Find the total energy stored in the capacitors in the given network.

Explanation

The equivaelent capacitance of `C_(1) and C_(2)` in series
`C'=(C_(1)C_(2))/(C_(1)+C_(2))=(2xx2)/(2+2)=1muF`
`C'` is in parallel with `C_(3)`, so equivalent capacitance of
`C_(1),C_(2) and C_(3)` is
`C"=1+1=2muF`
`C'` is in series with `C_(4)`, their equivalent capacitance
`C"'=(C_(4)C')/(C_(4)+C')=(2xx2)/(2+2)=1muF`
this is in parallel with `C_(5)`, so equivalent capacitance across AB is `C_(AB)=1+1=2muF`
Energy stored `U=(1)/(2)C_(AB)V^(2)=(1)/(2)xx2xx10^(-6)xx(6)^(2)=36xx10^(-6)J`