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.
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.