Three capacitors of capacitances 10 mu F,5 mu | vedclass

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

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$40$

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(B) When capacitors are connected in series,the equivalent capacitance $C_{	ext{eq}}$ is given by the formula: $\frac{1}{C_{	ext{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$.Substituting the given values: $\frac{1}{C_{	ext{eq}}} = \frac{1}{10} + \frac{1}{5} + \frac{1}{20} = \frac{2 + 4 + 1}{20} = \frac{7}{20} \mu F^{-1}$.Therefore,$C_{	ext{eq}} = \frac{20}{7} \mu F$.The total charge $Q$ supplied by the $14 	ext{ V}$ $DC$ source is $Q = C_{	ext{eq}} 	imes V = \frac{20}{7} \mu F 	imes 14 	ext{ V} = 40 \mu C$.In a series circuit,the charge on each capacitor is the same and equal to the total charge supplied by the source.Thus,the charge on the $5 \mu F$ capacitor is $40 \mu C$.

Three capacitors of capacitances $10 \mu F$,$5 \mu F$,and $20 \mu F$ are connected in series with a $14 \text{ V}$ $DC$ supply. The charge on the $5 \mu F$ capacitor is: (in $\mu C$)

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