Concept:The Transformer Utilization Factor (TUF) is the ratio of the DC power delivered to the load to the AC volt-ampere (VA) rating of the transformer secondary winding.
Formula:$$\text{TUF} = \frac{P_{\text{dc}}}{V_{\text{rms}} \cdot I_{\text{rms}}} = \frac{(I_m / \pi)^2 R_L}{(V_m / \sqrt{2}) \cdot (I_m / 2)} = \frac{2\sqrt{2}}{\pi^2} \approx 0.2865 \approx 0.287$$
Solution:- For a half-wave rectifier, \( P_{\text{dc}} = \frac{I_m^2 R_L}{\pi^2} \).
- The secondary volt-ampere rating is \( V_{\text{rms}} I_{\text{rms}} = \left(\frac{V_m}{\sqrt{2}}\right) \left(\frac{I_m}{2}\right) = \frac{I_m^2 R_L}{2\sqrt{2}} \).
- Dividing gives \( \text{TUF} = \frac{2\sqrt{2}}{\pi^2} \approx 0.287 \).
- This low utilization (\( 28.7\% \)) means the transformer must be substantially oversized for the DC power delivered.
Why other options are incorrect:- Option A: 0.812 is the TUF of a full-wave bridge rectifier.
- Option C: 0.672 is the TUF of a center-tapped full-wave rectifier.
- Option D: 1.000 represents an ideal transformer with zero reactive/harmonic losses.
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