Q. 234.1( 7 Votes )

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.

Answer :

Potential difference across the circuit = 1kV = 1000V

Capacitance of each capacitor = 1 μF

Potential difference each capacitor can withstand = 400V

Capacitance required across the circuit = 2 μF

Assume n number of capacitors are connected in series and further m number of such series circuits are connected in parallel to each other.

As the potential difference in the circuit is 1000V so the potential difference across each row of n capacitors is 1000V, as the potential difference each capacitor can withstand is 400V,

Therefore, 400V × n = 1000V

n = 1000V/400V = 2.5~3capacitors in each row.


Capacitance of each row = μF

As there are m column of three capacitors, which are connected in parallel.

Hence, equivalent capacitance of the circuit is

+ ……….m times =

Since, Required capacitance = 2 μF

m = 6

Total number of capacitors = 3 × 6 = 18.

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.