A network of four `10muF` capacitors is connected to a 500 V supply, as shown in figure. Determine (a) the equivalent capacitance of the network and (b) the charge on each capcitor.

Explanation

(a) In the given network, `C_(1),C_(2) and C_(3)` are connected in series. The effective capacitance C. of these three capacitors is given by `1/(C.)=1/C_(1)+1/C_(2)+1/C_(3)`
For, `C_(1)=C_(2)=C_(3)=10muF, C.=(10/3)muF`. The network has C. and `C_(n)` connected in parallel. Thus, the equivalent capacitance C of the network is
`C=C.+C_(4)=(10/3+10)muF=13.3muF`
(b) From the figure, the charge on each of the capacitors, `C_(1), C_(2) and C_(3)` is the same, say Q. Let the charge on `C_(4)` be Q.. Now, since the potential difference across AB is `Q/C_(1)` across
BC is `Q/C_(2)" across CD is "Q/C_(3)" we have "Q/C_(1)+Q/C_(2)+Q/C_(3)=500V`
Also, `(Q.)/C_(4)=500V`
This given for the given value of the capacitances,
`Q=500 xx 10/3 xx 10^(-6) =1.7 xx 10^(-3)C and`
`Q.=500 xx 10 xx 10^(-6) =5.0 xx 10^(-3)C`