The figure shows a circuit having eight resistances of $1 \, \Omega$ each, labelled $R_1$ to $R_8$, and two…

Point $A$ and $B$ are at same potential so they can be merged/folded.

Now, let $x$ and $y$ be the current flowing from $6 \mathrm{V}$ and $12 \mathrm{V}$ batteries.

The current distribution is shown below.

Applying Kirchhoff's voltage law,

For loop $1:$

$-\frac{1}{2}x-\frac{1}{2}\left(x+y\right)-1x+6=0$

$\Rightarrow -2x-\frac{y}{2}=-6$

$\Rightarrow -4 x-y=-12$

$\Rightarrow 4x+y=12 \cdots \left(1\right)$

For loop $2:$

$\frac{1}{2}y+1y+12+\frac{1}{2}\left(x+y\right)=0$

$\frac{3}{2}y+\frac{y}{2}+\frac{x}{2}=-12$

$2y+\frac{x}{2}=-12$

$4y+x=-24 \cdots \left(2\right)$

Using, $\left(2\right)-4\times \left(1\right)$, we have

$4y+x-16x-4y$$=-24-48$

$\Rightarrow -15x=-72$

$\Rightarrow x=\frac{72}{15}$

Putting in equation $\left(1\right)$, 

$4\left(\frac{72}{15}\right)+y=12$

$\Rightarrow y=12-\frac{288}{15}$$=\frac{180-288}{15}$$=\frac{-108}{15}=-7.2 \mathrm{A}$

Now, current in $R_1=7.2 \mathrm{A}$

Current in $R_2=\frac{x+y}{2}=\left(\frac{72}{15}-\frac{108}{15}\right)\frac{1}{2}$$=\frac{1}{2}\left(\frac{36}{15}\right)=\frac{2.4}{2} \mathrm{A}$

$=1.2 \mathrm{A}$

Current in $R_3=x=4.8 \mathrm{A}$

Current in $R_5=\frac{1}{2}x=2.4 \mathrm{A}$