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.


Now,


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.


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