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Good evening sir... Answer is 1 St option. Explain it.

2 years ago

Answers : (1)

Eshan
askIITians Faculty
2095 Points
							Dear student,

To obtain an output voltage asV_{net}, number of capacitors to be arranged in series=n_s=\dfrac{V_{net}}{V}

But the capacitance of such a series combination would be\dfrac{C}{n_s}.

Therefore number of such branches of series to be arranged in parallel to obtain the capacitance=n_p=\dfrac{C_{net}}{C/n_s}=n_s\dfrac{C_{net}}{C}
Therefore total number of such capacitors required=n_Sn_p=n_s^2\dfrac{C_{net}}{C}=\dfrac{C_{net}}{C}\dfrac{V_{net}^2}{V^2}
2 years ago
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