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

Equivalent resistance, $R_{\mathrm{eq}}=2 \mathrm{\Omega }+\frac{4\times 4}{4+4} \mathrm{\Omega }+1 \mathrm{\Omega }=5 \mathrm{\Omega }$

Current through battery, $i=\frac{V}{R_{eq}}=\frac{10}{5}=2 \mathrm{A}$

Current in resistance $R_3=2\times \left(\frac{4}{4+4}\right)$

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

$=1 \mathrm{A}$