Question

A capacitance of $2\mu F$ is required in an electrical circuit across a potential of $1.0kV$. A large number of $1\mu F$ capacitors are available which can withstand a potential difference not more than $300V$. The minimum number of capacitors required to achieve this is

• A. 24
• B. 32
• C. 8
• D. 16

Answer 1

Answer: (B)

Here, the required capacitance $=2\mu F$,
and potential difference, $V=1.0kV=1000V$
According to required the capacitance, we draw the following circuit,

Given, capacitance of each capacitor, $C_1=1\mu F$,
potential difference, $V_1=300V$
Now, from above the circuit potential difference across each capacitor,
$V_1=\frac{1000}{n}=300$
$\Rightarrow n=3.33$
So, for maximum bearable capacitance, we take $n=4$.
Hence, capacitance of each row in which all $4$ capacitors are in series,
$\frac{1}{C}=\frac{1}{1}+\frac{1}{1}+\frac{1}{1}+\frac{1}{1}=\frac{4}{1}$
$\Rightarrow C^{{}^{\prime }}=\frac{1}{4}\mu F$
The total capacitance of $m$ rows,
$m\times \frac{1}{4}=\frac{m}{4}\mu F$
Since, it is given that the total capacitance of circuit is $2\mu F$.
So, $2\mu F=\frac{m}{4}\mu F$
$\Rightarrow m=8$
Thus, the minimum number of capacitor,
$=m\times n=8\times 4=32$