Concept:A larger filter capacitor stores more charge during conduction peaks, slowing its discharge through \( R_L \) between cycles and reducing the output ripple voltage.
Formula:$$V_{\text{ripple (peak-to-peak)}} = \frac{I_{\text{dc}}}{2 f C} = \frac{V_{\text{dc}}}{2 f R_L C} \propto \frac{1}{C}$$
Solution:- The peak-to-peak ripple voltage \( V_r \) is inversely proportional to the filter capacitance \( C \).
- Increasing \( C \) reduces ripple, delivering a smoother, more stable DC voltage to the load.
Why other options are incorrect:- Option A: Ripple voltage decreases with increasing capacitance; it does not increase.
- Option C: Increasing capacitance damps fluctuations rather than doubling them.
- Option D: Filtering performance depends strongly on capacitance \( C \).
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