JEE MAIN - Physics (2024 - 29th January Evening Shift - No. 2)
In the given circuit, the current in resistance R$$_3$$ is :
_29th_January_Evening_Shift_en_2_1.png)
2 A
1.5 A
1 A
2.5 A
Explanation
_29th_January_Evening_Shift_en_2_2.png)
$$\mathrm{R}_{\mathrm{eq}}=2 \Omega+2 \Omega+1 \Omega=5 \Omega$$
$$\mathrm{i}=\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{eq}}}=\frac{10}{5}=2 \mathrm{~A}$$
Current in resistance
$$\begin{aligned} R_3 & =2 \times\left(\frac{4}{4+4}\right) \\ & =2 \times \frac{4}{8} \\ & =1 \mathrm{~A} \end{aligned}$$
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