P-N Junction Diode question from JEE Main 2021 (20 Jul, Shift 2)

Answer: 25

1. Dynamic resistance $r_d = \dfrac{\Delta V_D}{\Delta I_D}$ on the straight part of the curve around $3$ mA.
2. From the graph, $I_D$ goes from $3$ mA to $5$ mA while $V_D$ goes from $0.70$ V to $0.75$ V.
3. $r_d = \dfrac{0.75 - 0.70}{(5 - 3)\times 10^{-3}} = \dfrac{0.05}{2\times 10^{-3}} = 25\ \Omega$
4. So the dynamic resistance is $25\ \Omega$.