Concept:Efficiency of rectification is defined as the ratio of the DC output power delivered to the load to the total AC input power supplied to the circuit.
Formula:$$\eta = \frac{P_{\text{dc}}}{P_{\text{ac}}} = \frac{I_{\text{dc}}^2 R_L}{I_{\text{rms}}^2 (r_f + R_L)}$$
$$\text{For Half-Wave Rectifier: } I_{\text{dc}} = \frac{I_0}{\pi}, \quad I_{\text{rms}} = \frac{I_0}{2}$$
Solution:- Substituting the values (assuming ideal diode where \(r_f \ll R_L\)):
- $$\eta_{\text{max}} = \frac{(I_0 / \pi)^2 R_L}{(I_0 / 2)^2 R_L} = \frac{4}{\pi^2} \approx 0.406 = 40.6\%$$
Why other options are incorrect:- Option A: \(81.2\%\) is the theoretical maximum efficiency of a full-wave rectifier (\(8 / \pi^2\)).
- Option C: \(50.0\%\) is an incorrect estimate neglecting the sinusoidal integration factor \(4/\pi^2\).
- Option D: \(100.0\%\) is unattainable because half the AC waveform is blocked and power is dissipated.
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