Average DC output voltage for controlled rectifiers (single-phase) Given a peak input voltage Vm and firing angle α (in radians or degrees, as standard in power electronics), determine the expressions for the average output voltage for: a single-phase half-wave controlled rectifier with a resistive load, and a single-phase full-wave controlled bridge rectifier with a resistive load.

Introduction / Context:
In controlled rectifiers, the average (DC) output voltage depends on how long the sinusoidal source is allowed to conduct each half-cycle. The firing angle α delays conduction and therefore reduces the average output. Understanding these relationships is fundamental for designing DC supplies and drives.

Given Data / Assumptions:

• Single-phase sinusoidal input v = Vm sin(ωt).
• Purely resistive load to avoid phase-shift complications.
• Gate firing angle α measured from the start of each relevant half-cycle.

Concept / Approach:
The average value of a periodic waveform equals the integral of the instantaneous voltage over one period divided by the period. Conduction begins at ωt = α and ends at ωt = π for half-wave, while in a full-wave bridge the rectified conduction repeats twice per cycle.

Step-by-Step Solution:
Half-wave: Vdc = (1 / 2π) * ∫ from α to π of Vm sin θ dθ.Compute integral: ∫ sin θ dθ = −cos θ; evaluate → (Vm / (2π)) * (1 + cos α).Full-wave: Two identical conduction intervals per cycle yield Vdc = (1 / π) * ∫ from α to π of Vm sin θ dθ = (Vm / π) * cos α.

Verification / Alternative check:
At α = 0°, half-wave gives Vdc = Vm / (2π) * (1 + 1) = Vm / π, and full-wave gives Vdc = Vm / π, which matches standard results.

Why Other Options Are Wrong:
They misplace the 1/(2π) versus 1/π factor or ignore α, leading to incorrect scaling of the average value.

Common Pitfalls:
Confusing half-wave and full-wave constants or using RMS instead of average when sizing DC loads.

Final Answer:
Half-wave: Vdc = (Vm / (2π)) * (1 + cos α); Full-wave: Vdc = (Vm / π) * cos α