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

raju nagula , 6 Years ago

Grade

1 Answers

Eshan

Last Activity: 6 Years ago

Dear student,

To obtain an output voltage as $V_{net}$ , 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= $n_Sn_p=n_s^2\dfrac{C_{net}}{C}=\dfrac{C_{net}}{C}\dfrac{V_{net}^2}{V^2}$

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