A silicon diode is reverse biased by a voltage of 20 V. The current through the diode is 4 μA. The resistance of the diode under this reverse bias is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(5\times10^{6}\ \Omega\)
In reverse bias the diode passes only a very small current. Its resistance in this region is the reverse voltage divided by this current, as in Ohm’s law.
\( V=20\ \mathrm{V} \)
\( I=4\ \mu\mathrm{A}=4\times10^{-6}\ \mathrm{A} \)
\( \begin{aligned} r &= \frac{V}{I} \\ &= \frac{20}{4\times10^{-6}} \\ &= 5\times10^{6}\ \Omega \end{aligned} \)
NCERT (2026-27):
Example 14.4 – “… (b) From the curve at V = –10 V, I = –1 μA, Therefore, rrb = 10 V/1μA= 1.0 × 107 Ω”