The circuit shown in fig consists of the following $E_{1}=6, E_{2}=2, E_{3}=3$ Volt
$\mathrm{R}_{1}=6, \mathrm{R}_{4}=3 \mathrm{Ohm}$
$\mathrm{R}_{3}=4, \mathrm{R}_{2}=2$ 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$
$\mathrm{R}_{2}=2 \Omega \quad \mathrm{R}_{3}=4 \Omega \quad \mathrm{R}_{4}=3 \Omega$
The current through resistance $\mathrm{R}_{3}$ is.
(A) $1.5 \mathrm{~A}$
(B) $1.2 \mathrm{~A}$
(C) $0.9 \mathrm{~A}$
(D) $0.6 \mathrm{~A}$