A network of four each of `12 mu F` capacitance is connected to a `500 V` apply as shown in Fig.

(a) Equivalent capacitance of the network.
(b) Charge on each capacitor.

Explanation

The equivalent network of the circuit is
(i) The capacitors `C_(1) , C_(2)` and `C_(3)` are in series

`therefore (1)/(C_(s)) = (1)/(C_(1)) + (1)/(C_(2)) + (1)/(C_(3)) implies (1)/(C_(s)) implies (1)/(C_(s)) = (1)/(12) + (1)/(12) + (1)/(12)`
`(1)/(C_(s)) = (3)/(12) implies C_(s) = 4 mu F`
Now `C_(s) ` and `C_(4)` are in parallel combination . Therefore , the equivalent capacitance of the network is
`C = C_(s) + C_(4) = 4 + 12 = 16 mu F `
(i) Total charge of capacitor `" "` q = CV
`q = 16 xx 10^(-6) xx 500 = 8000 xx 10^(-16) = 8000 mu C`
Charge on capacitor `C_(4)` is
`q = C_(4) V = 12 xx 10^(-6) xx 500`
= `6000 xx 10^(-6)C = 6000 mu C`
Charge on each capacitor `C_(1) , C_(2)` and `C_(3)` is
`q = C_(s) V = 4 xx 10^(-6) xx 500 [ because (1)/(C_(s)) = (1)/(C_(1)) + (1)/(C_(2)) + (1)/(C_(3))]`
`= 2000 xx 10^(-6) C = 2000 mu C`