Concept:Rectification efficiency is the ratio of DC output power to AC input power: \( \eta = (P_{\text{dc}} / P_{\text{ac}}) \times 100\% \).
Formula:$$\eta = \frac{P_{\text{dc}}}{P_{\text{ac}}} \times 100\%$$
$$\eta_{\text{max, full-wave}} \approx 81.2\% \quad \text{and} \quad \eta_{\text{max, half-wave}} \approx 40.6\%$$
Solution:- Given: \( P_{\text{dc}} = 80.0\text{ W} \) and \( P_{\text{ac}} = 100.0\text{ W} \).
- Efficiency: \( \eta = \frac{80.0}{100.0} \times 100\% = 80.0\% \).
- This is close to the theoretical maximum for full-wave rectifiers (\( 81.2\% \)) and roughly double that of half-wave rectifiers (\( 40.6\% \)).
Why other options are incorrect:- Option A: 40.0% is typical for half-wave rectifiers.
- Option B: 50.0% underestimates the efficiency of this circuit.
- Option D: 100.0% is physically impossible due to non-zero harmonic ripple and diode losses.
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