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(D) Step $1$: Calculate the equivalent capacitance of the two $2\,\mu F$ capacitors connected in parallel.For parallel connection,$C_p = C_1 + C_2 = 2

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(D) Step $1$: Calculate the equivalent capacitance of the two $2\,\mu F$ capacitors connected in parallel.For parallel connection,$C_p = C_1 + C_2 = 2\,\mu F + 2\,\mu F = 4\,\mu F$.Step $2$: Calculate the equivalent capacitance of the system by connecting this parallel combination in series with the $12\,\mu F$ capacitor.For series connection,$\frac{1}{C_{eq}} = \frac{1}{C_p} + \frac{1}{C_3} = \frac{1}{4} + \frac{1}{12} = \frac{3+1}{12} = \frac{4}{12} = \frac{1}{3}$.Therefore,$C_{eq} = 3\,\mu F$.

Two capacitors each of capacity $2\,\mu F$ are connected in parallel. This system is connected in series with a third capacitor of $12\,\mu F$ capacity. The equivalent capacity of the system will be......$\mu F$

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