Given C_1 = 2 mu F,C_2 = 6 mu F,and C_3 = 4 m | vedclass

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(B) From the circuit diagram,$C_1$ and $C_2$ are in series,and their combination is in parallel with $C_3$.First,calculate the equivalent capacitance

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$11 	imes 10^{-6}\ J$

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(B) From the circuit diagram,$C_1$ and $C_2$ are in series,and their combination is in parallel with $C_3$.First,calculate the equivalent capacitance of $C_1$ and $C_2$ in series:$C_{12} = \frac{C_1 C_2}{C_1 + C_2} = \frac{2 	imes 6}{2 + 6} = \frac{12}{8} = 1.5\ \mu F$.Now,$C_{12}$ is in parallel with $C_3$,so the total equivalent capacitance $C_{eq}$ is:$C_{eq} = C_{12} + C_3 = 1.5 + 4 = 5.5\ \mu F$.The total energy $U$ stored in the system is given by:$U = \frac{1}{2} C_{eq} V^2 = \frac{1}{2} 	imes 5.5 	imes 10^{-6} 	imes (2)^2$$U = \frac{1}{2} 	imes 5.5 	imes 10^{-6} 	imes 4 = 11 	imes 10^{-6}\ J$.

Given $C_1 = 2\ \mu F$,$C_2 = 6\ \mu F$,and $C_3 = 4\ \mu F$. What is the total energy of the system?

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