JEE Main · 2024 · 01 Feb 2024 · Shift 1 · Semiconductors

The voltage across $R_s$ is $V_1=8-5=3$ V. Current through the load resistor is $I=\frac{5}{1000}=5$ mA. Maximum Zener current is $I_{z,max}=\frac{10\text{ mW}}{5\text{ V}}=2$ mA. Therefore, maximum series current is $I_{s,max}=5+2=7$ mA, giving $R_{s,min}=\frac{3}{7}$ kΩ ≈ 0.43 kΩ. Minimum series current is $I_{s,min}=5$ mA (when Zener current is zero), giving $R_{s,max}=\frac{3}{5}$ kΩ = 0.6 kΩ. The required resistance must satisfy $0.43\text{ kΩ}<R_s<0.6\text{ kΩ}$. None of the given options fall in this range.