Two capacitors of capacities 2 mu F and 4 mu | vedclass

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(A) Let $C_1 = 2 \mu F$,$C_2 = 4 \mu F$,and $C_3 = 6 \mu F$.$C_1$ and $C_2$ are in parallel,so their equivalent capacitance is $C_p = C_1 + C_2 = 2 +

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$12$

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(A) Let $C_1 = 2 \mu F$,$C_2 = 4 \mu F$,and $C_3 = 6 \mu F$.$C_1$ and $C_2$ are in parallel,so their equivalent capacitance is $C_p = C_1 + C_2 = 2 + 4 = 6 \mu F$.Now,$C_p$ and $C_3$ are in series. The equivalent capacitance of the whole circuit is $C_{eq} = \frac{C_p 	imes C_3}{C_p + C_3} = \frac{6 	imes 6}{6 + 6} = 3 \mu F$.The total charge supplied by the battery is $Q = C_{eq} 	imes V = 3 \mu F 	imes 12 	ext{ V} = 36 \mu C$.Since $C_p$ and $C_3$ are in series,the charge on the parallel combination $(C_p)$ is also $36 \mu C$.This charge $Q$ is divided between $C_1$ and $C_2$ in proportion to their capacitances:$Q_1 = Q 	imes \left( \frac{C_1}{C_1 + C_2} ight) = 36 \mu C 	imes \left( \frac{2}{2 + 4} ight) = 36 	imes \frac{2}{6} = 12 \mu C$.Thus,the charge on the $2 \mu F$ capacitor is $12 \mu C$.

Two capacitors of capacities $2 \mu F$ and $4 \mu F$ are connected in parallel. $A$ third capacitor of $6 \mu F$ capacity is connected in series with this combination. $A$ battery of $12 \text{ V}$ is connected across this combination. The charge on the $2 \mu F$ capacitor is: (in $\mu C$)

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