A capacitance of $2\ μF$ is required in an electrical circuit across a potential difference of $1.0\ kV$. A large number of $1\ μF$ capacitors are available which can withstand a potential difference of not more than $300\ V$. The minimum number of capacitors required to achieve this is
JEE MAIN 2017, Medium
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To get a capacitance of $2\, \mu \mathrm{F}$ arrangement of capacitors of capacitance $1\, \mu \mathrm{F}$ as shown in figure $8$ capacitors of $1\, \mu \mathrm{F}$ in parallel with four such branches in series i.e., $32$ such capacitors are required.
$\frac{1}{C_{e q}}=\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8} \quad \therefore C_{e q}=2\, \mu F$
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