The charge in an LC circuit with negligible r | vedclass

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Question

From equation $\omega^{2}=16 \pi^{2} \quad ightarrow \omega=4 \pi$

$	herefore \mathrm{q}=\mathrm{q}_{\mathrm{o}} \cos (\omega \mathrm{t})$

$=

Vedclass · Accepted Answer

$12$

Vedclass · Answer

From equation $\omega^{2}=16 \pi^{2} \quad ightarrow \omega=4 \pi$

$	herefore \mathrm{q}=\mathrm{q}_{\mathrm{o}} \cos (\omega \mathrm{t})$

$=24 \cos \left(4 \pi 	imes \frac{1}{12}ight)$

$=24 \cos \left(\frac{\pi}{3}ight)$

$=24 	imes \frac{1}{2}=12$

The charge in an $LC$ circuit with negligible resistance oscillates as given by equation $\frac{{{d^2}q}}{{d{t^2}}} + 16{\pi ^2}q = 0$. If the charge is maxiumum equal to $24\,\mu C$ at $t = 0$, find the charge at $t = \frac{1}{{12}}\,s$............$\,\mu C$

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