Concept:A shunt capacitor filter charges to the peak voltage \( V_m \) during conduction and slowly discharges through the load resistance \( R_L \) between peaks, reducing voltage fluctuations.
Formula:$$V_{\text{ripple}} = \frac{I_{\text{dc}}}{2 f C} = \frac{V_{\text{dc}}}{2 f R_L C} \quad (\text{for full-wave})$$
Solution:- The capacitor is connected in parallel (shunt) with the load resistor \( R_L \).
- It provides a low-impedance path to ground for AC ripple frequencies while blocking DC, delivering a smoothed DC voltage across \( R_L \).
Why other options are incorrect:- Option A: A series primary capacitor acts as a capacitive reactance dropper, not a DC filter.
- Option B: Connecting a capacitor in series with the diode blocks the DC output current entirely.
- Option D: A capacitor across the transformer primary only provides AC power factor correction.
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