Three capacitors of 2,mu F,3,mu F,and 6,mu F | vedclass

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Question

(A) The equivalent capacitance $C_{eq}$ for capacitors in series is given by $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$.Substi

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$4$

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(A) The equivalent capacitance $C_{eq}$ for capacitors in series is given by $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$.Substituting the values: $\frac{1}{C_{eq}} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3+2+1}{6} = \frac{6}{6} = 1\,\mu F$.Thus,$C_{eq} = 1\,\mu F$.The total charge $Q$ supplied by the battery is $Q = C_{eq} 	imes V = 1\,\mu F 	imes 24\,V = 24\,\mu C$.In a series combination,the charge on each capacitor is the same.Therefore,the potential difference $V'$ across the $6\,\mu F$ capacitor is $V' = \frac{Q}{C_3} = \frac{24\,\mu C}{6\,\mu F} = 4\,V$.

Three capacitors of $2\,\mu F$,$3\,\mu F$,and $6\,\mu F$ are joined in series,and the combination is charged by means of a $24\,V$ battery. The potential difference across the plates of the $6\,\mu F$ capacitor is.......$V$.

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