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 $I _{0}=1 A$ and $\omega=500\, rad / s$ 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 $50\, V$, 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. $Q =2 \times 10^{-3} C$

Answer 1

Answer: (C), (D)

As current leads voltage by $\pi / 2$ in the given circuit initially,
then $ac$ voltage can be represent as
$V = V _{0} \sin \omega t$
$\therefore q = CV _{0} \sin \omega t = Q \sin \omega t$
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
$i =\frac{ V _{ c }+50}{ R }=10 \,A$
$\bullet$ Charge flow $=Q_{\text {final }}-Q_{(7 \pi / 6 \omega)}$
$=2 \times 10^{-6} C$