	ext{ s} = 2 \pi 	imes 10^{-4} 	ext{ s}$.

If maximum energy is stored in a capacitor at $t=0$,then the time after which the current in the circuit will be maximum is:

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