$\therefore$ its equivalent $R^{\prime}=\frac{4 \times 8}{12}=\frac{8}{3} \Omega$

Circuit can be redrawn as

$$\begin{aligned} \& \mathrm{R}_{\mathrm{eq}}=\frac{8}{3}+\frac{1}{3}+1.5+5.5 \\ \& =10 \Omega \\ \& i=\frac{V}{\mathrm{R}_{\mathrm{eq}}}=\frac{5}{10}=0.5 \mathrm{~A} \end{aligned}$$