In the circuit shows (Fig.) switches `S_(1)` and `S_(3)` have been closed for `1 s` and `S_(2)` remained open. Just after `1 s`, switch `S_(2)` is closed and `S_(1)` and `S_(3)` are opened. Find after that instant `(t = 0)`:

the maximum charge on the capacitor

Options

A. `sqrt(3) (1 + (1)/(e))`
B. `sqrt(3) (1 - (1)/(e))`
C. `sqrt(2) (1 + (1)/(e))`
D. `sqrt(2) (1 - (1)/(e))`

Answer

D

Explanation

`S_(1)` and `S_(2)` are closed for `1 s`.
Charge on capacitor,
`q = CE (1 - e^(-t//RC)) = 1 - (1)/(e)` (i)
The current in inductor,
`I = (E)/(R ) (1 - e^(-t//RC)) = 1 - (1)/(e)`(ii)
Now `S_(1)` and `S_(3)` are opened and `S_(2)` is closed,
It is `LC` circuit,
`q = q_(max) sin (omegat + phi)` (iii)
`I = (q_(max)) omega cos(omegat + phi)` (iv)
As total energy (Magnetic + electrical) is constant
`(1)/(2) LI_(max)^(2) = (1)/(2)(q_(max)^(2))/(C) = (1)/(2) LI^(2) + (1)/(2) (q^(2))/(C)` (v)
`q_(max) = sqrt(2) (1 - (1)/(e))` (vi)
`I_(max) = sqrt(2) (1 - (1)/(e))` (vii)
From (iii) at `t = 0`, we get
`(1 - (1)/(e)) = sqrt(2) (1 - (1)/(e)) sin phi rArr = (pi)/(4)` or `(3pi)/(4)`
After closing `S_(2)`, circuit will be as shows. Direction of current shows that charge on capacitor will be decreaseing. Hence `phi = 3 pi//4`
`q = sqrt(2)(1 - (1)/(e)) sin(omegat + (3pi)/(4))`
where, `omega = (1)/(sqrt(LC)) = 1`
`q = sqrt(2) (1 - (1)/(e))sin(t + (3pi)/(4))`