In the following circuit find $I_{1}and{I}_2$

The equivalent circuit can be redrawn as From figure it is clear that current drawn from the battery.

There will be no current passing through $14\mathrm{k\Omega }$ and $12\mathrm{k\Omega }$ resistors.

Only resistance is $2\mathrm{k\Omega }$

$\mathrm{i}={\mathrm{i}}_2=\frac{10}{2\mathrm{x}{10}^3}=5\mathrm{mA}$ and $i_1=0$