Concept:Because a full-wave rectifier conducts during both half-cycles, its average (DC) output voltage is twice that of a half-wave rectifier.
Formula:$$V_{\text{dc}} = \frac{1}{\pi} \int_0^\pi V_m \sin(\theta) d\theta = \frac{2 V_m}{\pi} \approx 0.636 V_m$$
Solution:- In a full-wave rectifier, the waveform repeats every \( \pi \) radians with \( v(t) = V_m \sin(\theta) \).
- Evaluating the average value:
- $$V_{\text{dc}} = \frac{1}{\pi} \int_0^\pi V_m \sin(\theta) d\theta = \frac{V_m}{\pi} [-\cos(\theta)]_0^\pi = \frac{V_m}{\pi} (1 - (-1)) = \frac{2 V_m}{\pi}$$
Why other options are incorrect:- Option A: \( \frac{V_m}{\pi} \) is the DC output voltage of a half-wave rectifier.
- Option B: \( \frac{V_m}{2} \) does not represent the DC average of a full-wave rectified sine wave.
- Option D: \( \sqrt{2} V_m \) exceeds the peak input voltage and is not possible for an unamplified passive rectifier.
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