At time t = 0,terminal A in the circuit shown | vedclass

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(C) The charge on the capacitor is $q(t) = \int_0^t I(t) dt = \int_0^t I_0 \cos(\omega t) dt = \frac{I_0}{\omega} \sin(\omega t)$.Given $I_0 = 1 	ext

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$(C, D)$

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(C) The charge on the capacitor is $q(t) = \int_0^t I(t) dt = \int_0^t I_0 \cos(\omega t) dt = \frac{I_0}{\omega} \sin(\omega t)$.Given $I_0 = 1 	ext{ A}$ and $\omega = 500 	ext{ rad s}^{-1}$,$q(t) = \frac{1}{500} \sin(500t) = 2 	imes 10^{-3} \sin(500t) 	ext{ C}$.Maximum charge $Q_{\max} = 2 	imes 10^{-3} 	ext{ C}$. Thus,$(A)$ is incorrect.At $t = \frac{7\pi}{6\omega}$,$I(t) = I_0 \cos(\frac{7\pi}{6}) = 1 	imes (-\frac{\sqrt{3}}{2}) At $t = \frac{7\pi}{6\omega}$,the charge on the capacitor is $q = 2 	imes 10^{-3} \sin(\frac{7\pi}{6}) = -1 	imes 10^{-3} 	ext{ C}$.When connected to the battery $(50 	ext{ V})$,the initial charge is $q_i = -1 	imes 10^{-3} 	ext{ C}$.Applying $KVL$: $50 - \frac{q_i}{C} - IR = 0 \Rightarrow 50 - \frac{-1 	imes 10^{-3}}{20 	imes 10^{-6}} - I(10) = 0 \Rightarrow 50 + 50 - 10I = 0 \Rightarrow I = 10 	ext{ A}$. Thus,$(C)$ is correct.Final charge $q_f = CV = 20 	imes 10^{-6} 	imes 50 = 1 	imes 10^{-3} 	ext{ C}$.Charge flown $Q = q_f - q_i = 1 	imes 10^{-3} - (-1 	imes 10^{-3}) = 2 	imes 10^{-3} 	ext{ C}$. Thus,$(D)$ is correct.

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