Four capacitors each of capacity \(2 \mu \mathrm{F}\) are connected as shown in the figure \(\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}}=10 \mathrm{~V}\). The energy stored in the system is

The combination of parallel resistors is
\(
\mathrm{C}_{\|}=2 \mu \mathrm{F}+2 \mu \mathrm{F}=4 \mu \mathrm{F}
\)
To find the equivalent capacitance of the circuit, we consider series combination:
\(\frac{1}{\mathrm{C}_{\mathrm{eq}}}=\frac{1}{2 \mu \mathrm{F}}+\frac{1}{4 \mu \mathrm{F}}+\frac{1}{2 \mu \mathrm{F}}=\frac{2+1+2}{4 \mu \mathrm{F}} \)
\( \Rightarrow \mathrm{C}_{\mathrm{eq}}=\frac{4}{5} \mu \mathrm{F}\)
\(\text {Energy stored }=\frac{\mathrm{c}_{\mathrm{eq}} \mathrm{v}^2}{2}=\) \(\frac{4 \times 10^{-6} \mathrm{~F} \times(10)^2 \mathrm{~V}^2}{5 \times 2} \)
\( =40 \times 10^{-6} \mathrm{~J}=400 \times 10^{-7} \mathrm{~J}\)