Concept:In a half-wave rectifier, current flows through the secondary winding in one direction for only half of each cycle, resulting in low transformer utilization.
Formula:$$\text{TUF} = \frac{P_{\text{dc}}}{V_{\text{rms}} I_{\text{rms}}} = \frac{(I_m / \pi)^2 R_L}{(V_m / \sqrt{2})(I_m / 2)} = \frac{2\sqrt{2}}{\pi^2} \approx 0.287$$
Solution:- Because DC current flows only during alternate half-cycles, significant transformer core saturation and reactive losses occur.
- The resulting \( \text{TUF} \) is approximately \( 0.287 \).
Why other options are incorrect:- Option A: \( 0.693 \) is the \( \text{TUF} \) of a center-tapped full-wave rectifier.
- Option C: \( 0.406 \) is the rectification efficiency (\( \eta = 40.6\% \)) of a half-wave rectifier.
- Option D: \( 0.500 \) is not the correct derived utilization factor.
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