An electrical technician requires a capacitance of 2 µF in a circuit across a potential difference of 1 kV.
Question:

An electrical technician requires a capacitance of 2 µF in a circuit across a potential difference of 1 kV. A large number of 1 µF 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.

Solution:

Total required capacitance, C = 2 µF

Potential difference, V = 1 kV = 1000 V

Capacitance of each capacitor, C1 = 1µF

Each capacitor can withstand a potential difference, V1 = 400 V

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 $=\frac{1}{1+1+1}=\frac{1}{3} \mu \mathrm{F}$

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

$\frac{1}{3}+\frac{1}{3}+\frac{1}{3}+$

$=\frac{n}{3}$

However, capacitance of the ciruit is given as $2 \mu \mathrm{F}$.

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

$n=6$

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

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