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 = ( 7pi)/(6 omega)` is `1 xx 10^(-3) C`
B. This current in the left part of the circuit just before `t = (7 pi)/(6 omega)` is clockwise
C. Immediatley after `A` is connected to `D`, the current in `R` is `10 A`
D. `Q = 2 xx 10^(-3) C`

Answer

C, D

Explanation

Due to capacitor in ac circuit, the voltage lags behind the current by phase `pi//2`
As `I = I_(0) cos omega t`
So `V = V_(0) cos (omega t = pi//2) = V_(0) sin omega t`
Change in circuit, at a given instatn `t`
`Q = int_(0)^(t) I dt int_(0)^(t) I_(0) cos omega t dt = (I_(0))/(omega) sin omega t`
Max. charge on capacitor,
`Q_(0) = (I_(0))/(omega) = (1)/(500) = 2 xx 10^(-3) C`
The current in the left part of the circuit at time
`t = (7 pi)/(6 omega)`
`I = I_(0) cos omega xx (7 pi)/(6 omega) = I_(0) "cos" (7 pi)/(6)`
`= I_(0) "cos" (pi + (pi)/(6))`
`= I_(0) [ cos pi "cos" (pi)/(6) - sin pi "sin" (pi)/(6)]`
`= I_(0) [- 1 xx (sqrt3)/(2) - 0 ] = (sqrt3)/(2) I_(0)`
Thus current in the left part of the circuit is anticlockwise as value of current is negative.
At time `t = (7 pi)/(6 omega)`, change on capacitor is
`Q = 2 xx 10^(-3) sin omega xx (7 pi)/(6 omega)`
`2 xx 10^(-3) "sin" (7 pi)/(6) = 2 xx 10^(-3) ["sin" (pi + (pi)/(6))]`
`2 xx 10^(-3) [sin pi "cos" (pi)/(6) + cos pi "sin" (pi)/(6)]`
`2 xx 10^(-3) [0 + (-1) xx (1)/(2)] = - 10^(-3) C`
Potential difference across capacitor due to
charge `Q V_(C ) = (Q)/(C ) = (10^(-3))/(20 xx 10^(-6)) = 50 V`
Final charge on capacitor after shifting the switch
`Q' = CV_(C ) = (20 xx 10^(-6)) xx 50 = 10^(-3) C`
Charge flown from battery ` = 10^(-3) + 10^(-3)`
`= 2 xx 10^(-3) C`
Thus options (c ) and (d) are ture.