Answer
current through $10\Omega $ is $ 0.36A$
Work Step by Step
let $I_1$, $I_2$ and $I_3$ be the currents through $5\Omega$, $10\Omega$ and $5\Omega$ respectively (from above to below). let the direction of all currents is from left to right.
Now applying kirchoffs laws, we have
Junction law
$I_1+I_2+ I_3 =0 \space ....................(i)$
1st loop, loop law
$ 12V - (I_2 *10)V +3V +(I_1*5) =0 \space ....................(ii)$
2nd loop, loop law
$9V -3V +(I_2*10)V -(I_3* 5)V=0 \space ....................(iii)$
Subtracting $(iii)$ from $(ii)$ we get
$9V -(20I_2)V +5(I_1+I_3) V =0$
but $I_1+I_3 =-i_2$
hence
$9V -20I_2 V +5(-I_2) V = 0$
$ 25 I_2 = 9V$
$ I_2 = 0.36A$
hence current through $10\Omega $ is $ 0.36A$
