A network of four capacitors of 6 `muF` each is connected to a 240 V supply. Determine the charge on each capacitor.

Explanation

The capacitors `C_(1),C_(2)` and `C_(3)` are connected in series. The equivalent capacitance C' is given by
`(1)/(C')=(1)/(6)+(1)/(6)+(1)/(6)=(3)/(6)=(1)/(2)`
`implies" "C'=2muF`
Now, Total Capacitance,
`C=6+2=8 muF`
In series the charge remains same. Therefore, the charge on `C_(1),C_(2)` and `C_(3)` is given by
`q="C' V"`
`implies" "=2xx10^(-6)xx240`
`implies" "q=4.8xx10^(-4)"coulomb"`
The charge on capacitance `C_(4)` is given by
`q_(4)=C_(4)V`
`=6xx10^(-6)xx240`
`=1.44xx10^(-3)"coulomb".`