The capacitor 'C' is initially unchanged. Switch `S_(1)` is closed for a long time while `S_(2)` remains open. Now at `t = 0 S_(2)` is closed while `S_(1)` is opened. All the batteries are ideal and connecting wires are resistanceless

Options

A. At time t = 0, reading of ammeter is `(E)/(5R)`
B. At time t =0, reading of ammeter is zero
C. Heat developed till time `t = 5 RC l n 2` in resistance 3 R is `(9)/(40) CE^(2)`
D. After time `t gt 0` charge on the capacitor follows the equation `CEe^(-t//5RC)`

Answer

A, C, D

Explanation

`I = (E)/(5R)`
equation of charge of capacitor
`q = CEe^(-t//tau)`
`q = Ece^(-t//5RC)`
at `t = 5 RC l n2`
`q = (EC)/(2)`
Change in energy of capacitor
`= (1)/(2)CE^(2)-(1)/(8) CE^(2) = (3)/(8) CE^(2)`
Heat in `3R = (3)/(8) CE^(2) xx (3)/(5) = (9)/(40) CE^(2)`