Shown in the figure is a parallel R , L,C circuit with key `K_(1)` closed and `K_(2)` opened . When `K_(1)` is opened and `K_(2)` is closed simultaneously find the maximum potential difference across the capacitor and maximum charge stored in it .

Explanation

The current flowing through the inductor L in steady state before closing the switch `K_(2)` is given as
`i=(E )/ (R ) ( because ` the inductor offers no resistance at steady state ).
The maximum energy stored in the inductor
`U_(L) = (1)/(2) Li^(2)= (L)/(2) ((E)/(2))^(2) = (LE^(2))/(2R^(2))`
Since the maximum energy `U_(c)` stored in the capacitor after closing the switch `K_(2)` is equal to `U_(L)` according to conservation of total energy.
`U_(c)= (1)/(2) CV^(2)= U_(L)`
`implies (1)/(2) CV^(2)= (LE^(2))/(2R^(2))`
`implies V =(E)/(R) sqrt((L)/(C))` ,where V = maximum potential difference across the capacitor
(Maximum charged stored in the capacitor )
`q_("max")= CV = (Esqrt(LC))/(R)`