Concept:Full-wave rectifiers convert both half-cycles of the AC input into DC, yielding twice the conversion efficiency of a half-wave rectifier.
Formula:$$\eta_{\text{max}} = \frac{P_{\text{dc}}}{P_{\text{ac}}} = \frac{(2I_0 / \pi)^2 R_L}{(I_0 / \sqrt{2})^2 R_L} = \frac{8}{\pi^2}$$
Solution:- Evaluating the mathematical constant:
- $$\eta_{\text{max}} = \frac{8}{(3.14159)^2} = \frac{8}{9.8696} \approx 0.8116 = 81.2\%$$
- Thus, the maximum theoretical efficiency of a full-wave rectifier is \(81.2\%\).
Why other options are incorrect:- Option A: \(40.6\%\) is the maximum efficiency of a half-wave rectifier.
- Option B: \(50.0\%\) is a common misconception assuming linear ratio instead of sinusoidal integration.
- Option D: \(90.5\%\) is mathematically incorrect for an unfiltered full-wave sine output.
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