JEE Main · 2022 · 29 Jul 2022 · Shift 2 · Semiconductors

In the breakdown region the supply divides between the series resistor and the Zener, so $20=IR+8$ and the drop across $R$ is $12\text{V}$. The current grows as $R$ shrinks, and the Zener current may rise only to its maximum of $25\text{mA}$, so the least permissible resistance is $R=\frac{12}{25\times 10^{-3}}=480\Omega$.