The figure shows a circuit having eight resistances of 1Ω...

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

Now, let& $x$ and& $y$ be the current flowing from& $6V$ and& $12V$ batteries.

The current distribution is shown below.

Applying Kirchhoff's voltage law,

For loop& $1:$

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

$⇒ - 2x - \dfrac{y}{2} = - 6$

$⇒ - 4x - y = - 12$

$⇒ 4x + y = 12 ⋯ \left(1\right)$

For loop $2:$

$\dfrac{1}{2}$ y + 1y + 12 + \dfrac{1}{2}\left( x + y\right) = 0$</p><p>\dfrac{3}{2}$ y + $\dfrac{y}{2} + \dfrac{x}{2} = -12$

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

$4y + x = - 24 ⋯ \left(2\right)$

Using, \left(2\right) - 4 \times\left(1\right)$, we have</p><p>$4y + x - 16x - 4y= - 24 - 48$</p><p>$⇒ - 15x = - 72$</p><p>$⇒ x = \dfrac{72}{15}$</p><p>Putting in equation \left(1\right)$,

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

$⇒ y = 12 - \dfrac{288}{15}= \dfrac{180 - 288}{15}= \dfrac{- 108}{15} = -7 . 2A$

Now, current in ${R}_{1} = 7 . 2A$

Current in ${R}_{2} = \dfrac{x + y}{2} = \left(\dfrac{72}{15} - \dfrac{108}{15}\right)\dfrac{1}{2}= \dfrac{1}{2}\left(\dfrac{36}{15}\right) = \dfrac{2 . 4}{2}A$

$= 1 . 2A$

Current in ${R}_{3} = x = 4 . 8A$

Current in ${R}_{5} = \dfrac{1}{2}x = 2 . 4A$