Obtain the equivalent capacitance of the network in Fig. For a 300 V supply. Determin the change and voltage across each capacitor.

Explanation

The equivalent circuit for the given figure is :

Equivalent capacitance for `C_(2) and C_(3)` connected in series
`rArr C_(23)=(C_(2)C_(3))/(C_(2)+C_(3))=(200xx200)/(200+200)`
`C_(23)=(40,000)/(400)=100pF`
Equivalent capacitance for `C_(23) and C_(1)` connected in parallel
`rArr C.=C_(23)+C_(1)=100pF+100pF=200pF`
Equivalent capacitance of series combination of `C. and C_(4)`
`rArr C_(eq)=(C_(4)C.)/(C_(4)+C.)=(100xx200)/(100+200)=(200)/(3)`
Let the charges on the capacitors `C_(1),C_(2),C_(3) and C_(4)` be `q_(1), q_(2),q_(3) and q_(4)`, respectively.
Let the potential difference across capacitors `C_(1), C_(2), C_(3) and C_(4)` be `V_(1), V_(2), V_(3) and V_(4)`, respectively.
Let the total charge in the circuit be q
`q=C_(eq)V=((200)/(3)xx10^(-12))xx300=2xx10^(-8)C`
The charge on the capacitor `C_(4)(q_(4))` will also be equal to q. Thus, `q_(4)=q=2xx10^(-8)C`
So, `V_(4)=(q_(4))/(C_(4))=(2xx10^(-8))/(100xx10^(-12))=200V`
Potential difference across B and D = 100 V
`q_(4)=C_(4)V_(4)=(100xx10^(-12))xx200=2xx10^(-8)C`
`q_(1)=C_(1)V_(1)=(100xx10^(-12))xx100=10^(-8)C`
Since, capacitance of `C_(2) and C_(3)` are equal, they will have same potential difference and charge on them.
So, `V_(2)=V_(3)=(100)/(2)=50V`
Also, `q_(2)=q_(3)=C_(2)V_(2)=C_(3)V_(2)=10^(-8)C`