Fig, shows a network of five capacitors connected to a `100 V` supply. Calculate the total charge and energy stored in the network.

Answer

`4xx10^(-4)C ; 0.02 J`

Explanation

The equivalent circuit diagram of given circuit is shown in Fig.

Two capacitors each of `3 muF` are in parallel.
`:. C_(1) = 3+3 = 6 muF`
Two capacitors of `1 muF` and `2 muF` are in parallel.
`:. C_(2) = 1+2 = 3 muF`
As `C_(1)` and `C_(2)` are in series, therefore
`C_(12) = (6xx3)/(6+3) = 2 mu F`
`C_(12)` is in parallel with `2 muF` capacitor.
`:.` Net capacitance, `C = C_(12) + 2 + 2 = 4 muF`,
Total charge , `q = CV = (4xx10^(-4)) xx 100`
`= 4xx10^(-4) C`
Total energy, `U = (1)/(2) CV^(2)`
`= (1)/(2) (4xx10^(-6)) (100)^(2) = 0.02 J`