Question: In a reversed-biased p-n junction diode, the applied voltage mostly drops across:
Step 1 (Governing Principle): [0.25 Mark]
In a p-n junction diode, the depletion region is a region depleted of mobile charge carriers (electrons and holes), resulting in a very high electrical resistance compared to the p-type and n-type regions which are doped to be conductive.
Step 2 (Physical Reasoning): [0.50 Mark]
According to Ohm's Law, $V = IR$. In a reverse-biased condition, the external voltage $V_{ext}$ is applied in the same direction as the built-in potential barrier $V_B$. Because the depletion region acts as an insulator with high resistance ($R_{depletion} \gg R_{p} + R_{n}$), the potential drop across the junction is governed by:
$$V_{total} \approx V_{depletion} = I \cdot R_{depletion}$$Since the current $I$ is negligible (reverse saturation current), the entire external potential difference appears across the region of high resistance, which is the depletion layer.
Step 3 (Final Concluded Answer): [0.25 Mark]
The applied voltage mostly drops across the depletion region.
Correct Option: (C) depletion region
💡 EL10 Concept Insight: Think of the depletion region like a narrow, dry bridge in a circuit; because it has no "traffic" (charge carriers), it acts like a roadblock that forces all the electrical pressure (voltage) to pile up right at that spot instead of flowing through the rest of the wire.