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 rads^(-1)` starts flowing in it with the initial direction shown in the figure. At `t = (7pi//6omega)`, 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).
.

Options

A. Magnitude of the maximum charge on the capacitor before `t=(7pi)/(6omega)` is `1xx10^(-3) C`.
B. The current in the left part of the circuit just before `t=(7pi)/(6omega)` is clockwise
C. Immediately after `A` is connected to `D`.the current in `R` is `10A`
D. `Q=2xx10^(-3) C`

Answer

C,D

Explanation

Charge on capacitor will be maximum at `t=pi/(2omega)`
`Q_(max)=2xx10^(-3)C`
`(A)` charge supplied by source from `t=omegat~~(7pi)/(6omega)`
`Q=underset(0)overset((7pi)/(6omega))intcos(500t)dt=[(sin 500t)/500]_(0)^((7pi)/(6omega))=("sin"(7pi)/6)/500=-1mC`
Apply `KVL` just after switching `50+Q_(1)/C-IR=0 rArr I=10A`
In steady state=`Q_(2)=1 mC`
net charge flown from battery =`2mC`