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

(i) Here, `V = 12V`
and `C_(1) = C_(2) = C_(3) = 6mu F = 6 xx 10^(-6)F`
Charge on capacitor `C_(3)` is
`q_(3) = C_(3)V = 6 xx 10^(-6) xx 12 = 72 xx 10^(-6) = 72 mu C`
Since, capacitor `C_(1) and C_(3)` are in series
`:.` Equivalent capacitance
`(1)/(C_(S)) = (1)/(C_(1)) + (1)/(C_(2)) = (1)/(C_(S)) = (1)/(6) + (1)/(6) = (2)/(6) = (1)/(3)`
`:. C_(S) = 3 mu F`
Charge on capacitor `C_(1) and C_(2)` is
`q = C_(S) V = 3 xx 10^(-6) xx 12`
`= 36 xx 10^(-6) = 36 mu C`
`:.` Charge on each capacitor `C_(1) and C_(2) " is " 36 mu C`
(ii) Since, `C_(1) and C_(2)` is in series
`:.` Equivalent capacitance `C_(S) = 3mu F`
Now `C_(3) and C_(S)` are in parallel
`:.` Equivalent capacitance `C = C_(3) + C_(S)`
`= 6 + 3 = 9mu F`
(iii) Energy stored
`= (1)/(2) CV^(2) = (1)/(2) xx 9 xx 10^(-6) xx (12)^(2) = (1)/(2) xx 9 xx 10^(-6) xx 144`
`= 648 xx 10^(-6) = 6.48 xx 10^(-4) J`