JEE MAIN - Physics (2023 - 10th April Evening Shift - No. 2)
If each diode has a forward bias resistance of $$25 ~\Omega$$ in the below circuit,
_10th_April_Evening_Shift_en_2_1.png)
Which of the following options is correct :
$$\frac{I_{3}}{I_{4}}=1$$
$$\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=2$$
$$\frac{I_{2}}{\mathrm{I}_{3}}=1$$
$$\frac{I_{1}}{I_{2}}=1$$
Explanation
_10th_April_Evening_Shift_en_2_2.png)
$$ \begin{aligned} & \mathrm{R}_{\mathrm{eq}}=\frac{150 \times 150}{300}+25=100 \Omega \\\\ & \mathrm{I}_1=\frac{5}{10}=0.05 \mathrm{~A} \\\\ & \mathrm{I}_2=\mathrm{I}_4=\frac{0.05}{2}=0.025 \mathrm{~A} \\\\ & \frac{\mathrm{I}_1}{\mathrm{I}_2}=2 \end{aligned} $$
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