Concept:In a center-tapped rectifier, only one diode conducts in series with the load during each half-cycle, introducing a single forward drop \( V_B \).
Formula:$$V_{\text{load, peak}} = V_m - V_B$$
Solution:- The peak voltage per half-secondary is \( V_m = 15.7\text{ V} \).
- A single diode conducts per half-cycle with a voltage drop of \( V_B = 0.7\text{ V} \).
- Calculate peak load voltage:
- $$V_{\text{load, peak}} = 15.7\text{ V} - 0.7\text{ V} = 15.0\text{ V}$$
Why other options are incorrect:- Option B: \( 14.3\text{ V} \) subtracts two diode drops (\( 1.4\text{ V} \)), which applies to bridge rectifiers.
- Option C: \( 15.7\text{ V} \) neglects the forward barrier drop.
- Option D: \( 30.0\text{ V} \) uses the total end-to-end secondary voltage rather than the voltage relative to the center tap.
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.