In the following circuit `C_1 = 900 mu F` which is charged to a p.d of 100V and `C_2 = 100 mu F` is an unchanged capacitor L = 10 henry . Explain how will you change `C_2` to a.p.d of 300V by adjusting the switched `S_1` and `S_2`

Explanation

Energy of `C_1 = 1/2 C_1V^2 = 1/2 xx 900 xx 10^(-6) xx 100 xx 100`
` = 4.5 J`
To charge `C_2` to a p.d. of 300 V, the energy needed ` = 1//2 C_2V^2`
` = 1//2 xx 100 xx 10^(-6) xx 300 xx 300 = 4.5 J`
Now first close `S_1` and open `S_2` . allow the charge on `C_1` to discharge completely and the energy is stored in the inductor . this happens during the first `1/4` th of the period of oscillation
i.e., `t_1 = T/4 = (2pi sqrt(LC_1) )/(4) = 0.15 sec `
upto 0.15 sec , ` S_1` should be closed and `S_2` should be opened . Now close `S_2` and open `S_1` so that the inductor begins to discharge and `C_2` gets completely charged during the next ` 1/4 th ` of period of oscillation .
i.e., `t_2 = T/4 = (2pi sqrt(LC_2) )/(4) = 0.05 sec `
Upto 0.05 sec , `S_2` is to closed and `S_1`is to be opened