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).
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Options

A. Magnitude of the maximum charge on the capacitor before `t = (7 pi)/(6 omega)` is `1 xx 10^(-3) C`
B. The current in the left part of the circuit just before `t = (7 pi)/(6 omega)` 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

B

Explanation

`I = cos 500t`
Till `t = (7 pi)/(6 omega)`, The charge will be maximum at `(pi)/(2 omega)`
`Q' = underset(0)overset(pi//2 omega)(int) cos 500t dt = [(sin 500 t)/(500)]_(0)^(pi // 2 omega)`
`= (1)/(500) sin (500 xx (pi)/(2 xx 500)) = (1)/(500) C`
`:.` (a) is correct
From the graph it is clear that just before `t = (7 pi)/(6 omega)`, the current is in anticlockwise direction
`:.` (b) is incorrect
At `t = (7 pi)/(6 omega)` the charge on the upper plate of capacitor is
`underset(0)overset(7 pi // 6 omega)(int) cos 500 omega t dt = (1)/(500) sin (500 xx (7 pi)/(6 xx 500))`
`= - (1)/(500) xx (1)/(2) = 10^(-3) C`
`50 + (10^(-3))/(20 xx 10^(-6)) - i xx 10 = 0 implies i = 10 A`
`:.` (c) is correct option
The maximum charge on `C` is `Q = CV`
`= 20 xx 10^(-6) xx 50 = 10^(-3) C`
Therefore, the total charge flown `= 2 xx 10^(-3) C`
`:.` (d) is the correct option.