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.

Total required capacitance, C = 2 pF Potential difference, V = 1, kV = 1000 V

Capacitance of each capacitor, C₁ = lμF

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 1000/4 = 2.5

Hence, there are three capacitors in each row.

Capacitance of each row = 1/(1+1+1) = 1/3 μF

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

1/3 +1/3 + 1/3 +……..n terms = n/3

However, capacitance of the circuit is given as 2 μF.

Therefore n/3 =2 ,therefore n=6

Hence, 6 rows of three capacitors are present in the circuit. A minimum of 6 x3 i.e.,18 capacitors are required for the given arrangement.