RMS output voltage of a single-phase full-wave converter (M-2 connection) with R–L load For a single-phase full-wave controlled rectifier (M-2 connection) supplied by v(t) = Vm sin(ωt) and operating with continuous current into an R–L load, what is the expression for the RMS value of the output voltage in terms of Vm and firing angle α?

Introduction / Context:
The RMS output voltage of a phase-controlled full-wave rectifier is needed for thermal ratings and ripple calculations. Unlike the average output, the RMS value depends on the square of the instantaneous waveform over its conduction intervals.

Given Data / Assumptions:

• Input: v(t) = Vm sin(ωt).
• Full-wave, single-phase controlled rectifier (M-2 connection, two SCRs, center-tapped secondary equivalent).
• R–L load with continuous conduction; overlap neglected.
• Firing angle α (0 ≤ α ≤ π).

Concept / Approach:

With continuous current, the output follows the source magnitude during conduction windows: from α to π on the positive half and from π+α to 2π on the negative half (polarity corrected by rectification). RMS is computed from the square of the waveform over one full period, averaged, and square-rooted.

Step-by-Step Solution:

Verification / Alternative check:

For α = 0, V_rms = Vm/√2 (as expected for a full-wave rectified sine). For α → π, RMS tends to 0, consistent with vanishing conduction interval.

Why Other Options Are Wrong:

(b) and (c) are average or fundamental component relations, not true RMS of the chopped waveform; (d) omits the sin 2α term and overestimates RMS for intermediate α.

Common Pitfalls:

Mixing average and RMS expressions; forgetting to account for both conduction intervals in a full period.

Final Answer:

V_rms = Vm * sqrt( (π − α)/(2π) + (sin 2α)/(4π) )