Concept:Adding a shunt capacitor filter smooths the output voltage, making the ripple factor inversely proportional to ripple frequency, capacitance, and load resistance.
Formula:$$\gamma = \frac{V_{\text{rms, ripple}}}{V_{\text{dc}}} = \frac{1}{4\sqrt{3} f C R_L} \quad (\text{Full-wave rectifier with capacitor filter})$$
Solution:- For a full-wave rectifier with a capacitor filter, the peak-to-peak ripple voltage is \( V_r = \frac{I_{\text{dc}}}{2 f C} = \frac{V_{\text{dc}}}{2 f C R_L} \).
- Assuming a triangular ripple waveform, the RMS ripple voltage is \( V_{\text{rms, ripple}} = \frac{V_r}{2\sqrt{3}} \).
- Substituting \( V_r \) yields the ripple factor:
- $$\gamma = \frac{V_{\text{rms, ripple}}}{V_{\text{dc}}} = \frac{1}{4\sqrt{3} f C R_L}$$
Why other options are incorrect:- Option B: Ripple factor is inversely proportional to filter capacitance; larger capacitance reduces ripple rather than increasing it.
- Options C & D: These expressions are dimensionally incorrect for the dimensionless ripple factor.
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