Concept:A filter capacitor acts as a local energy reservoir, charging to peak voltage during diode conduction and discharging slowly through the load during non-conduction intervals.
Formula:$$V_{\text{ripple}} \approx \frac{I_{\text{dc}}}{f C} \quad (\text{or } \frac{I_{\text{dc}}}{2f C} \text{ for full-wave})$$
Solution:- As the rectified voltage rises to its peak, the shunt capacitor charges rapidly to \( V_m \).
- When the rectified voltage drops below the capacitor voltage, the diodes turn off and the capacitor discharges slowly into the load resistor.
- This reduces voltage fluctuations (ripple), producing a steadier DC output.
Why other options are incorrect:- Option B: Capacitors store and release energy; they cannot act as active voltage amplifiers.
- Option C: Passive capacitors do not alter the input line frequency.
- Option D: Capacitors cannot alter the intrinsic PN junction physics of reverse-biased diodes.
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.