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 `1 xx 10^(-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 10 A
D. Q = `2 xx 10^(-3)` C

Answer

C, D

Explanation

(c,d) Charge on capacitor will be maximum at t`=pi/(2omega)`
`Q_(max)=2 xx 10^(-3)`C
A) Charge supplied by source from t=0 to t=`(7pi)/(6omega)`
`Q=int_(0)^(7pi)/(6Omega) cos(500)dt = [(sin500t)/(500)]_(0)=(7pi)/(6omega)= (sin(7pi)/6)/(500)=-1mC`

Apply KVL, just after switching.
`50 _(Q_(t))/C=IR=0`
`rArr I=10A`
In steady state `Q_(2)= 1mC`
Net charge flown from battery `Q=2mc`