Question

The minimum number of capacitors, each of $8\mu F$ and $250V$, required to make a composite capacitor of $16\mu F$ and $1000V$ is

Answer 1

Answer: 32

Minimum number of capacitors in each row
$=\frac{1000}{250}=4$
Therefore, $4$ capacitors are connected in series.
If there are $m$ such rows, then total capacity
$=m\times 2=16$
$\Rightarrow m=\frac{16}{2}=8$
$\therefore$ Minimum number of capacitors
$=4\times 8=32$