Question
Three capacitors of capacitances $6\mu\text{F}$ each are available. The minimum and maximum capacitances, which may be obtained are:
✓
Answer
- $2\mu\text{F},\ 18\mu\text{F}$
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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