Concept:The smoothing performance of a shunt capacitor filter depends on the discharge time constant \( \tau = R_L C \). A larger load resistance \( R_L \) (lower load current) gives a longer discharge time constant, reducing voltage droop between peaks.
Formula:$$V_{\text{ripple (peak-to-peak)}} = \frac{I_{\text{dc}}}{2 f C} = \frac{V_{\text{dc}}}{2 f R_L C}$$
Solution:- When load current \( I_{\text{dc}} \) is low (large \( R_L \)), the capacitor discharges slowly between charging peaks.
- This minimizes the peak-to-peak ripple voltage and keeps the output voltage near \( V_m \).
- At high load currents, the capacitor discharges rapidly, increasing ripple.
Why other options are incorrect:- Option A: High load current discharges the capacitor too quickly, causing large ripple.
- Option C: Fast load current fluctuations degrade capacitor filtering.
- Option D: An open diode prevents any rectification from occurring.
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