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JEE Main 20248 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
8 April
Shift
Evening Shift
Subject
Physics
Chapter
Semiconductors

JEE Main 2024 · 8 April · Evening Shift · Physics · Semiconductors

A potential divider circuit is connected with a dc source of 20 ~V...

A potential divider circuit is connected with a dc source of 20 V20 \mathrm{~V}, a light emitting diode of glow in voltage 1.8 V1.8 \mathrm{~V} and a zener diode of breakdown voltage of 3.2 V3.2 \mathrm{~V}. The length (PR) of the resistive wire is 20 cm20 \mathrm{~cm}. The minimum length of PQ to just glow the LED is _____cm\mathrm{cm} A potential divider circuit is connected with a dc source of 20 ~V , a light emitting diode of glow in voltage 1.8 ~V

Correct answer

0

Step-by-step solution

A potential divider circuit is connected with a dc source of 20 ~V , a light emitting diode of glow in voltage 1.8 ~V
PR=20 cm VPQ=14×RPRℓmin⁡(PQ)=14×20=5 cm\begin{aligned} & \mathrm{PR}=20 \mathrm{~cm} \\ & \mathrm{~V}_{\mathrm{PQ}}=\frac{1}{4} \times \mathrm{R}_{\mathrm{PR}} \\ & \ell_{\min }(\mathrm{PQ})=\frac{1}{4} \times 20 \\ & =5 \mathrm{~cm}\end{aligned}
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