Physics Electronics PMDC Conceptual Practice
PMDC Verified Question 6 of 494
The theoretical maximum rectification efficiency (\(\eta_{\text{max}}\)) of a half-wave rectifier without a smoothing filter is equal to:
A
\(81.2\%\)
B
\(40.6\%\)
C
\(50.0\%\)
D
\(100.0\%\)
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option B: \(40.6\%\)
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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