Q. 234.0( 12 Votes )
An electrical tec
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.
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