An inductor and two capacitors are connected in the circuit as show in Fig Initially capacitor `A` has no charge and capacitor `B` has `CV` charge. Assume that the circuit has no resistance at all. At `t = 0`, switch `S` is closed, then [given
`LC = (2)/(pi^(2) xx 10^(4)) S^(2)` and `Cv = 100 mC]`

Options

A. when current in the circuit is maximum, charge on each capacitor is same
B. when current in the circuit is maximum, charge on capacitor `A` is twice the charge on capacitor `B`
C. `q = 50(1 + cos 100 pi t) mC`, where `q` is the charge on capacitor `B` at time `t`
D. `q = 50(1 - cos 100 pi t) mC`, where `q` is the charge on capacitor `B` at time `t`

Answer

A, C

Explanation

Let at any time, charge and current in the circuit wire are as shows.

`I = (dq_(1))/(dt)`
Applying kirchoff's law,
`(CV - q_(1))/(C ) - (q_(1))/(C ) - L(dI)/(dt) = 0 rArr (CV - 2q_(1))/(LC) = L (d^(2)q_(1))/(dt^(2))`
`rArr (d^(2)q_(1))/(dt^(2)) = - (2)/(LC) (q_(1) - (CV)/(2))`
`rArr q_(1) - (CV)/(2) = A sin (omega t + delta)` (i)
where `omega = sqrt((2)/(LC)) = 100 pi rad//s`
At `t = 0`, `q_(1) = 0 rArr 0 - (CV)/(2) = A sin delta` (ii)
Differentiating (i), `(dq_(1))/(dt) - 0 = A omega cos (omega t + delta)`
`rArr I = A omega cos (omega t + delta)`
At `t = 0, I = 0 rArr 0 = A omega cos delta rArr delta = (pi)/(2)`
From (ii), `A = - (CV)/(2)`
From (i), `q_(1) = (CV)/(2) - (CV)/(2) sin (omega t + (pi)/(2))` (iv)
`rArr q_(1) = (CV)/(2) (1 cos 100 pi t) mC`
Now `q_(2) = CV - q_(1) = (CV)/(2) (1 + cos omega t)` (v)
`= 50 (1 + cos omega pi t)`
From (ii), `I = - A omega sin omega t`
Current in the circuit is maximum for `omega t = (pi)/(2)*`
And for `omegat = (pi)/(2)`, we see from (iv) and (v) that `q_(1) = q_(2) = (CV)/(2)`