Initially the `900 mu F` capacitor is charged to `100 V` and the `100 mu F`capacitor is uncharged in Fig. Then switch `S_(2)` is closed for time `t_(1)`, after which it is opened and at the same instant switch `S_(1)` is closed for time `t_(2)` and then opened. It is now found that `100 mu F` capacitor is charged to `300 mu F` capacitor is charged to `300 V`. Find the minimum possible value of the time interval `t_(1)` and `t_(2)`.

Explanation

Initial energy stored in the `900 mu F` capcitor is
`U_(1) = (1//2) xx 900 xx 10^(-6) xx (100)^(2) = 4.5 J`
Finally, energy stored in the `100 mu F` capacitor is
`U_(2) = (1//2) xx 100 xx 10^(-6) xx (300)^(2) = 4.5 J`
The entire energy of the `900 mu f` capacitor has been transferred to the `100 mu F` capacitor. First, electrical energy of the `900 mu F` capacitor is converted into magnetic energy in the inductor and then this energy is converted into electrical energy once again using `S_(2)` and `S_(1)` appropriately.
In an `LC` circuit, the transfer of electrical energy into magnetic energy and vice verse takes place in time `T//4` where `T = 2 pi sqrt(LC)` is time period of the electrical oscillations. Thus,
`T_(1) 2 pi sqrt(10 xx 900 xx 10^(-6)) = 0.6 s`
`T_(2) 2 pi sqrt(10 xx 100 xx 10^(-6)) = 0.2 s`
Therefore, switch `S_(2)` is first closed for time `0.6//4 = 0.15 s`, during which time the `900 mu F` capacitor gets fully discharged and the current in the inductor is fully established. Next switch `S_(2)` is opened and simultaneously switch `S_(1)` is closed for time `0.2//4 = 0.05` s during which the current in the inductor disappears and the `100 mu F` capacitor gets fully charged.
After this time, switch `S_(1)` is also opened. The `100 mu F` capacitor is now charged to `300 V`.
Thus, `t_(1) = 0.15s` and `t_(2) = 0.05 s`