Initially, both capacitors are uncharged. A long time after the switch S is closed, the potential energy stored in the capacitor of capacitance `2muF is __________xx10^(-6)J`.

Answer

64

Explanation

At steady, current through any branch that contains a capacitor becomes zero.
So, these branches can be removed while analyzing the circuit.
`implies `P.D. across `2muF` is 8V `implies U=(1)/(2)CV^(2)=(1)/(2)(2)(8)^(2)muJ,=64xx10^(-6)J`.