$A$ network of four $10 \; \mu F$ capacitors is connected to a $500 \; V$ supply,as shown in the figure. Determine:
$(a)$ the equivalent capacitance of the network and
$(b)$ the charge on each capacitor.

(N/A) In the given network,$C_{1}, C_{2}$ and $C_{3}$ are connected in series. The effective capacitance $C^{\prime}$ of these three capacitors is given by:
$\frac{1}{C^{\prime}} = \frac{1}{C_{1}} + \frac{1}{C_{2}} + \frac{1}{C_{3}}$
For $C_{1} = C_{2} = C_{3} = 10 \; \mu F$,$C^{\prime} = (10 / 3) \; \mu F$.
The network has $C^{\prime}$ and $C_{4}$ connected in parallel. Thus,the equivalent capacitance $C$ of the network is:
$C = C^{\prime} + C_{4} = \left(\frac{10}{3} + 10\right) \; \mu F = 13.33 \; \mu F$.
$(b)$ 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 $Q^{\prime}$.
The potential difference across the series branch containing $C_{1}, C_{2}, C_{3}$ is $500 \; V$. Thus:
$\frac{Q}{C_{1}} + \frac{Q}{C_{2}} + \frac{Q}{C_{3}} = 500 \; V$
$Q \left(\frac{1}{10} + \frac{1}{10} + \frac{1}{10}\right) \times 10^{6} = 500$
$Q = 500 \times \frac{10}{3} \; \mu C = 1.67 \times 10^{-3} \; C$.
For $C_{4}$,the potential difference is $500 \; V$:
$Q^{\prime} = C_{4} \times V = 10 \; \mu F \times 500 \; V = 5.0 \times 10^{-3} \; C$.