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14.Semiconductor Electronics
hard
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

A
$50 \mathrm{~mA}$
B
$0$
C
$30 \mathrm{~mA}$
D
$20 \mathrm{~mA}$
(JEE MAIN-2024)
Solution

Zener is in breakdown region.
$ 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 \ m A$
Standard 12
Physics