Three capacitors of capacitance 1 mu F,2 mu F | vedclass

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(D) In a series combination,the equivalent capacitance $C_{eq}$ is given by $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$.Substit

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$6$

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(D) In a series combination,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: $\frac{1}{C_{eq}} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} = \frac{6+3+2}{6} = \frac{11}{6}$.Thus,$C_{eq} = \frac{6}{11}\ \mu F$.The total charge $Q$ supplied by the battery is $Q = C_{eq} 	imes V = \frac{6}{11}\ \mu F 	imes 11\ V = 6\ \mu C$.In a series circuit,the charge on each capacitor is the same. Therefore,the charge on the $1\ \mu F$ capacitor is $6\ \mu C$.The potential difference $V_1$ across the $1\ \mu F$ capacitor is $V_1 = \frac{Q}{C_1} = \frac{6\ \mu C}{1\ \mu F} = 6\ V$.

Three capacitors of capacitance $1\ \mu F$,$2\ \mu F$,and $3\ \mu F$ are connected in series,and a potential difference of $11\ V$ is applied across the combination. The potential difference across the plates of the $1\ \mu F$ capacitor is......$V$.

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