Concept:When load resistance \(R_L\) decreases, the load draws more current, causing the filter capacitor to discharge more rapidly between rectifier conduction peaks. This lowers the average DC output voltage and increases peak-to-peak ripple voltage.
Formula:$$V_{\text{ripple (p-p)}} = \frac{I_{\text{dc}}}{f C} = \frac{V_{\text{dc}}}{f C R_L}$$
$$V_{\text{dc}} = V_m - \frac{V_{\text{ripple (p-p)}}}{2} = V_m - \frac{I_{\text{dc}}}{2 f C}$$
Solution:- Decreasing \(R_L\) increases the discharge rate: \(I_{\text{dc}} = V_{\text{dc}} / R_L\).
- Faster discharge deepens the voltage drop between charging cycles, increasing ripple voltage \(V_{\text{ripple}}\).
- As the valleys between cycles drop lower, the average DC output voltage \(V_{\text{dc}}\) decreases.
Why other options are incorrect:- Option B: Higher load current causes greater discharge, degrading rather than improving filtering.
- Option C: Without an active voltage regulator, the output voltage drops under increased load.
- Option D: Ripple frequency is set solely by the rectifier topology and line frequency (\(f\) or \(2f\)), not by \(R_L\).
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