For equal load resistance and filter capacitor values, why does a filtered full-wave rectifier produce a smaller ripple voltage than a filtered half-wave rectifier?

Introduction / Context:Ripple amplitude after a capacitor-input filter depends primarily on the load current and the time interval over which the capacitor must supply current between charging peaks. Full-wave rectification doubles the charging frequency, directly influencing ripple size.

Given Data / Assumptions:

• Same load resistance and same filter capacitance for both cases.
• Input line frequency f_line is fixed (e.g., 50/60 Hz).
• Ideal diodes; transformer regulation effects neglected.

Concept / Approach:For a capacitor-input filter, ripple voltage approximately follows ΔV ≈ I_load * Δt / C, where Δt is the time between consecutive charging peaks. In half-wave rectification, Δt ≈ 1/f_line; in full-wave, Δt ≈ 1/(2*f_line). Halving Δt roughly halves the ripple for the same load and C, hence a smaller ripple in the full-wave case.

Step-by-Step Solution:Half-wave: charging at every cycle → peak spacing T = 1/f_line.Full-wave: charging at every half-cycle → peak spacing T/2.Ripple relation: smaller Δt reduces discharge time, so ΔV decreases.Therefore, full-wave filtered outputs exhibit less ripple than half-wave.

Verification / Alternative check:Numerical example: with I_load = 100 mA, C = 1000 µF, f_line = 50 Hz, half-wave ΔV ≈ 0.1 * (1/50) / 0.001 = 2 V; full-wave ΔV ≈ 0.1 * (1/100) / 0.001 = 1 V.

Why Other Options Are Wrong:Longer time between peaks (b) increases ripple, opposite of observed behavior.

(c) is a misconception; more ripple indicates worse filtering.

(d) and (e) do not address the time-between-peaks mechanism that governs ripple.

Common Pitfalls:Ignoring load current in ripple estimates. If I_load increases, ripple increases even with full-wave rectification; always use ΔV ≈ I_load / (C * f_ripple).

Final Answer:there is a shorter time between peaks