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(D) The circuit consists of two capacitors in parallel connected in series with two other capacitors.First,calculate the equivalent capacitance of the

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$400 	imes 10^{-7} \,J$

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(D) The circuit consists of two capacitors in parallel connected in series with two other capacitors.First,calculate the equivalent capacitance of the two parallel capacitors:$C_{\|} = 2 \mu F + 2 \mu F = 4 \mu F$Now,the circuit is equivalent to three capacitors in series: $2 \mu F$,$4 \mu F$,and $2 \mu F$.The equivalent capacitance $C_{eq}$ is given by:$\frac{1}{C_{eq}} = \frac{1}{2 \mu F} + \frac{1}{4 \mu F} + \frac{1}{2 \mu F} = \frac{2 + 1 + 2}{4 \mu F} = \frac{5}{4 \mu F}$$C_{eq} = \frac{4}{5} \mu F = 0.8 	imes 10^{-6} \,F$The energy stored $U$ is given by:$U = \frac{1}{2} C_{eq} V^2$$U = \frac{1}{2} 	imes (0.8 	imes 10^{-6} \,F) 	imes (10 \,V)^2$$U = 0.4 	imes 10^{-6} 	imes 100 \,J = 40 	imes 10^{-6} \,J = 400 	imes 10^{-7} \,J$

Four capacitors each of capacity $2 \mu F$ are connected as shown in the figure. If $V_{A}-V_{B}=10 \,V$,the energy stored in the system is:

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