JEE MAIN - Physics (2023 - 11th April Evening Shift - No. 1)
_11th_April_Evening_Shift_en_1_1.png)
The current flowing through R$$_2$$ is :
$$\frac{1}{2} \mathrm{~A}$$
$$\frac{1}{3} \mathrm{~A}$$
$$\frac{1}{4} \mathrm{~A}$$
$$\frac{2}{3} \mathrm{~A}$$
Explanation
$$
\mathrm{R}_{\mathrm{eq}}=2 \Omega
$$
$$ \text { I through battery } \frac{8}{2}=4 \mathrm{~A} $$
$$ I \text { through } C D= $$$$ \frac{2 \times 3}{3+6}=\frac{2}{3} A $$
$$ I \text { through } R_2= $$$$ \frac{2 / 3}{2}=\frac{1}{3} \mathrm{~A} $$
$$ \text { I through battery } \frac{8}{2}=4 \mathrm{~A} $$
$$ I \text { through } C D= $$$$ \frac{2 \times 3}{3+6}=\frac{2}{3} A $$
$$ I \text { through } R_2= $$$$ \frac{2 / 3}{2}=\frac{1}{3} \mathrm{~A} $$
Comments (0)
