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Question

Initially the 900 μF capacitor is charged to 100 V and the 100μF capacitor is uncharged in the figure shown. Then the switch S2 is closed for a time t1after which it is opened and at the same instant switch S1 is closed for a time t2 and then opened. It is now found that the 100 μF capacitor to 300 V. If t1 and t2 minimum possible values of the time intervals, then find out t1t2
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Solution

Initial energy stored in the 900μF capacitor is
Ui(1/2)×900×106×(100)2=4.5J
Finally, energy stored in the 100μF capacitor is
U2=(1/2)×100×106×(300)2=4.5J
The entire energy of the 900μF capacitor has been transferred to the 100μF capacitor. First, the electrical energy of the 900μF capacitor is converted into magnetic energy in the inductor, and then this energy is converted into electrical energy once again using S2 and S1 appropriately.
In an LC circuit, the transfer of electrical energy into magnetic energy and vice versa takes place in time T4 where T=2πLC is the time period of the electrical oscillations. Thus,
T1=2π10×900×106=0.6s
T2=2π10×100×106=0.2s
Therefore, switch S2 is first closed for time 0.64=0.15s, during which time the 900μF capacitor gas fully discharged and the current in the inductor is fully established. Next, switch S2 is opened and simultaneously switch S1 is closed for time 0.24=0.05s during which the current in the inductor disappears and the 100μF capacitor gets fully charged.
After this time, switch S1 is also opened. The 100μF capacitor is now charged to 300V.
Thus, t1=0.15s and t2=0.05s
Thus, t1t2=0.150.5=3

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