Question

At time $t=0$, terminal $A$ in the circuit shown in the figure is connected to $B$ by a key and an alternating current $I(t)=I_0\cos (\omega t)$, with $I_0=1A$ and $\omega =500rad/s$ 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 only $A$ and $D$ are connected. $A$ total charge $Q$ flows from the battery to charge the capacitor fully. If $C=20\mu F,R=10\Omega$ and the battery is ideal with emf of $50V$, 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 $10A$
• D. $Q=2\times 10^{-3}C$

Answer 1

Answer: (C), (D)

As current leads voltage by $\pi /2$ in the given circuit initially,
then $ac$ voltage can be represent as
$V=V_0\sin \omega t$
$\therefore q=C{V}_0\sin \omega t=Q\sin \omega t$
where, $Q=2\times 10^{-3}C$
$\bullet$ At $t=7\pi /6\omega ;$
$I=-\frac{\sqrt{3}}{2}{I}_0$ and hence current is anticlockwise.
$\bullet$ Current ' $i$ ' immediately after $t=\frac{7\pi }{6\omega }$ is
$i=\frac{V_c+50}{R}=10A$
$\bullet$ Charge flow $=Q_{\text{final }}-Q_{(7\pi /6\omega )}$
$=2\times 10^{-6}C$