Minimum number of capacitors each of 8 μF and 250 V used to make a composite capacitor of 16 μF and 1000 V are

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(b) 32 Solution: (b) Minimum number of capacitors in each row = $\frac{1000}{250}$=4 Therefore, 4 capacitors connected in series. If there are m such rows, then total capacity = m x 2 = 16 $\therefore$ m = $\frac{16}{2}$ = 8 $\therefore$ minimum number of capacitors = 4 x 8 = 32

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