Three capacitors of 2.0;mu F,3.0;mu F,and 6.0 | vedclass

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(B) When capacitors are connected in series,the equivalent capacitance $C_{eq}$ is given by $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{

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$10$

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(B) When capacitors are connected in series,the equivalent capacitance $C_{eq}$ is given by $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$.Substituting the given values: $\frac{1}{C_{eq}} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3+2+1}{6} = \frac{6}{6} = 1\,\mu F^{-1}$.Thus,$C_{eq} = 1\,\mu F$.The total charge $Q$ supplied by the source is $Q = C_{eq} 	imes V = 1\,\mu F 	imes 10\,V = 10\,\mu C$.In a series combination,the charge on each capacitor is the same and equal to the total charge supplied by the source.Therefore,the charge on the $3.0\;\mu F$ capacitor is $10\,\mu C$.

Three capacitors of $2.0\;\mu F$,$3.0\;\mu F$,and $6.0\;\mu F$ are connected in series to a $10\,V$ source. The charge on the $3.0\;\mu F$ capacitor is ........ $\mu C$.

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