JEE MAIN - Physics (2023 - 11th April Morning Shift - No. 28)
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 ___________
_11th_April_Morning_Shift_en_28_1.png)
Answer
1
Explanation
_11th_April_Morning_Shift_en_28_2.png)
$$ \mathrm{E}_2-\mathrm{E}_1=8-4=4 \mathrm{~V} $$
$$ \begin{aligned} & \frac{1}{3}+\frac{1}{6}=\frac{1}{2}=\frac{1}{R} \\\\ & R=2 \Omega \end{aligned} $$
_11th_April_Morning_Shift_en_28_3.png)
_11th_April_Morning_Shift_en_28_4.png)
$$ I=\frac{4}{8}=0.5 \mathrm{~A} $$
_11th_April_Morning_Shift_en_28_5.png)
$$ \begin{aligned} & \mathrm{I}_1=\left(\frac{6}{3+6}\right) \times 0.5 \\\\ & \mathrm{I}_1=\frac{2}{3} \times 0.5=\frac{1}{3} \mathrm{~A} \end{aligned} $$
$$ I_1=\frac{x}{3}=\frac{1}{3} \therefore x=1 $$
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