Question

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. $24$

Answer 1

Answer: (A)

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
$V_{capacitor}=\frac{1000}{n}V\le 300V$
$n\ge \frac{1000}{300}=3.333$ .
$n_{min}=4$
This means we need to connect at least $4$ capacitors in series to make sure that the potential across each one is less than $300V$ .
$C_{eq}=\frac{m}{n}C=2\mu F$
$\frac{m}{4}=2\Rightarrow m=8$
$\therefore$ Minimum no. of capacitors $=8\times 4=32$