Concept:A diode's biasing state depends on the potential difference across its terminals: \( V_D = V_{\text{anode}} - V_{\text{cathode}} \). If \( V_D < 0 \), the cathode is at a higher potential than the anode, placing the diode in reverse bias.
Formula:$$V_D = V_A - V_K = (+2.0\text{ V}) - (+5.0\text{ V}) = -3.0\text{ V} < 0 \implies \text{Reverse Bias}$$
Solution:- The anode potential is \( +2.0\text{ V} \) and the cathode potential is \( +5.0\text{ V} \).
- Because the cathode is \( 3.0\text{ V} \) more positive than the anode, the junction is reverse-biased with \( V_R = 3.0\text{ V} \).
- In this state, the depletion layer widens and only a tiny reverse saturation leakage current flows.
Why other options are incorrect:- Option A: Forward bias requires \( V_A > V_K \) (anode more positive than cathode).
- Option C: Zero bias occurs only when \( V_A = V_K \).
- Option D: Standard diodes at \( -3.0\text{ V} \) are well below reverse breakdown and do not regulate voltage.
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