Question

A capacitance of $2 \,\mu F$ is required in an electrical circuit across a potential of $1.0 \,kV$. A large number of $1\,\mu F$ capacitors are available which can withstand a potential difference not more than $300\, V$. 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, $V=1.0 \,kV =1000\, V$
According to required the capacitance, we draw the following circuit,

Given, capacitance of each capacitor, $C_{1}=1\, \mu F$,
potential difference, $V_{1}=300\, V$
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^{'}=\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 $