Concept:Because squaring a waveform makes negative half-cycles positive, the RMS value of a full-wave rectified sine wave equals the RMS value of the original AC sine wave.
Formula:$$V_{\text{rms}} = \sqrt{\frac{1}{\pi} \int_0^\pi V_m^2 \sin^2(\theta) d\theta} = \frac{V_m}{\sqrt{2}} \approx 0.707 V_m$$
Solution:- Evaluating the mean-square integral for full-wave rectification:
- $$V_{\text{rms}}^2 = \frac{1}{\pi} \int_0^\pi V_m^2 \left(\frac{1 - \cos(2\theta)}{2}\right) d\theta = \frac{V_m^2}{2\pi} [\theta]_0^\pi = \frac{V_m^2}{2}$$
- Taking the square root gives:
- $$V_{\text{rms}} = \frac{V_m}{\sqrt{2}} \approx 0.707 V_m$$
Why other options are incorrect:- Option A: \( \frac{V_m}{2} \) is the RMS value of a half-wave rectified waveform.
- Option B: \( \frac{V_m}{\pi} \) is the average DC value of a half-wave rectifier.
- Option D: \( \frac{2 V_m}{\pi} \) is the average DC value of a full-wave rectifier.
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