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At time $t =0$, terminal $A$ in the circuit shown in the figure is connected to $B$ by a key and alternating current $I ( t )= I _0 \cos (\omega t$,$) , with I _0=1 A$ and $\omega=500 \ rad s ^{-1}$ starts flowing in it with the initial direction shown in the figure.
At $t=\frac{7 \pi}{6 \omega}$, the key is switched from $B$ to $D$. Now onwards on ly $A$ and $D$ are connected. $A$ total charge $Q$ flows from the battery to charge the capacitor fully. If $C =20 \mu, R =10 \Omega$ and the battery is ideal with emf of $50 \ V$, identify the correct statement $(s)$
$(A)$ Magnitude of the maximum charge on the capacitor before $t=\frac{7 \pi}{6 \omega}$ is $1 \times 10^{-3} C$.
$(B)$ The current in the left part of the circuit just before $t=\frac{7 \pi}{6 \omega}$ is clockwise
$(C)$ Immediately after $A$ is connected to $D$. the current in $R$ is $10 A$.
$(D)$ $Q =2 \times 10^{-3} C$.

$(B,C)$
$(B,D)$
$(C,D)$
$(A,D)$
Solution

Charge on capacitor will be maximum at $t =\frac{\pi}{2 \omega}$
$Q _{\max }=2 \times 10^{-3} C$
$(A)$ charge supplied by source from $t=0$ to $t=\frac{7 \pi}{6 \omega}$
$Q=\int_0^{\frac{7 \pi}{6 \omega}} \cos (500 t) d t=\left[\frac{\sin 500 t}{500}\right]_0^{\frac{7 \pi}{6 \omega}}=\frac{\sin \frac{7 \pi}{6}}{500}=-1 m C$
$Image$
Apply KVL just after switching
$50+\frac{ Q _1}{ C }- IR =0 \Rightarrow I =10 A$
In steady state $Q_2=1 mC$ net charge flown from battery $=2 mC$