Question

Three capacitors of capacitances 6 µF each are available. The minimum and maximum capacitances, which may be obtained are
(a) 6 µF, 18 µF
(b) 3 µF, 12 µF
(c) 2 µF, 12 µF
(d) 2 µF, 18 µF.

Answer 1

(d) 2 µF, 18 µF

The minimum capacitance can be obtained by connecting all capacitors in series. It can be calculated as follows:
$\frac{1}{C}=\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=\frac{1}{2}$
$\Rightarrow$C = 2 µF

The maximum capacitance can be obtained by connecting all capacitors in parallel. It can be calculated as follows:
C = 6 + 6 + 6 = 18 µF