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) = `underset(o)(I)`cos(`omega`t), with `underset(o)(I)` = 1 A and `omega` = 500 `rad s^(-1)` starts flowing in it with the initial direction shown in the figure . At t = `7pi`/`6omega` , the keys 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` , R = 10 `Omega` and the battery is deal 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`