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=(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. Immidiately after A is connected to D, the current in R is 10 A.
D. `Q=2 xx 10^(-3)C`

Answer

C, D

Explanation

If q represents the charge on capacitor's upper plate:
`I(t)=(I_0)cos (omega t)=(dq)/(dt) implies q(t)=(I_0)/(omega) sin (omega t)`
Max charge `=(1 A)/(500 rad s^(-1)) = 2 xx 10^(-3) C`
Charge on upper plate at `t=(7 pi)/(6 omega)=(I_0)/(omega) sin ((7 pi)/(6 omega))=-(I_0)/(2 omega)`
When capacitor is fully charged, charge on upper plate `=50V`
`xx20 (mu)F=1 xx 10^(-3)C`
`:. Q= 1 xx 10^(-2)C-(-(I_0)/(2 omega)) = 2 xx 10^(-3)C`
Voltage across capacitor when A and D are connected
`=(1 xx 10^(-3)C)/(20 xx 10^(-6)mu F)=50 V`
Total voltage resistor `= 100 V`
`implies `Current `=(100 V)/(10 Omega)` 10 A
Current is anticlockwise.