In the given circuit, the current in resistance $\mathrm{R}_3 $ is :
JEE MAIN 2024, Diffcult
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$\mathrm{R}_{\mathrm{eq}}=2 \Omega+2 \Omega+1 \Omega=5 \Omega$
$\mathrm{i}=\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{eq}}}=\frac{10}{5}=2 \mathrm{~A}$
Current in resistance $\mathrm{R}_3=2 \times\left(\frac{4}{4+4}\right)$
$ =2 \times \frac{4}{8} $
$ =1 \mathrm{~A}$
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