Three capacitors each of 4 μ F are to be connected in such a way that the effective capacitance is 6 μ F. This can be done by connecting them: (a) all in series (b) two in series and one in parallel (c) all in parallel (d) two in parallel and one in series

Answer: (b). Two 4 in series: (4×4)/(8) = 2. Add the third in parallel: 2 + 4 = 6. So the answer is (b) two in series and one in parallel.

Three capacitors each of 4 μ F are to be connected in such a way that… | JEE Main 2016 (9 Apr, Shift 1)

Q83JEE Main 2016 · 9 APR · SHIFT 1Easy

Combination of Capacitors

Three capacitors each of 4 $\mu$ F are to be connected in such a way that the effective capacitance is 6 $\mu$ F. This can be done by connecting them:

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Answer: (b)

Two $4\ \mu\text{F}$ in series: $\frac{4\times4}{8} = 2\ \mu\text{F}$ .
Add the third in parallel: $2 + 4 = 6\ \mu\text{F}$ .
So the answer is (b) two in series and one in parallel.

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Combination of Capacitors question from JEE Main 2016 (9 Apr, Shift 1)

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