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(C) 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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(C) 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: $\frac{1}{C_{eq}} = \frac{1}{1} + \frac{1}{2} + \frac{1}{8} = \frac{8+4+1}{8} = \frac{13}{8}$.Thus,$C_{eq} = \frac{8}{13}\,\mu F$.The total charge $Q$ supplied by the battery is $Q = C_{eq} 	imes V = \frac{8}{13} 	imes 13 = 8\,\mu C$.In a series circuit,the charge on each capacitor is the same and equal to the total charge $Q$.Therefore,the potential difference $V_2$ across the $2\,\mu F$ capacitor is $V_2 = \frac{Q}{C_2} = \frac{8\,\mu C}{2\,\mu F} = 4\,V$.

$A$ series combination of three capacitors of capacities $1\,\mu F$,$2\,\mu F$,and $8\,\mu F$ is connected to a battery of $e.m.f.$ $13\,V$. The potential difference across the plates of the $2\,\mu F$ capacitor will be.....$V$

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