Question

Initially the 900 $\mu F$ capacitor is charged to 100 V and the 100$\mu F$ capacitor is uncharged in the figure shown. Then the switch $S_2$ is closed for a time $t_1$after which it is opened and at the same instant switch $S_1$ is closed for a time $t_2$ and then opened. It is now found that the 100 $\mu F$ capacitor to 300 V. If $t_1$ and $t_2$ minimum possible values of the time intervals, then find out $\frac{t_1}{t_2}$

Answer 1

Initial energy stored in the $900\mu F$ capacitor is
$U_i\infty \left(1/2\right)\times 900\times {10}^{-6}\times (100{)}^2=4.5J$
Finally, energy stored in the $100\mu F$ capacitor is
$U_2=\left(1/2\right)\times 100\times {10}^{-6}\times (300{)}^2=4.5J$
The entire energy of the $900\mu F$ capacitor has been transferred to the $100\mu F$ capacitor. First, the 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 versa takes place in time $\frac{T}{4}$ where $T=2\pi \sqrt{LC}$ is the time period of the electrical oscillations. Thus,
$T_1=2\pi \sqrt{10\times 900\times {10}^{-6}}=0.6s$
$T_2=2\pi \sqrt{10\times 100\times {10}^{-6}}=0.2s$
Therefore, switch $S_2$ is first closed for time $\frac{0.6}{4}=0.15s$, during which time the $900\mu F$ capacitor gas 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 $\frac{0.2}{4}=0.05s$ 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 $300V.$
Thus, $t_1=0.15s$ and $t_2=0.05s$

Thus, $\frac{t_1}{t_2}=\frac{0.15}{0.5}=3$