Question 59 of 180

A capacitance of $2\mu F$ is required in an electrical circuit across a potential difference of $1\text{.}0kV$. A large number of $1\mu F$ capacitors are available which can withstand a potential difference of not more than $300V$. The minimum number of capacitors required to achieve this is:

A$32$B$2$C$16$D$24$

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Answer: (A) $32$

Let us assume that we connect $n$(whole number) capacitors in series such that the potential difference across the combination is $1000V$. This means the potential difference across each capacitor is
${V}_{capacitor}=\dfrac{1000}{n}V\leq 300V$
$n⩾\dfrac{1000}{300}=3.333$.
${n}_{min}=4$
This means we need to connect at least $4$ capacitors in series to make sure that the potential across each one is less than $300V$.
${C}_{eq}=\dfrac{m}{n}C=2\mu F$
$\dfrac{m}{4}=2 \Rightarrow m=8$
$∴$ Minimum no. of capacitors $=8\times 4=32$