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

Eshan
4 years ago
Dear student,

To obtain an output voltage as$\dpi{80} V_{net}$, number of capacitors to be arranged in series=$\dpi{80} n_s=\dfrac{V_{net}}{V}$

But the capacitance of such a series combination would be$\dpi{80} \dfrac{C}{n_s}$.

Therefore number of such branches of series to be arranged in parallel to obtain the capacitance=$\dpi{80} n_p=\dfrac{C_{net}}{C/n_s}=n_s\dfrac{C_{net}}{C}$
Therefore total number of such capacitors required=$\dpi{80} n_Sn_p=n_s^2\dfrac{C_{net}}{C}=\dfrac{C_{net}}{C}\dfrac{V_{net}^2}{V^2}$