The current ${\text{I}}_{\text{1}}$ (in A) flowing through $1\Omega$ resistor in the following circuit is

$$\begin{gathered} Resis\tan ce in parallel=\frac{R_1{R}_2}{R_1+R_2}=\frac{1\left(1\right)}{1+1}=\frac{1}{2}\Omega \\ \Rightarrow 1/2\Omega is in sereies with 2\Omega \\ \Rightarrow Resis\tan ce in series= R\text{'}=\frac{1}{2}+2=\frac{5}{2}=2.5\Omega \\ \Rightarrow 2.5\Omega is in parallel with 2\Omega \\ \Rightarrow Resis\tan ce in parallel=R\text{'}\text{'}=\frac{R_1{R}_2}{R_1+R_2}=\frac{2(2.5)}{2+2.5}=\frac{5}{4.5}=\frac{10}{9}\Omega \\ \Rightarrow V=iR\text{'}\text{'} \\ \Rightarrow 10=i\frac{10}{9}\Rightarrow i=9A \\ Current through upper branch=I\text{'}=\frac{V}{R\text{'}}=\frac{10}{2.5}=4A \\ current through 1\Omega =\frac{i\text{'}}{2}=\frac{4}{2}=2A \end{gathered}$$