Three capacitors of capacitance 3,mu F,10,mu | vedclass

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

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$200$

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(C) When capacitors are connected in series,the equivalent capacitance $C_{eq}$ is given by the formula: $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$.Substituting the given values: $\frac{1}{C_{eq}} = \frac{1}{3} + \frac{1}{10} + \frac{1}{15}$.Finding a common denominator $(30)$: $\frac{1}{C_{eq}} = \frac{10 + 3 + 2}{30} = \frac{15}{30} = \frac{1}{2}$.Therefore,$C_{eq} = 2\,\mu F$.The total charge $Q$ supplied by the source is $Q = C_{eq} 	imes V = 2\,\mu F 	imes 100\,V = 200\,\mu C$.In a series combination,the charge on each capacitor is the same and is equal to the total charge supplied by the source.Thus,the charge on the $15\,\mu F$ capacitor is $200\,\mu C$.

Three capacitors of capacitance $3\,\mu F$,$10\,\mu F$,and $15\,\mu F$ are connected in series to a voltage source of $100\,V$. The charge on $15\,\mu F$ is ....... $\mu C$.

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