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

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