Concept:Because a full-wave rectifier operates at twice the ripple frequency (\( 2f_{\text{in}} \)), the time between charging peaks is halved, reducing the time available for the filter capacitor to discharge into the load.
Formula:$$V_{\text{ripple}} = \frac{I_{\text{dc}}}{2 f_{\text{in}} C} \quad \text{vs} \quad V_{\text{ripple, half-wave}} = \frac{I_{\text{dc}}}{f_{\text{in}} C}$$
Solution:- Doubling the ripple frequency halves the discharge interval between peaks.
- This means a capacitor of half the physical size can achieve the same ripple reduction as in a half-wave circuit.
Why other options are incorrect:- Option A: Full-wave rectifiers deliver twice the DC voltage and power of half-wave rectifiers.
- Option B: Full-wave rectifiers use two or four semiconductor diodes.
- Option C: Full-wave rectifiers reach the same peak voltage \( V_m \) as half-wave rectifiers.
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