Answer
$0.8175\times 10^{-3}A$
Work Step by Step
We can determine the current through resistor $R_2$ as follows:
As $R_1$ and $R_2$ are parallel therefore $R_2=\frac{R_1R_2}{R_1+R_2}$
$\implies R_{12}=\frac{6000\times 4000}{6000+4000}=2400\Omega$
Similarly $R_{12}$ and $R_3$ are in series
$\implies R_{123}=R_3+R_{12}$
$\implies R_{123}=2400+2000=4400\Omega$
Now current through $R_3$ and $R_{12}$
$i=\frac{6V}{4400\Omega}=\frac{3}{2200}A$
Similarly, the voltage though $R_1$ and $R_2$ is given as
$V_{12}=iR_{12}$
We plug in the known values to obtain:
$V_{12}=(\frac{3}{2200})(2400)$
$\implies V_{12}=3.27V$
Finally, the required current is given as
$i_2=\frac{V_{12}}{R_2}$
We plug in the known values to obtain:
$i_2=\frac{3.27}{4000}$
This simplifies to:
$i_2=0.8175\times 10^{-3}A$
