Concept:In a full-wave rectifier, the ripple frequency is \( 2f = 100\text{ Hz} \). The required filter capacitance is calculated from \( C = \frac{I_{\text{dc}}}{2 f V_r} \).
Formula:$$C = \frac{I_{\text{dc}}}{2 f V_r}$$
Solution:- Given: \( I_{\text{dc}} = 1.0\text{ A} \), \( f = 50\text{ Hz} \) (so \( 2f = 100\text{ Hz} \)), and \( V_r = 1.0\% \times 12.0\text{ V} = 0.12\text{ V} \).
- Required capacitance: \( C = \frac{1.0\text{ A}}{100\text{ Hz} \times 0.12\text{ V}} = \frac{1.0}{12} = 0.08333\text{ F} = 83,333\ \mu\text{F} \).
Why other options are incorrect:- Option A: 1000 µF would allow a large ripple voltage of 10 V.
- Option B: 41,667 µF would be used if the ripple frequency were 200 Hz.
- Option D: 500 µF is too small and would cause massive voltage droop.
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.