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, *C*_{1} = 1µF

Each capacitor can withstand a potential difference, *V*_{1} = 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

`1000/400=2.5`

Hence, there are three capacitors in each row.

Capacitance of each row `=1/(1+1+1)=1/3mu F`

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 `muF`

`therefore 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.