Concept:The ripple factor (\( \gamma \)) measures the proportion of residual AC ripple voltage relative to the DC output voltage in a rectified waveform.
Formula:$$\gamma = \sqrt{\left(\frac{V_{\text{rms}}}{V_{\text{dc}}}\right)^2 - 1}$$
Solution:- For a half-wave rectifier:
- $$V_{\text{rms}} = \frac{V_m}{2}, \quad V_{\text{dc}} = \frac{V_m}{\pi} \implies \gamma_{\text{HW}} = \sqrt{\left(\frac{\pi}{2}\right)^2 - 1} = \sqrt{1.467} \approx 1.21$$
- For a full-wave rectifier:
- $$V_{\text{rms}} = \frac{V_m}{\sqrt{2}}, \quad V_{\text{dc}} = \frac{2V_m}{\pi} \implies \gamma_{\text{FW}} = \sqrt{\left(\frac{\pi}{2\sqrt{2}}\right)^2 - 1} = \sqrt{0.233} \approx 0.482$$
Why other options are incorrect:- Option A: The values are inverted; half-wave rectifiers produce more ripple than full-wave rectifiers.
- Option B: Ripple factors are determined by waveform geometry and are not unity for standard rectifiers.
- Option C: A ripple factor of zero corresponds to pure, unrippled DC, which requires extensive filtering.
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