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# Good evening sir.Answer is 2nd option.Explain it sir.

Arun
25763 Points
3 years ago

Total required capacitance, C = 2 µF

Potential difference, V = 1 kV = 1000 V

Capacitance of each capacitor, C1 = 1 µF

Each capacitor can withstand a potential difference, V1 = 300 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 300 V. Hence, number of capacitors in each row is given as

1000/300 = 10/3 = 3.33

Hence, there are 4 capacitors in each row.

Capacitance of each row   = 1 / 1 + 1 + 1 + 1 = ¼ mF

Let there are n rows, each having four capacitors, which are connected in parallel. Hence, equivalent capacitance of the circuit is given as

¼  + ¼ + ¼ + ¼ + ….....n times

= n/4

How ever capacitance of circuit = 2

n/4 = 2

n = 8

Hence, 8 rows of 4 capacitors are present in the circuit. A minimum of 8*4 i.e., 32 capacitors are required for the given arrangement.