# A capacitance $2\mu F$ is required in an electrical circuit across a potential difference $1kV$ . 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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**Hint:**It is given that a large number of $1\mu F$ capacitors are available that can withstand a maximum potential $300V$ . We have to arrange these $1\mu F$ capacitors that can handle a maximum potential in series. Then we have to calculate the total number of parallel connections of this series connection of capacitors. And then finally we will calculate the number of capacitors required.

**Complete step by step solution:**

As it is given that a capacitor is required in an electrical circuit of $1kV$ potential, we will divide by the maximum potential that a $1\mu F$ capacitor can withstand, to find the number of capacitors that should be connected in series.

Let $n$ be the number of capacitors that are connected in a single series row, then

$n=\dfrac{1000}{300}$

$\Rightarrow n=3.33$

As the value $n$ is not a whole number, we will estimate the value to the nearest whole number.

Therefore,

$n=4$

Therefore, equivalent capacitance after connecting four capacitors in series is

$\dfrac{1}{{{C}_{s}}}=\dfrac{1}{{{C}_{1}}}+\dfrac{1}{{{C}_{2}}}+\dfrac{1}{{{C}_{3}}}+\dfrac{1}{{{C}_{4}}}$

Substituting the values, we get

$\dfrac{1}{{{C}_{s}}}=4$

$\Rightarrow {{C}_{s}}=\dfrac{1}{4}\mu F$

Now, let $m$ be the number of rows of $n$ capacitors connected in series. Then,

${{C}_{p}}={{C}_{1}}+{{C}_{2}}+{{C}_{3}}+...+{{C}_{m}}$

As the total capacitance of the arrangement should be $2\mu F$ ,

$m\times {{C}_{s}}=2$

By substituting the value of the equivalent capacitance in series connection, we get

$m=4\times 2$

$\Rightarrow m=8$

Therefore, the total number of $1\mu F$ capacitance arranged is given by

$N=n\times m$

By substituting the values of $n$ and $m$ , we get

$N=4\times 8$

$\Rightarrow N=32$

That is, there has to be a total $32$ number of $1\mu F$ capacitors to constitute an equivalent capacitance of $2\mu F$ operated at a potential $1kV$.

**$\therefore $ option (A) is the correct option.**

**Note:**

In this question, the calculation was based on the operating voltage of the capacitors. If the value of the maximum voltage that the capacitor can withstand, or the potential difference across the circuit is changed, then the final value of the number of capacitors required also changes. But the procedure will remain the same.