QJEEM24Phacea: JEE Main 2024 Physics - Class 12 JEEM24Phacea Solution: In the given circuit, the current in resistance R is :

$\mathrm{R}_{\mathrm{eq}}=2 \Omega+2 \Omega+1 \Omega=5 \Omega$

$\mathrm{i}=\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{eq}}}=\frac{10}{5}=2 \mathrm{~A}$

Current in resistance

$\begin{aligned}R_3 & =2 \times\left(\frac{4}{4+4}\right) \\& =2 \times \frac{4}{8} \\& =1 \mathrm{~A}\end{aligned}$