c
The circuit can be redrawn as:
$\varepsilon_{\mathrm{eq}}=\frac{\frac{\varepsilon_{1}}{\mathrm{r}_{1}}+\frac{\varepsilon_{2}}{\mathrm{r}_{2}}}{\frac{1}{\mathrm{r}_{1}}+\frac{1}{\mathrm{r}_{2}}}=\frac{\frac{5}{2}+\frac{5}{2}}{\frac{1}{2}+\frac{1}{2}}$
$\varepsilon_{\mathrm{eq}}=5 \mathrm{\,V}$
$\frac{1}{r_{e q}}=\frac{1}{r_{1}}+\frac{1}{r_{2}}=\frac{1}{2}+\frac{1}{2} \Rightarrow r_{e q}=1 \,\Omega$
$\therefore \mathrm{i}=\frac{\varepsilon}{\mathrm{r}+\mathrm{R}}=\frac{5}{10+1}=0.45 \mathrm{\,A}$
