JEE MAIN - Physics (2023 - 15th April Morning Shift - No. 4)
In the given circuit, the current (I) through the battery will be
_15th_April_Morning_Shift_en_4_1.png)
1A
$2.5 \mathrm{~A}$
$2 \mathrm{~A}$
$1.5 \mathrm{~A}$
Explanation
In the circuit $\mathrm{D}_1$ and $\mathrm{D}_3$ are forward biased and $\mathrm{D}_2$ is reverse biased.
_15th_April_Morning_Shift_en_4_2.png)
$$ \therefore \mathrm{I}=\frac{10}{20 / 3}=\frac{3}{2} \mathrm{~A}=1.5 \mathrm{~A} $$
$$ \therefore \mathrm{I}=\frac{10}{20 / 3}=\frac{3}{2} \mathrm{~A}=1.5 \mathrm{~A} $$
Comments (0)
