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  4. The minimum number of capacitors, each of 8 μ F and 250 V, required to make a composite capacitor of 16 μ F and 1000 V is

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

Electrostatic Potential and Capacitance

Solution:

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
$=4 \times 8=32$