The charge deposited on the 4 mu F capacitor | vedclass

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(C) The capacitors $4 \mu F$ and $2 \mu F$ are connected in parallel. Their equivalent capacitance is $C_p = 4 \mu F + 2 \mu F = 6 \mu F$.This paralle

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$24 	imes 10^{-6} C$

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(C) The capacitors $4 \mu F$ and $2 \mu F$ are connected in parallel. Their equivalent capacitance is $C_p = 4 \mu F + 2 \mu F = 6 \mu F$.This parallel combination is in series with the $6 \mu F$ capacitor. The total equivalent capacitance $C_{eq}$ is given by:$\frac{1}{C_{eq}} = \frac{1}{6 \mu F} + \frac{1}{6 \mu F} = \frac{2}{6 \mu F} = \frac{1}{3 \mu F} \Rightarrow C_{eq} = 3 \mu F$.The total charge $Q$ supplied by the $12 	ext{ V}$ source is:$Q = C_{eq} 	imes V = 3 \mu F 	imes 12 	ext{ V} = 36 \mu C$.This total charge $Q$ flows through the $6 \mu F$ capacitor and then divides between the $4 \mu F$ and $2 \mu F$ capacitors connected in parallel.Let $Q_1$ be the charge on the $4 \mu F$ capacitor and $Q_2$ be the charge on the $2 \mu F$ capacitor.Since they are in parallel,the potential difference across them is the same:$V_p = \frac{Q_1}{C_1} = \frac{Q_2}{C_2} \Rightarrow \frac{Q_1}{4 \mu F} = \frac{Q_2}{2 \mu F} \Rightarrow Q_1 = 2 Q_2$.Also,$Q_1 + Q_2 = Q = 36 \mu C$.Substituting $Q_1 = 2 Q_2$ into the equation:$2 Q_2 + Q_2 = 36 \mu C \Rightarrow 3 Q_2 = 36 \mu C \Rightarrow Q_2 = 12 \mu C$.Then,$Q_1 = 2 	imes 12 \mu C = 24 \mu C = 24 	imes 10^{-6} C$.

The charge deposited on the $4 \mu F$ capacitor in the circuit is

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