Concept:A capacitor has capacitive reactance \( X_C = \frac{1}{2\pi f C} \). It acts as an open circuit to DC while providing a low-impedance path to ground for AC ripple frequencies.
Formula:$$X_C(\text{dc}) = \infty, \quad X_C(\text{ac}) = \frac{1}{2\pi f C} \ll R_L$$
Solution:- For DC (\( f = 0 \)), capacitive reactance is infinite (\( X_C = \infty \)), so DC current flows entirely through the parallel load resistor \( R_L \).
- For AC ripple frequencies, \( X_C \ll R_L \), providing a low-impedance bypass path that diverts AC ripple away from the load.
- The capacitor also stores energy at peak voltages and releases it as the voltage falls, reducing output ripple.
Why other options are incorrect:- Option B: The capacitor blocks DC from shunting to ground, ensuring DC flows through the load resistor.
- Option C: A capacitor filter affects PIV moderately (\( \text{PIV} \approx 2V_m \) in half-wave), not infinitely.
- Option D: Capacitive reactance decreases at higher frequencies, conducting AC more readily rather than acting as an open circuit.
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.