If three capacitors each of capacity 1,mu F a | vedclass

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(D) The equivalent capacitance of two capacitors of $1\,\mu F$ connected in series is given by $\frac{1}{C_s} = \frac{1}{1} + \frac{1}{1} = 2$,so $C_s

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Two of them are in series and then connected in parallel to the third

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(D) The equivalent capacitance of two capacitors of $1\,\mu F$ connected in series is given by $\frac{1}{C_s} = \frac{1}{1} + \frac{1}{1} = 2$,so $C_s = 0.5\,\mu F$.When this combination is connected in parallel with the third capacitor of $1\,\mu F$,the total equivalent capacitance is $C_{eq} = C_s + 1\,\mu F = 0.5\,\mu F + 1\,\mu F = 1.5\,\mu F$.Therefore,the correct configuration is that two capacitors are in series and then connected in parallel to the third.

If three capacitors each of capacity $1\,\mu F$ are connected in such a way that the resultant capacity is $1.5\,\mu F$,then

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