The charge deposited on the 4,mu F capacitor | vedclass

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(C) First,find the equivalent capacitance of the parallel combination of the $4\,\mu F$ and $2\,\mu F$ capacitors:$C_p = 4\,\mu F + 2\,\mu F = 6\,\mu

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$24	imes10^{-6}\,C$

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(C) First,find the equivalent capacitance of the parallel combination of the $4\,\mu F$ and $2\,\mu F$ capacitors:$C_p = 4\,\mu F + 2\,\mu F = 6\,\mu F$.Now,the circuit consists of this equivalent capacitor $C_p$ in series with the $6\,\mu F$ capacitor and the $12\,V$ battery.The total equivalent capacitance $C_{eq}$ of the circuit is given by:$\frac{1}{C_{eq}} = \frac{1}{C_p} + \frac{1}{6\,\mu F} = \frac{1}{6\,\mu F} + \frac{1}{6\,\mu F} = \frac{2}{6\,\mu F} = \frac{1}{3\,\mu F}$.Thus,$C_{eq} = 3\,\mu F$.The total charge $Q_{total}$ supplied by the battery is:$Q_{total} = C_{eq} 	imes V = 3\,\mu F 	imes 12\,V = 36\,\mu C$.Since the $6\,\mu F$ capacitor and the parallel combination $(C_p)$ are in series,they both carry the same charge $Q_{total} = 36\,\mu C$.The voltage across the parallel combination $C_p$ is:$V_p = \frac{Q_{total}}{C_p} = \frac{36\,\mu C}{6\,\mu F} = 6\,V$.Since the $4\,\mu F$ and $2\,\mu F$ capacitors are in parallel,the voltage across the $4\,\mu F$ capacitor is also $6\,V$.The charge $Q$ on the $4\,\mu F$ capacitor is:$Q = C 	imes V = 4\,\mu F 	imes 6\,V = 24\,\mu C = 24 	imes 10^{-6}\,C$.

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

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