An electric circuit requires a total capacitance of 2µF across a potential of 1000 V . Large number of 1µF   capacitances are available each of which would breakdown if the potential is more than 350V . Then, the number of capacitors required to make the circuit is __________.

Here, capacitance of each capacitor,$C=1\mu F$Voltage rating of each.................capacitor  $=350\mathrm{V}$
Supply voltage  = 1000 V 
Total capacitance 2µF
Let  capacitors of 1µF
Each be connected in series in a row and  such rows be connected in parallel as shown in the figure
.
$\therefore$each capacitor can withstand  $350V$
$\therefore \mathrm{n}=\frac{1000}{350}=2.8$
As n cannot be fraction, therefore n = 3
Capacitance of each row of 3 capacitors of 1µFeach in series is ${\mathrm{C}}_{\mathrm{s}}=\frac{1}{3}\mathrm{\mu F}$
Total capacitance of m such rows in parallel $=\frac{\mathrm{m}}{3}\mathrm{\mu F}$
$\therefore \frac{\mathrm{m}}{3}\mathrm{\mu F}=2\mathrm{\mu F}\mathrm{or}\text{\hspace{0.17em}}\mathrm{m}=6$
Total number of capacitors $=\mathrm{n}\times \mathrm{m}=3\times 6=18$