Special Purpose Diodes question from JEE Main 2024 (1 Feb, Shift 1)

Answer: (d)

1. Maximum Zener current: $I_{Z,max} = \frac{10\text{ mW}}{5\text{ V}} = 2$ mA; load current $I_L = \frac{5}{1\text{ k}\Omega} = 5$ mA.
2. Drop across $R_s$: $8 - 5 = 3$ V, and $I_s = I_Z + I_L$ must lie between $5$ mA and $7$ mA.
3. $R_s = \frac{3}{I_s}$: from $\frac{3}{7}$ k$\Omega$ (at $7$ mA) to $\frac{3}{5}$ k$\Omega$ (at $5$ mA).
4. So the answer is (d) $\frac{3}{7}\text{ k}\Omega < R_s < \frac{3}{5}\text{ k}\Omega$.