A network of four `10 muF` capacitors is connected to a `500V ` Supply , as shown in Fig. Determine (a) the equivalent capacitance of the network and (b) the charge on each capacitor, (Note , the charge on a capacitor is the charge on the plate with higher potential , equal and opposite to the charge on the plate with lower potential. )

Explanation

In the given network, `C_(1), C_(2)` and `C_(3)` are connected in series. The effective capacitive 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) = 10 mu F. C' = (10//3) nu F`. The network has C' and `C_(4)` connected in parallel. Thus, the equivalent capacitance C of the network is
`C = C' + C_(4) = ((10)/(3) + 10) mu F = 13.3 nu F`
(b) Clearly, 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)) = 500 V`
also `Q'//C_(4) = 500 V`. The gives for the given value of the capacitances,
`Q = 500 V xx (10)/(3) mu F = 1.7 xx 10^(-3) C` and
`Q = 500 V xx 10 mu F = 5.0 xx 10^(-3) C`