b
Taking potential of point $\mathrm{x}$ is zero potential of $\mathrm{y}$ would be $10$ while of $\mathrm{z}$ would be $20 .$ Current will flow from hight to low potential.
$I_{y}=\frac{10-0}{2}=5 \,A$
$I_{2}=\frac{20-0}{5}=4 \,A$
applying $\mathrm{KCL}$ at $\mathrm{X}$
$I_{z}+I_{x}=I_{y}$
$4+I_{x}=I_{y}=5$
$ \Rightarrow \boxed{{{\text{r}}_{\text{x}}} = 1\,{\text{A}}}$
