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)=I₀ cos (ω t),with I₀=IA and ω=500rad/s starts flowing in it with the initial direction shown in the figure.At t=7π/6ω, 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{F},R=10 and the battery is ideal with emf of 50 V, identify the correct statement(s). [JEE Adv 2014 P1]

Magnitude of the maximum charge on the capacitor before $t=\frac{7\pi}{6\omega}$ is $1\times10^{-3}$ C
The current in the left part of the circuit just before $t=\frac{7\pi}{6\omega}$ is clockwise
Immediately after A is connected to D,the current in R is $10$ A
Q $=2\times10^{-3}$ C

Solution:

As current leads voltage by

\frac{\pi}{2}

in the given circuit initially, then ac voltage can be represented as

V=V_0

sin

\omega t

\therefore q=CV_0

sin

\omega t=Q

sin

\omega t

where, Q

=2\times10^{-3}

•At

t=\frac{7\pi}{6\omega};I=-\frac{\sqrt{3}}{2}I_0

and hence current is anticlockwise

• Current 'i' immediately after

t=\frac{7\pi}{6\omega}

i=\frac{V_c+50}{R}=10

•Charge flow

=Q_{\text{final}}-Q_{\frac{7\pi}{6\omega}}=2\times10^{-6}

Hence C & D are correct options

Go to Question:

Prev Question

Next Question