Three capacitors each of 4,mu F are to be con | vedclass

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(D) To get an effective capacitance of $6\,\mu F$,two capacitors of $4\,\mu F$ each are connected in series,and one capacitor of $4\,\mu F$ is connect

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two in series and one in parallel

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(D) To get an effective capacitance of $6\,\mu F$,two capacitors of $4\,\mu F$ each are connected in series,and one capacitor of $4\,\mu F$ is connected in parallel with them.First,calculate the equivalent capacitance of the two capacitors in series:$\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2}$$	herefore C_s = 2\,\mu F$Now,connect this equivalent capacitance in parallel with the third capacitor $(C_3 = 4\,\mu F)$:$C_{eq} = C_s + C_3 = 2\,\mu F + 4\,\mu F = 6\,\mu F$Thus,the correct configuration is two capacitors in series and one in parallel.

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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