Concept:During the negative half-cycle, an ideal diode is reverse-biased and acts as an open circuit, with the entire peak secondary voltage \( V_m \) appearing across it.
Formula:$$\text{PIV}_{\text{half-wave}} = V_m$$
Solution:- During the negative half-cycle, no current flows through the load resistor (\( V_{\text{load}} = 0 \)).
- Applying Kirchhoff's Voltage Law around the loop shows that the full peak secondary voltage \( V_m \) appears across the reverse-biased diode.
- Therefore, \( \text{PIV} = V_m \).
Why other options are incorrect:- Option A: \( 2V_m \) is the PIV for a center-tapped full-wave rectifier.
- Option B: \( V_m / 2 \) would cause reverse breakdown because the full voltage \( V_m \) appears across the diode.
- Option D: \( V_m / \sqrt{2} \) is the RMS voltage, not the peak reverse voltage.
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.