In the circuit diagram shown in figure given below, the current flowing through resistance $3\, \Omega$ is $\frac{ x }{3}\,A$. The value of $x$ is $...........$
JEE MAIN 2023, Diffcult
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$\frac{1}{3}+\frac{1}{6}=\frac{1}{2}=\frac{1}{R}$
$R =2\,\Omega$
$I _1=\left(\frac{6}{3+6}\right) \times 0.5$
$I _1=\frac{2}{3} \times 0.5=\frac{1}{3}\,A$
$I _1=\frac{ x }{3}=\frac{1}{3} \therefore x =1$
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