QJEEM23Ph985f: JEE Main 2023 Physics - Class 12 JEEM23Ph985f Solution: In the circuit diagram shown in figure given below, the current …

Step-by-Step Verified Solution

$\mathrm{E}_2-\mathrm{E}_1=8-4=4 \mathrm{~V}$

$\begin{aligned}& \frac{1}{3}+\frac{1}{6}=\frac{1}{2}=\frac{1}{R} \\\\& R=2 \Omega\end{aligned}$




$I=\frac{4}{8}=0.5 \mathrm{~A}$



$\begin{aligned}& \mathrm{I}_1=\left(\frac{6}{3+6}\right) \times 0.5 \\\\& \mathrm{I}_1=\frac{2}{3} \times 0.5=\frac{1}{3} \mathrm{~A}\end{aligned}$

$I_1=\frac{x}{3}=\frac{1}{3} \therefore x=1$