In the circuit shown switches `S_(1)` and `S_(2)` have been closed for `1 sec` and `S_(2)` remained open.Just after `1` second is over switch `S_(2)` is closed and `S_(1),S_(3)` are opened.Find after that instant The charge on the upper plate of the capacitor as function of time taking the instant of switching on of `S_(2)` and switching off all time switches to be `t=0` is `q=x xx10^(-2)cos (t+pi/4)`.Findout value of `x`.

Explanation

`tau_(C)=tau_(L)=1 sec` ltbr. `Q_(C)=(1-e^(-t)),i_(L)=(1-e^(-t))`
`U_(i)=U_(f) rArr 1/2Li^(2)+1/2q^(2)/C=1/2 Li_(m)^(2)`
`rArr 1(1-e^(-1))^(2)+1(1-e^(-1))^(2)=i_(m)^(2)`
`rArr i_(m)=sqrt2(1-1/e)A`
`1/2 q_(m)^(2)/c=1/2 Li_(m^(2)) rArr q_(m)=sqrt2(1-1/e)`Coul.
`q=q_(m) cos (omega t + phi) rArr (1-w^(-1))=sqrt2(1-e^(-1))cos phi rArr phi=pi/4`
Hence `q=sqrt2(1-e^(-1))cos (t+pi/4)`
`sqrt2(1-e^(-1))cos (t+pi/4)=0.89 cos(t+pi/4)`