Obtain equivalent capacitance of the following network, Fig. For a 300V supply determine the charge and voltage across each capacitor.

Explanation

Capacitance of capacitor `C_(1)` is 100 pF.
Capacitance of capacitor `C_(2)` is 200 pF
Capacitance of capacitor `C_(3)` is 200 pF
Capacitance of capacitor `C_(4)` is 100 pF
Supply potential, V=300 V
Capacitor `C_(2)` and `C_(3)` are connected in series. let their equivalent capacitance be C''.
`:. 1/(C')=1/200+1/200=2/200`
C'=100 pF
Capacitors `C_(1)` and C' are in parallel. let their equivalent capacitances be C''
`:. C''=C'+C_(1)`
`=100+100=200 pF`
`C''` and `C_(4)` are connected in series . let their equivalent capacitance be C.
`:. 1/C=1/(C'')+1/(C_(4))`
`=1/200+1/100=(2+1)/200`
`C=200/3 pF`
Hence, the equivalent capacitance of the circuit is `200/3 pf`
potential difference across C''=V''
potential difference across `C_(4)=V_(4)`
`:. V''+V_(4)=V=300V`
Charge on `C_(4)` is given by
`Q_(4)=CV`
`=200/3xx10^(-12) xx300`
`=2xx10^(-8)C`
`:. V_(4)=(Q_(4))/(C_(4))`
`=(2xx10^(-8))/(100xx10^(-12)) =200 V`
`:.` Voltage across `C_(1)` is given by
`V_(1)=V-V_(4)`
=300-200=100 V
Hence , potential difference `V_(1)` across `C_(1)` is 100 V
charge on `C_(1)` is given by
`Q_(1)=C_(1)V_(1)`
`=100xx10^(-12)xx100`
`=10^(-8)C`
`C_(2)` and `C_(3)` having same capacitances have a potential difference of 100 V together. since `C_(2)` adn `C_(3)` are in series, the potential difference across `C_(2)` adn `C_(3)` given by
`V_(2)=V_(3)=50 V`
Therefore, charge on `C_(4)` is given by
`Q_(2)=C_(2)V_(2)`
`=200xx10^(-12) xx50 `
`=10^(-8)C`
And charge on `C_(3)` is given by
`Q_(3)=C_(3)V_(3)`
`=200xx10^(-12)xx50`
`=10^(-3)C`
Hence, the equivalent capacitance of the given circuit is `200/3 pF` with .