The total energy stored in the capacitor syst | vedclass

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(C) The $6 \mu F$ and $3 \mu F$ capacitors are connected in series.Therefore,the equivalent capacitance $C_1$ is given by $\frac{1}{C_1} = \frac{1}{6}

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$8$

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(C) The $6 \mu F$ and $3 \mu F$ capacitors are connected in series.Therefore,the equivalent capacitance $C_1$ is given by $\frac{1}{C_1} = \frac{1}{6} + \frac{1}{3} = \frac{1+2}{6} = \frac{3}{6} = \frac{1}{2}$.Thus,$C_1 = 2 \mu F$.This equivalent capacitor $C_1$ is in parallel with the $2 \mu F$ capacitor.Therefore,the total equivalent capacitance of the system is $C_{eq} = C_1 + 2 \mu F = 2 \mu F + 2 \mu F = 4 \mu F$.The total energy $U$ stored in the system is given by $U = \frac{1}{2} C_{eq} V^2$.Given $V = 2 	ext{ V}$,we have $U = \frac{1}{2} 	imes 4 \mu F 	imes (2 	ext{ V})^2 = 2 	imes 4 = 8 \mu J$.

The total energy stored in the capacitor system shown in the figure will be (in $\mu J$)

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