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).
.

Options

A. Magnitude of the maximum charge on the capacitor before `t=(7 pi)/(6 omega) is 1xx10^(_3)C`
B. The current in the left part of the circuit just before `t=(7 pi)/(6 omega)` is clockwise
C. Immediatly after A is connected to D, the current R is 10A.
D. `Q=2xx10^(03)C`

Answer

C, D

Explanation

`I=cos 500t`


Till `t=(7 pi)/(6 omega)`, the charge will be maximum at `(pi)/(2 omega)`
`(Q')=int_(0)^(pi//2omega) cos 500t (dt)=[(sin 500t)/(500)]_(0)^(pi/2 omega)`
`=1/500 sin(500xx(pi)/(2xx500)=(1)/(500)C`
:. (a) is incorrect
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
`int_(0)^((7pi)/(6 omega)) cos 500t dt =(1)/(500) sin(500xx(7 pi)/(6xx500))`
`=-1/500xx1/2=-10^(-3)C`
Now appliying KVL (when A is just connected to D)
`50+(10^(-3))/(20xx10^(-6))-ixx10=0 impliesi-10A`
:. (c) is the correct option.
The maximum charge on C is `Q=CV=20xx10^(-6)xx50`
`=10^(-3)C`.
Therefore, the total charge flown `=2xx10^(-3)C`.
`:. (d) is the correct option.