Minimum number of capacitors that must be connected in series in a row are n = $\frac{100\mathrm{V}}{400\mathrm{V}}$ = 2.5 ~ 3 capacitance of 3 capacitors in series in a row is C' = $\frac{1}{3}$ µF Minimum number of rows of 3 capacitors each to be connected in parallel to obtain net capacitance of 2 µF are m = $\frac{2\mu\mathrm{F}}{\frac{1}{3}\mu\mathrm{F}}$ = 6 So minimum number of capacitors required are m × n = 6 × 3 = 18