Concept:If one diagonal diode pair has a higher forward resistance or voltage drop, the load voltage peak \( V_m - 2V_{D1} \) on one half-cycle will differ from \( V_m - 2V_{D2} \) on the alternate half-cycle, creating an asymmetric output.
Formula:$$V_{\text{peak,1}} = V_m - 2V_{D1} \quad \ne \quad V_{\text{peak,2}} = V_m - 2V_{D2}$$
Solution:- During positive half-cycles, current passes through the lower-resistance pair, producing larger output peaks.
- During negative half-cycles, current passes through the higher-resistance pair, producing smaller output peaks.
- This amplitude mismatch creates an asymmetry with a period equal to the original AC cycle, introducing an unwanted \( f_{\text{in}} \) ripple component in addition to the normal \( 2f_{\text{in}} \) ripple.
Why other options are incorrect:- Option A: Current still flows through the load in a single direction, so the output remains DC rather than AC.
- Option B: Both pairs still conduct, so output appears during both half-cycles.
- Option C: Asymmetry introduces lower-frequency ripple components, not higher harmonics.
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.