The circuit shown in the figure is used to transfer energy from one capacitor to another. Initially capacitor of capacitance `C_(1) = C_(0)` is charged to a potential difference of `V_(0)`. Switch `S_(1)` is closed at time t = 0. After some time `S_(1)` is opened and `S_(2)` is closed simultaneously. At time `t = T, S_(2)` was opened and it was found that the potential difference across capacitor of capacitance `C_(2) = (C_(0))/(9)` was `3V_(0)`. Find the smallest possible value of time T. The coil has inductance L. Assume no resistance.

Answer

`T_(min) = (2pi)/(3) sqrt(LC_(0))`