Question

A capacitance of$2\mu F$ is required in an electrical circuit across a potential difference of$1.0kV.$ 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. $2$
• B. $16$
• C. $24$
• D. $32$

Answer 1

The correct option is D

$32$

Step 1. Given data:

Required capacitance, $C_r=2\mu F$ across potential of $V_r=1kV$

The capacitance of a single capacitor is $1\mu F$

Step 2. Finding the number of capacitors:

To get the required capacitance of $2\mu F$ , we can connect the eight $1\mu F$ in parallel

Using $C_p=C_1+C_2+......C_8$

It gives us capacitance of $\left(1+1....8times\right)\mu F=8\mu F$,

Since, the potential of this parallel connection need not be more than $300V$,

Now when this combination of $8\mu F$ is connected in series $4$ times,

Using $\frac{1}{C_s}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}+\frac{1}{C_4}$

We get $C=\frac{8}{4}\mu F=2\mu F$

Since, the potential of this parallel connection need not be more than $300V$, a potential of each parallel combination is $250V$ making a total voltage of $1kV$ across the obtained $2\mu F$

The number of required capacitors are number of capacitors connected in parallel times the number of capacitors connected in series

$\Rightarrow 8\times 4=32$

Hence, option D is correct.