Concept:The Transformer Utilization Factor (\( \text{TUF} \)) measures how effectively a transformer secondary winding delivers DC power to the load.
Formula:$$\text{TUF} = \frac{P_{\text{dc}}}{P_{\text{ac(rated)}}} = \frac{V_{\text{dc}} I_{\text{dc}}}{V_{\text{rms}} I_{\text{rms}}}$$
Solution:- For a half-wave rectifier:
- $$V_{\text{dc}} = \frac{V_m}{\pi}, \quad I_{\text{dc}} = \frac{I_m}{\pi}, \quad V_{\text{rms}} = \frac{V_m}{\sqrt{2}}, \quad I_{\text{rms}} = \frac{I_m}{2}$$
- $$\text{TUF} = \frac{\left(\frac{V_m}{\pi}\right)\left(\frac{I_m}{\pi}\right)}{\left(\frac{V_m}{\sqrt{2}}\right)\left(\frac{I_m}{2}\right)} = \frac{2\sqrt{2}}{\pi^2} \approx \frac{2.8284}{9.8696} \approx 0.287$$
- This indicates that the transformer must be sized \( 3.5 \) times larger than the delivered DC load power.
Why other options are incorrect:- Option B: \( 0.693 \) is the \( \text{TUF} \) of a center-tapped full-wave rectifier.
- Option C: \( 0.812 \) is the \( \text{TUF} \) of a full-wave bridge rectifier.
- Option D: \( 0.406 \) (\( 40.6\% \)) is the maximum rectification efficiency of a half-wave rectifier, not its \( \text{TUF} \).
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