Physics · Semiconductor and Electronic Devices

JEE Main 2024 — 8 April, Shift 2 — Question 55

A potential divider circuit is connected with a dc source of 20 V , a light emitting diode of glow in voltage 1.8 V and a zener diode of breakdown voltage of 3.2 V . The length ( PR ) of the resistive wire is 20 cm . The minimum length of PQ to just glow the LED is $\qquad$ cm .

Answer: 5

Numerical answer — enter this value.

Step-by-step solution

$\mathrm{PR}=20 \mathrm{~cm}$

$\mathrm{V}_{\mathrm{PQ}}=\frac{1}{4} \times \mathrm{R}_{\mathrm{PR}}$

$\ell_{\text {min }}(\mathrm{PQ})=\frac{1}{4} \times 20$

$=5 \mathrm{~cm}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 8 April, Shift 2
Subject: Physics
Chapter: Semiconductor and Electronic Devices
Topic: p-n Diode and its Applications