Obtain the equivalent capacitance of the netwrok in the figure given below. For a 300V supply, determine the charge and voltage across each capacitor.

Explanation

`C_(S)=(C_(2)C_(3))/(C_(2)+C_(3))` for a series combination of `C_(2) and C_(3)`.
i.e., `C_(S)=(200pF)/(2)=100pF`
Capacitors `C_(1) and C_(S)` are parallel to each other.
`C_(P)=C_(1)+C_(S)=100pF+100pF=200pF`
the equivalent circuit is written as,

The equivalent capacitance of the circuit
`C_(S)=(200xx100)/(200+100)pF=(200xx1cancel(00))/(3cancel(00))pF`
`therefore C_(S)=(200)/(3)pF`
Total charge, `Q=C_(S)'V=(200)/(3)xx10^(-12)xx300`
i.e., `Q=2xx10^(-8)C`
The charges on `C_(p) and C_(4)` are the same
i.e., `Q_(p)=Q_(4)=2xx10^(-8)C`
The charge `2xx10^(-8)C` branches out across the `C_(1),C_(2) and C_(3)` combination
`therefore Q_(1)=Q_(2)=Q_(3)=1xx10^(-8)C and Q_(4)=2xx10^(-8)C`
Voltage drop in `C_(1)(=100pF)=(Q_(1))/(C_(1))=(1xx10^(-8))/(100xx10^(-12))=100V`
Voltage drop in 200 pF `(C_(2)&C_(3))=(1xx10^(-8))/(200xx10^(-12))V=50V`
Voltage drop in `C_(4)=(2xx10^(_8))/(100xx10^(-12))=200V`.