Three identical capacitors `C_(1) , C_(2)` and `C_(3)` of capacitance `6 mu F` each are connected to a `12 V` battery as shown in Fig. Find
(i) charge on each capcitor
(ii) equivalent capacitance of the network.
(iii) energy stored in the network of capacitors.

Explanation

Let the total charge be Q.
Charge across `6muF=(Q)/(3)`
Charges across `12muF=(2Q)/(3)`
Charges across `3muF=Q`
(i) Given, energy stored in `6muF` is E,
Energy stored in `12muF` will be `(Q^(2))/(2C)=(((2Q)/(3))^(2))/(2xx12)`
`=(4Q^(2))/(9xx2xx12)=2E`
(ii) Energy stored in `3muF` is `(Q^(2))/(2xx3)=(18xxQ^(2))/(18xx2xx3)=18E`
(iii) Total energy stored `=E+2E+18E=21E`
Total energy dissipated = 21E
Total energy drawn from battery `=21E+21E=42E`.