- A$6\mu\text{F},\ 18\mu\text{F}$
- B$3\mu\text{F},\ 12\mu\text{F}$
- C$2\mu\text{F},\ 12\mu\text{F}$
- D$2\mu\text{F},\ 18\mu\text{F}$
Explanation:
The minimum capacitance can be obtained by connecting all capacitors in series. It can be calculated as follows:
$\frac{1}{\text{C}}=\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=\frac{1}{2}$
$\Rightarrow\text{C}=2\mu\text{F}$
The maximum capacitance can be obtained by connecting all capacitors in parallel. It can be calculated as follows:
$\text{C}=6+6+6=18\mu\text{F}$
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