Concept:Under reverse bias, the depletion layer widens and blocks majority carrier flow. The only current is a very small minority leakage current (\( \text{nA} \) to \( \mu\text{A} \)), resulting in a very high static resistance.
Formula:$$R_{\text{reverse}} = \frac{V_R}{I_0} \approx \frac{10\text{ V}}{10^{-8}\text{ A}} = 10^9\ \Omega = 1000\text{ M}\Omega$$
Solution:- Because the reverse saturation current \( I_0 \) is extremely small in silicon (order of \( \text{nA} \)), the ratio of applied reverse voltage to reverse current is on the order of several megaohms (\( \text{M}\Omega \)) to gigohms.
Why other options are incorrect:- Option A: Resistances below \( 1\ \Omega \) occur in forward-biased power diodes or superconductors.
- Option B: \( 10\text{–}50\ \Omega \) is typical for the forward dynamic resistance of a conducting diode.
- Option C: \( 100\text{–}500\ \Omega \) is typical for standard fixed circuit resistors.
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