Three capacitors,each of 3,mu F,are provided. | vedclass

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(D) We are given three capacitors,each with a capacitance of $C = 3\,\mu F$.Possible combinations are:$1$. All three in series: $\frac{1}{C_{eq}} = \f

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(D) We are given three capacitors,each with a capacitance of $C = 3\,\mu F$.Possible combinations are:$1$. All three in series: $\frac{1}{C_{eq}} = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = 1 \implies C_{eq} = 1\,\mu F$.$2$. All three in parallel: $C_{eq} = 3 + 3 + 3 = 9\,\mu F$.$3$. Two in parallel and one in series: $C_p = 3 + 3 = 6\,\mu F$. Then,$\frac{1}{C_{eq}} = \frac{1}{6} + \frac{1}{3} = \frac{1+2}{6} = \frac{3}{6} = \frac{1}{2} \implies C_{eq} = 2\,\mu F$.$4$. Two in series and one in parallel: $C_s = \frac{3 	imes 3}{3 + 3} = 1.5\,\mu F$. Then,$C_{eq} = 1.5 + 3 = 4.5\,\mu F$.Comparing these results with the given options,$6\,\mu F$ cannot be obtained.

Three capacitors,each of $3\,\mu F$,are provided. These cannot be combined to provide the resultant capacitance of.........$\mu F$

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