In an oscillating LC-circuit,L = 3 mH and C | vedclass

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(B) The charge $q$ as a function of time $t$ is given by $q = q_0 \sin(\omega t)$,where $q_0$ is the maximum charge and $\omega$ is the angular freque

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$18 	imes 10^{-5} \ C$

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(B) The charge $q$ as a function of time $t$ is given by $q = q_0 \sin(\omega t)$,where $q_0$ is the maximum charge and $\omega$ is the angular frequency.Differentiating with respect to $t$,the current $I$ is given by $I = \frac{dq}{dt} = \omega q_0 \cos(\omega t)$.At $t = 0$,the current is $I = \omega q_0 \cos(0) = \omega q_0$.Given $I = 2 \ A$ at $t = 0$,we have $2 = \omega q_0$.Since $\omega = \frac{1}{\sqrt{LC}}$,we substitute this into the equation: $q_0 = I \sqrt{LC}$.Given $L = 3 	imes 10^{-3} \ H$ and $C = 2.7 	imes 10^{-6} \ F$,we calculate:$q_0 = 2 	imes \sqrt{3 	imes 10^{-3} 	imes 2.7 	imes 10^{-6}}$$q_0 = 2 	imes \sqrt{8.1 	imes 10^{-9}} = 2 	imes \sqrt{81 	imes 10^{-10}}$$q_0 = 2 	imes 9 	imes 10^{-5} = 18 	imes 10^{-5} \ C$.

In an oscillating $LC$-circuit,$L = 3 \ mH$ and $C = 2.7 \ \mu F$. At $t = 0$,the charge on the capacitor is zero and the current is $2 \ A$. The maximum charge that will appear on the capacitor will be

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