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Three capacitors each of capacity 4µF are to be connected in such a way that the effective capacitance is 6 µF. This can be done by

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connecting two in parallel and one in series

connecting two in series and one in parallel

connecting all of them in series

connecting them in parallel

answer is D.

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Detailed Solution

For $C^{'} = \frac{C_{1} \times C_{2}}{C_{1} + C_{2}} = \frac{4 \times 4}{4 + 4} = 2 μ F$

For $C_{e q} = C^{'} + C_{3} = 2 + 4 = 6 μ F$

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