JEE MAIN - Physics (2024 - 30th January Morning Shift - No. 14)
A Zener diode of breakdown voltage $$10 \mathrm{~V}$$ is used as a voltage regulator as shown in the figure. The current through the Zener diode is :
_30th_January_Morning_Shift_en_14_1.png)
0
30 mA
20 mA
50 mA
Explanation
_30th_January_Morning_Shift_en_14_2.png)
Zener is in breakdown region.
$$\begin{aligned} & I_3=\frac{10}{500}=\frac{1}{50} \\\\ & I_1=\frac{10}{200}=\frac{1}{20} \\\\ & I_2=I_1-I_3 \\\\ & I_2=\left(\frac{1}{20}-\frac{1}{50}\right)=\left(\frac{3}{100}\right)=30 \mathrm{~mA} \end{aligned}$$
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