Concept:For an ideal bridge rectifier, the peak voltage across the load equals the peak secondary voltage \( V_m \). The corresponding RMS secondary voltage is \( V_{\text{rms}} = V_m / \sqrt{2} \).
Formula:$$V_{\text{rms}} = \frac{V_m}{\sqrt{2}} \approx 0.7071 \times V_m$$
Solution:- Given: Peak output voltage \( V_m = 40.0\text{ V} \).
- RMS secondary voltage: \( V_{\text{rms}} = \frac{40.0\text{ V}}{\sqrt{2}} = \frac{40.0}{1.414} \approx 28.28\text{ V} \approx 28.3\text{ V} \).
Why other options are incorrect:- Option A: 14.1 V is \( 20 / \sqrt{2} \), which uses half the required peak voltage.
- Option B: 56.6 V incorrectly multiplies the peak voltage by \( \sqrt{2} \) instead of dividing.
- Option D: 40.0 V is the peak voltage, not the RMS 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.