Concept:The \( \text{TUF} \) of a bridge rectifier is higher than that of center-tapped or half-wave circuits because the entire secondary winding carries symmetrical AC current during both half-cycles.
Formula:$$\text{TUF} = \frac{P_{\text{dc}}}{V_{\text{rms}} I_{\text{rms}}}$$
Solution:- For a bridge rectifier:
- $$V_{\text{dc}} = \frac{2V_m}{\pi}, \quad I_{\text{dc}} = \frac{2I_m}{\pi}, \quad V_{\text{rms}} = \frac{V_m}{\sqrt{2}}, \quad I_{\text{rms}} = \frac{I_m}{\sqrt{2}}$$
- $$\text{TUF} = \frac{\left(\frac{2V_m}{\pi}\right)\left(\frac{2I_m}{\pi}\right)}{\left(\frac{V_m}{\sqrt{2}}\right)\left(\frac{I_m}{\sqrt{2}}\right)} = \frac{8}{\pi^2} \approx 0.812$$
Why other options are incorrect:- Option A: \( 0.287 \) is the \( \text{TUF} \) of a half-wave rectifier.
- Option C: \( 0.693 \) is the \( \text{TUF} \) of a center-tapped full-wave rectifier.
- Option D: \( 0.482 \) is the ripple factor of a full-wave rectifier.
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