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QJEEM22Ph1f08: JEE Main 2022 Physics - Class 12 JEEM22Ph1f08 Solution: A Zener diode along with a series resistance is connected…

A 8 V8 \mathrm{~V} Zener diode along with a series resistance R\mathrm{R} is connected across a 20 V20 \mathrm{~V} supply (as shown in the figure). If the maximum Zener current is 25 mA25 \mathrm{~mA} , then the minimum value of R will be _______ Ω\Omega .

JEE Main 2022 (Online) 29th July Evening Shift Physics - Semiconductor Question 58 English

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JEE Main 2022 (Online) 29th July Evening Shift Physics - Semiconductor Question 58 English Explanation
Applying voltage rule in loop ABCD.

E−IR−VZamp;=020−IR−8amp;=02.5×10−3×Ramp;=12\begin{aligned}\mathrm{E}-\mathrm{IR}-\mathrm{V}_{\mathrm{Z}} & =0 \\\\20-\mathrm{IR}-8 & =0 \\\\2.5 \times 10^{-3} \times \mathrm{R} & =12\end{aligned}

R=1225×10−3=480Ω\mathrm{R}=\frac{12}{25 \times 10^{-3}}=480 \Omega

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