MCQ
Calculate the current in wire $BD$ ................ $A$


- A$0$
- ✓$1$
- C$2$
- D$5$

$I_{y}=\frac{10-0}{2}=5 \,A$
$I_{2}=\frac{20-0}{5}=4 \,A$
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}}}$
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Reason : The rest mass energy of the products must be less than that of the parent
