A network of four 10 µF capacitors is connected to a 500 V supply, as shown in Fig. 2.29. 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

(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) = 10 muF, C. = (10//3) muF`.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) muF = 13.3 muF`
(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`
This gives for the given value of the capacitances.
`Q = 500 V xx 10/3 muF = 1.7 xx 10^(-3)` C and `Q. = 500 V xx 10 muF = 5.0 xx 10^(-3)` C