Bridge rectifier with capacitor filter at 50 Hz: For good smoothing, the ripple period is 1/(2f). What is the minimum recommended time constant R_L C so that the output waveform remains acceptably smooth?

Introduction / Context:
In a full-wave (bridge) rectifier, the charging pulses arrive at twice the mains frequency. The reservoir capacitor must be large enough relative to the load resistance so that the discharge between peaks is small, giving low ripple. Designers often target a time constant at least an order of magnitude larger than the ripple period.

Given Data / Assumptions:

• Supply frequency f = 50 Hz → ripple frequency f_r = 2f = 100 Hz.
• Ripple period T_r = 1/f_r = 10 ms.
• Good waveform shape typically requires R_L C ≥ 10 * T_r.

Concept / Approach:

Capacitor discharge between charging peaks is approximately ΔV ≈ I_load * T_r / C. To make ΔV small for a given I_load, choose C large (or equivalently R_L C large). A common rule-of-thumb is R_L C ≥ 10 T_r so that the exponential decay over T_r is modest.

Step-by-Step Solution:

Verification / Alternative check:

With R_L C = 100 ms, the fractional voltage drop over 10 ms is approximately 1 − exp(−T_r/(R_L C)) ≈ 1 − exp(−0.1) ≈ 9.5%, which is acceptable for many supplies; larger time constants reduce ripple further.

Why Other Options Are Wrong:

• 10 ms equals only one ripple period; ripple would be high.
• 20 ms and 50 ms are improvements but typically still not smooth for many loads.

Common Pitfalls:

• Using the mains period (20 ms) instead of the ripple period (10 ms) for full-wave rectification.

Final Answer:

100 ms.