When the forward voltage across a silicon diode is raised from 0.70 V to 0.75 V, the current through the diode rises from 5 mA to 15 mA. The dynamic resistance of the diode in this range is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 5\,\Omega \)
Above the threshold voltage the forward current of a diode rises steeply. It is not proportional to the voltage. So over a small range the diode is described by its dynamic resistance. This is the ratio of a small change in voltage to the change in current it produces.
The changes in voltage and current are:
\( \begin{aligned} \Delta V &= 0.75\,\mathrm{V}-0.70\,\mathrm{V} \\ &= 0.05\,\mathrm{V} \end{aligned} \)
\( \begin{aligned} \Delta I &= 15\,\mathrm{mA}-5\,\mathrm{mA} \\ &= 10\,\mathrm{mA}=0.01\,\mathrm{A} \end{aligned} \)
The dynamic resistance is:
\( \begin{aligned} r_d &= \frac{\Delta V}{\Delta I} \\ &= \frac{0.05\,\mathrm{V}}{0.01\,\mathrm{A}} \\ &= 5\,\Omega \end{aligned} \)
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:”
\( \displaystyle r_d=\frac{\Delta V}{\Delta I} \qquad (14.6) \)