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Initially the 900muF capacitor is change...

Initially the `900muF` capacitor is changed to `100 V` and the `100 muF` 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 muF` capacitor is charged to `300 V`.If `t_(1)` and `t_(2)` minimum possible values of the time intervals, then findout `t_(1)/t_(2)(pi^(2)=10)`

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