A network of four capacitors each of `15 mu F` four capacitors each of the `muF` capacitance is connected to a 500 V supply as shown in the figure . Determine (a) equivalent capacitance of the network and (b) charge on each capacitor .

Explanation

(a) 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)) = (1)/(15) + (1)/(15) + (1)/(15)`
`(1)/(C_(s)) = (3)/(15) implies C_(S) = 5 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) = 5 + 15 = 20 mu F`.

(b) Charge on capacitor `C_(4)` is = `C_(4) V = 15 xx 10^(-6) xx 500 = 7500 xx 10^(-6) C = 7500 mu C`
Charge on each capacitor , `C_(1) , C_(2)` and `C_(3)` is q = `C_(s) V = 5 xx 10^(-6) xx 500`
`= 2500 xx 10^(-6) C = 2500 mu C`