The I-V characteristics of a p-n junction diode in forward bias is shown in the figure. The ratio of dynamic resistance, corresponding to forward bias voltage of 2 V and 4 V respectively, is :

Solution

Physics sir · 4 steps

Dynamic resistance is the slope Δ V/Δ I near each point, so read the small step on the graph.

1. What to find. Dynamic resistance is r_d = (Δ V)/(Δ I), taken over a small interval near the operating point. It is not V/I at that point.
2. Near 2 V. From 2 V to 2.1 V the current rises from 5 mA to 10 mA. So r₁ = (0.1 V)/(5 × 10⁻³ A) = 20 Ω.
3. Near 4 V. From 4 V to 4.2 V the current rises from 200 mA to 250 mA. So r₂ = (0.2 V)/(50 × 10⁻³ A) = 4 Ω.
4. So. r₁ : r₂ = 20 : 4 = 5 : 1, which is option (b). The curve is steeper at 4 V, so the diode offers less resistance there.

The idea to keep. For a diode, dynamic resistance means Δ V/Δ I from a small step on the curve. The steeper the I-V curve, the smaller the dynamic resistance.

Answer from the NTA final answer key. Solution by PrepWiser.