In the given circuit, the current in resistance R _ 3 is:

Equivalent resistance, ${R}_{eq} = 2Ω + \dfrac{4 \times 4}{4 + 4}Ω + 1Ω = 5Ω$

Current through battery, $i = \dfrac{V}{{R}_{eq}} = \dfrac{10}{5} = 2A$

Current in resistance ${R}_{3} = 2 \times \left(\dfrac{4}{4 + 4}\right)$

$= 2 \times \dfrac{4}{8}$

$= 1A$