JEE MAIN - Physics (2024 - 8th April Evening Shift - No. 24)
A potential divider circuit is connected with a dc source of $$20 \mathrm{~V}$$, a light emitting diode of glow in voltage $$1.8 \mathrm{~V}$$ and a zener diode of breakdown voltage of $$3.2 \mathrm{~V}$$. The length (PR) of the resistive wire is $$20 \mathrm{~cm}$$. The minimum length of PQ to just glow the LED is _________ $$\mathrm{cm}$$.
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Answer
5
Explanation
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For minimum length of $$P Q$$,
$$\begin{aligned} & \mathrm{V}_{\text {diodes }}=1.8+3.2=5 \mathrm{~V} \\ & \therefore \quad I_{P Q}=\frac{20}{20} \times 5 \\ & =5 \mathrm{~cm} \\ & \end{aligned}$$
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