Consider the circuit shown in the figure. Capacitors `A` and `B`, each have capacitance `C=2 F`. The plates of capacitor `A` are shorted using a wire of resistance `R=1 Omega` while the left plate of capacitor `B` is given an initial charge `Q=+4C`. The switch is closed at a time `t=0`. What will be the initial current (in ampere) drawn from the battery immediately after the switch is closed ?

Explanation

The resistance shortng the capacitor plates may be considered in parallel with the capacitor, and at `t=0`, the capacitor behaves as a short circuit.