A silicon diode under forward bias shows a change of current from 10 mA to 15 mA when the bias voltage is changed from 0.8 V to 0.9 V. What is the resistance of the diode?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 20 Ω
The question gives a change of current for a change of voltage, so the resistance asked for is the dynamic (a.c.) resistance of the diode, the ratio of the small changes.
\( \Delta V = 0.9\,\mathrm{V} - 0.8\,\mathrm{V} = 0.1\,\mathrm{V} \)
\( \Delta I = 15\,\mathrm{mA} - 10\,\mathrm{mA} = 5\times10^{-3}\,\mathrm{A} \)
\( \displaystyle r_d = \frac{\Delta V}{\Delta I} = \frac{0.1\,\mathrm{V}}{5\times10^{-3}\,\mathrm{A}} \)
\( \displaystyle r_d = 20\,\Omega \)
Note: NTA picked this question from an NCERT (2026-27) example and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
p-n Junction Diode Under Reverse Bias – “… For diodes, we define a quantity called dynamic resistance as the ratio of small change in voltage ΔV to a small change in current ΔI:
\( \displaystyle r_d = \frac{\Delta V}{\Delta I} \qquad (14.6) \)”
p-n Junction Diode Under Reverse Bias – “… Example 14.4 … Calculate the resistance of the diode at (a) ID = 15 mA and (b) VD = –10 V.”
p-n Junction Diode Under Reverse Bias – “… (a) From the curve, at I = 20 mA, V = 0.8 V; I = 10 mA, V = 0.7 V
\( \displaystyle r_{fb} = \Delta V/\Delta I = 0.1\,\mathrm{V}/10\,\mathrm{mA} = 10\,\Omega \)”