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Good evening sir. Answer is 2nd option. Explain it sir.
Total required capacitance, C = 2 µFPotential difference, V = 1 kV = 1000 VCapacitance of each capacitor, C1 = 1 µFEach capacitor can withstand a potential difference, V1 = 300 VSuppose 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 = 3.33Hence, there are 4 capacitors in each row.Capacitance of each row = 1 / 1 + 1 + 1 + 1 = ¼ mFLet there are n rows, each having four capacitors, which are connected in parallel. Hence, equivalent capacitance of the circuit is given as¼ + ¼ + ¼ + ¼ + ….....n times = n/4How ever capacitance of circuit = 2n/4 = 2n = 8Hence, 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.
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.
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