Concept:The DC power delivered by a full-wave rectifier is determined by its average DC current, which is twice that of a half-wave rectifier.
Formula:$$P_{\text{dc}} = I_{\text{dc}}^2 R_L$$
Solution:- For a full-wave rectifier, the average DC current is \( I_{\text{dc}} = \frac{2I_m}{\pi} \).
- Calculating DC power:
- $$P_{\text{dc}} = I_{\text{dc}}^2 R_L = \left(\frac{2I_m}{\pi}\right)^2 R_L = \frac{4 I_m^2}{\pi^2} R_L$$
Why other options are incorrect:- Option A: \( \left(\frac{I_m}{\pi}\right)^2 R_L \) is the DC load power for a half-wave rectifier.
- Option C: \( \left(\frac{I_m}{\sqrt{2}}\right)^2 R_L \) is the total RMS power delivered to the load.
- Option D: \( \left(\frac{I_m}{2}\right)^2 R_L \) corresponds to the RMS power of a half-wave rectifier.
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.