The circuit shown in fig consists of the following $E_{1}=6, E_{2}=2, E_{3}=3$ Volt
$\mathrm{R}_{1}=6, \mathrm{R}_{4}=3 \mathrm{Ohm}$
$\mathrm{R}_{3}=4, \mathrm{R}_{2}=2 \mathrm{Ohm}$
$\mathrm{C}=5 \mu \mathrm{F}$$E_{1}=6 \mathrm{~V} \quad E_{2}=2 \mathrm{~V} \quad E_{3}=3 \mathrm{~V} \quad \mathrm{R}_{1}=6 \Omega$
$R_{2}=2 \Omega \quad R_{3}=4 \Omega \quad R_{4}=3 \Omega$
The energy stored in the capacitor is.
(A) $4.8 \times 10^{-6} \mathrm{~J}$
(B) $9.6 \times 10^{-6} \mathrm{~J}$
(C) $1.44 \times 10^{-5} \mathrm{~J}$
(D) $1.92 \times 10^{-5} \mathrm{~J}$