Question

An electrical technician requires a capacitance of $2\mu F$ in a circuit across a potential difference of 1 kV. A large number of $1\mu F$ capacitors are available to him each of which can withstand a potential difference of not more than 400 V. Suggest a possible arrangement that requires the minimum number of capacitors.

Answer 1

Total required capacitance, $C=2\mu F$

Potential difference, $V=1kV=1000V$

Capacitance of each capacitor, $C_1=1\mu F$

Each capacitor can withstand a potential difference, $V_1=400V$

Suppose a number of capacitors are connected in series and these series circuits are connected in parallel (row) to each other. The potential difference across each row must be 1000 V and potential difference across each capacitor must be 400 V. Hence, number of capacitors in each row is given as

$\frac{1000}{400}=2.5$

Hence, there are three capacitors in each row.

Capacitance of each row $=\frac{1}{1+1+1}=\frac{1}{3}\mu F$

Let there are n rows, each having three capacitors, which are connected in parallel. Hence, equivalent capacitance of the circuit is given as

$\frac{1}{3}+\frac{1}{3}+\frac{1}{3}+.......n$ terms

$=\frac{n}{3}$

However, capacitance of the circuit is given as $2\mu F$.

$\therefore \frac{n}{3}=2$

$n=6$

Hence, 6 rows of three capacitors are present in the circuit. A minimum of $6\times 3$ i.e., 18 capacitors are required for the given arrangement.