Question

Several 10 pF capacitors are given, each capable of withstanding 100 V. How would you construct a unit possessing a capacitance of 20 pF and capable of withstanding 300 V?

Answer 1

For standing $300V$ three capacitors should be in series in each parallel branch.

$C_{branch}=\frac{C}{3}=\frac{10}{3}pF$

For a total of $20pF$ of capacitance, $n$ branches should be connected in parallel.

Total capacitance = $n\frac{10}{3}=20\Rightarrow n=6$

6 rows of branches having 3 capacitors in series is required.