Physics · Semiconductor and Electronic Devices

JEE Main 2024 — 30 January, Shift 1 — Question 40

A Zener diode of breakdown voltage 10 V is used as a voltage regulator as shown in the figure. The current through the Zener diode is

Answer: C

Zener is in breakdown region. $I_{3}=\frac{10}{500}=\frac{1}{50}$

$I_{1}=\frac{10}{200}=\frac{1}{20}$

$I_{2}=I_{1}-I_{3}$

$I_{2}=\left(\frac{1}{20}-\frac{1}{50}\right)=\left(\frac{3}{100}\right)=30 m A$

Answer key and solution verified before publishing.

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