Concept:During the negative half-cycle, an ideal diode is reverse-biased, acting as an open circuit with zero current through the load.
Formula:$$V_L = I_L R_L = (0\text{ A}) \times R_L = 0\text{ V}$$
Solution:- During the negative half-cycle, the diode is reverse-biased and presents infinite resistance (\( R_{\text{diode}} = \infty \)).
- Because the circuit is open, current through the load resistor is zero (\( I_L = 0\text{ A} \)).
- Applying Ohm's law gives a load voltage of \( V_L = I_L R_L = 0\text{ V} \).
- The full peak input voltage \( V_m \) appears across the reverse-biased diode.
Why other options are incorrect:- Option A: \( -V_m \) appears across the reverse-biased diode as its peak inverse voltage, not across the load resistor.
- Options C & D: Positive voltages appear across the load during forward conduction in the positive half-cycle, not during the negative half-cycle.
Quality & Fidelity Assurance:
Every question on BeambePrep is rigorously curated against the official PMDC syllabus with zero filler, zero out-of-syllabus content, and zero typos. When an authentic past paper originally contained a historical mistake or ambiguity from the examining board (such as UHS or NUMS), BeambePrep faithfully reflects the original paper while detailing the nuance and scientific consensus in the autopsy above.