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

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

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$4$

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(A) For capacitors 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 values: $\frac{1}{C_{eq}} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3+2+1}{6} = \frac{6}{6} = 1\ \mu F$.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 circuit,the charge on each capacitor is the same,so $Q = 24\ \mu C$ for the $6\ \mu F$ capacitor.The potential difference $V'$ across the $6\ \mu F$ capacitor is $V' = \frac{Q}{C} = \frac{24\ \mu C}{6\ \mu F} = 4\ V$.

Three capacitors of $2\ \mu F, 3\ \mu F$,and $6\ \mu F$ are connected in series with a $24\ V$ battery. The potential difference across the $6\ \mu F$ capacitor is .......... $V$.

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