Question

A capacitance of $2\mu F$ is required in an electrical circuit across a potential difference of $1kV$ . 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) $24$

Answer 1

Hint:We are given that we have to place $2\mu F$ across a potential difference of $1kV$. Also, we are given that a number of $1\mu F$ capacitors can withstand a maximum potential difference of $300V$. Thus, we will place the number of capacitors which can handle the maximum potential in series. Then we will try to configure and calculate the number of such series connection rows in parallel and then finally we will calculate the total number of capacitors needed.
Formulae Used:
$\dfrac{1}{{{C_S}}} = \dfrac{1}{{{C_1}}} + \dfrac{1}{{{C_2}}} + \cdot \cdot \cdot + \dfrac{1}{{{C_n}}}$
Where, ${C_s}$ is the net series capacitance and ${C_1},{C_2},...,{C_n}$ are the individual capacitors.
${C_P} = {C_1} + {C_2} + ... + {C_n}$
Where, ${C_P}$ is the net parallel capacitance and ${C_1},{C_2},...,{C_n}$ are the individual capacitors.

Step By Step Solution
Here,
Given,
Potential Difference to withstand, ${V_{ws}} = 300V$
Net Potential Difference to go across, ${V_{net}} = 1kV = {10^3}V$
Now,
We will calculate the number of capacitors in each row of series connection.
$n = \dfrac{{{V_{net}}}}{{{V_{ws}}}} = \dfrac{{{{10}^3}}}{{300}} = 3.3$
But $3.3$ is not a whole number, so we cannot place these number of capacitors in series
Thus, we will take the smallest whole number greater than this calculated number. Thus, we will take $n = 4$ .
Now,
Calculating the net series capacitance after placing $n$ number of $1\mu F$ capacitors in series,
$\dfrac{1}{{{C_S}}} = \dfrac{1}{1} + \dfrac{1}{1} + \dfrac{1}{1} + \dfrac{1}{1}$
After calculation, we get
${C_S} = \dfrac{1}{4}\mu F$
Now,
We are told that the net capacitance of the circuit should be $2\mu F$.
Let us place $m$ number of series rows with ${C_S} = \dfrac{1}{4}\mu F$ ,
${C_P} = {C_S} + {C_S} + ...(m - terms)$
Thus, we get
${C_P} = m \times {C_S}$
Putting the values ${C_P} = 2\mu F$ and ${C_S} = \dfrac{1}{4}\mu F$, we get
$m = 8$
Thus, total number of capacitors needed is the number of capacitors in each series multiplied by the number of rows,
Thus,
$N = n \times m = 4 \times 8 = 32$

Hence, the option is (A).

Note: In this case, the net potential difference across is $1kV$ and the withstand potential difference is $300V$ and we got the number of capacitors needed is $32$ . But if anyone or both of the values is changed, the final number of capacitors needed also changes. But, the calculations remain the same.