Concept:In a capacitor-input filter, the capacitor discharges through the load between recharge peaks. A smaller load resistance increases the discharge rate, lowering the average DC output voltage.
Formula:$$V_{\text{dc}} = V_m - \frac{I_{\text{dc}}}{4 f C} = V_m - \frac{V_{\text{dc}}}{4 f C R_L}$$
Solution:- As load current \( I_{\text{dc}} \) increases (smaller \( R_L \)), the capacitor discharges more deeply during non-conduction intervals.
- This increases the peak-to-peak ripple voltage and decreases the average DC output voltage (\( V_{\text{dc}} \)).
- Consequently, simple capacitor filters have poorer voltage regulation than active regulators or choke-input filters.
Why other options are incorrect:- Option B: Capacitance is a fixed physical value that does not increase with load current.
- Option C: Heavy loads increase forward conduction currents; they do not cause reverse breakdown.
- Option D: AC line frequency is set by the utility supply and is independent of rectifier load current.
Quality & Fidelity Assurance:
Every question on BeambePrep is rigorously curated against the official PMDC syllabus with zero filler, zero out-of-syllabus content, and zero typos. When an authentic past paper originally contained a historical mistake or ambiguity from the examining board (such as UHS or NUMS), BeambePrep faithfully reflects the original paper while detailing the nuance and scientific consensus in the autopsy above.