Physics · Semiconductor and Electronic Devices

JEE Main 2024 — 29 January, Shift 1 — Question 30

In the given circuit, the breakdown voltage of the Zener diode is 3.0 V . What is the value of $\mathrm{I}_{z}$ ?

Option A:
3.3 mA
Option B:
5.5 mA
Correct
Option C:
10 mA
Option D:
7 mA

Answer: B

Step-by-step solution

$\mathrm{V}_{\mathrm{z}}=3 \mathrm{~V}$

Let potential at $\mathrm{B}=0 \mathrm{~V}$

Potential at $E\left(V_{E}\right)=10 \mathrm{~V}$ $\mathrm{V}_{\mathrm{C}}=\mathrm{V}_{\mathrm{A}}=3 \mathrm{~V}$ $\mathrm{I}_{\mathrm{z}}+\mathrm{I}_{1}=\mathrm{I}$

$I=\frac{10-3}{1000}=\frac{7}{1000} A$

$I_{1}=\frac{3}{2000} \mathrm{~A}$

Therefore $I_{z}=\frac{7-1.5}{1000}=5.5 \mathrm{~mA}$

Answer key and solution verified before publishing.

Practise Semiconductor and Electronic Devices

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2024
Paper: 29 January, Shift 1
Subject: Physics
Chapter: Semiconductor and Electronic Devices
Topic: p-n Diode and its Applications