Question

An electrical technician requires a capacitance of $2\mu F$ in a circuit across the potential difference $1kV$. A large number of $1\mu F$ capacitors are available to him each of which can withstand a potential difference of not more than $300$ volt. How many minimum numbers of capacitors are required to get $2\mu F$ capacitor?

• A. $32$
• B. $18$
• C. $16$
• D. $2$

Answer 1

The correct option is A $32$

Let a number of capacitors connected in series and these series circuits are connected in parallel row to each other.

The potential difference across each row must be 1kV = 1000 V

Potential difference across each capacitor= 300 V

Number of capacitors in each row is thus = 1000/300 = 3.33

So, there are capacitors in each row.

Capacitance of each row = $\frac{1}{1+1+1+1}=\frac{1}{4}\mu F$

Let there are n rows each having four capacitors connected in parallel, having equivalent capacitance = $\frac{n}{4}$

Capacitance of the circuit = 2 $\mu F$

$\therefore \frac{n}{4}=2\Rightarrow n=8$

Hence, 8 rows, each having 4 capacitors is the arrangement.

So, total number of capacitors = 8 * 4 = 32