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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.

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Solution

(d) 2 µF, 18 µF

The minimum capacitance can be obtained by connecting all capacitors in series. It can be calculated as follows:
1C = 16+16+16=12
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


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