Three capacitors 1,mu F,2,mu F,and 4,mu F are | vedclass

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(B) When capacitors are connected in series,the equivalent capacitance $C_{eq}$ is given by $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{

Vedclass · Accepted Answer

$\frac{40}{7}\,\mu C$

Vedclass · Answer

(B) When capacitors are connected 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: $\frac{1}{C_{eq}} = \frac{1}{1} + \frac{1}{2} + \frac{1}{4} = \frac{4+2+1}{4} = \frac{7}{4}\,\mu F^{-1}$.Thus,$C_{eq} = \frac{4}{7}\,\mu F$.The total charge $Q$ drawn from the source is $Q = C_{eq} 	imes V = \frac{4}{7} 	imes 10 = \frac{40}{7}\,\mu C$.In a series combination,the charge on each capacitor is the same and equal to the total charge supplied by the source.Therefore,the charge on the middle capacitor $(2\,\mu F)$ is $\frac{40}{7}\,\mu C$.

Three capacitors $1\,\mu F$,$2\,\mu F$,and $4\,\mu F$ are connected in series to a $10\,V$ source. The charge on the plates of the middle capacitor is:

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