In Fig, the energy stored in `C_(4)` is `27 J`. Calculate the total energy in the system.

Answer

`594.0 J`

Explanation

`U_(4) = (1)/(2) C_(4) V^(2) = 27 J`
`V^(2) = (2xx27)/(C_(4)) = (54)/(6xx10^(-6)) = 9xx10^(6)` Energy stored in `C_(2)`.
`U_(2) = (1)/(2) xx2xx10^(-6) xx9xx10^(6) = 9J`
energy stored in `C_(3)`,
`U_(3) = (1_/(2) xx3xx10^(-6) xx9xx10^(6) = 13.5 J`
Total energy stored in `C_(2), C_(3), C_(4)`
`= U_(2) + U_(3) + U_(4) = 9 + 13.5 + 27 = 49.5 J`
Equivalent capacitance of `C_(2), C_(3), C_(4)`
`= 2 + 3 + 6 = 11 muF`
As `U = (q^(2))/(2C) = (q^(2))/(2xx11xx10^(-6)) = 49.5`
`:. q^(2) = 49.5xx22xx10^(-6) C^(2)`
`U_(1) = (q^(2))/(2 C_(1)) = (49.5xx22xx10^(-6))/(2xx1xx10^(-6)) = 544.5 J`
Total energy stored in the arrangement
`= 54.5 + 49.5 = 594.0 J`