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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(a) 10–7 K |
(b) 107 K |
(c) 10–13 K |
(d) 5.86 × 107 K |
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(a) |
(b) |
|
(c) |
(d) |
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(a) 6.56 × 10–8 eV |
(b) 2 × 10–8 eV |
(c) 10–8 eV |
(d) 8 × 10–8 eV |
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(a) AB |
(b) BC |
(c) CD |
(d) DE |
