Single-phase half-wave converter with resistive load (α = 0) For a single-phase half-wave controlled rectifier feeding a purely resistive load with input v = Vm sin(ωt) and firing angle α = 0°, determine the average DC output voltage Vdc and the RMS output voltage Vrms.
-
AVdc = Vm/π, Vrms = Vm/2
-
BVdc = 2Vm/π, Vrms = Vm/√2
-
CVdc = Vm/2, Vrms = Vm/π
-
DVdc = Vm/π, Vrms = Vm/√2
-
EVdc = 2Vm/π, Vrms = Vm/2
Answer
Correct Answer: Vdc = Vm/π, Vrms = Vm/2
Explanation
Introduction / Context:Basic rectifier metrics like average (DC) and RMS output voltages are foundational for sizing filters, devices, and loads.
Given Data / Assumptions:
- Half-wave controlled rectifier, resistive load.
- Input v = Vm sin(ωt), firing angle α = 0° (i.e., diode-like conduction during 0 to π).
- Conduction interval: one half-cycle out of each period.
Concept / Approach:For α = 0° and R-load, output waveform is the positive half-sine. Standard results apply: Vdc equals the average of half-sine over a full period; Vrms equals RMS of half-sine over a full period.
Step-by-Step Solution:Average of half-wave rectified sine: Vdc = (1/2π) ∫₀^π Vm sinθ dθ = Vm/π.RMS of half-wave rectified sine: Vrms = √[(1/2π) ∫₀^π (Vm sinθ)^2 dθ] = Vm/2.Therefore, Vdc = Vm/π and Vrms = Vm/2.
Verification / Alternative check:Numerically, for Vm = 1: Vdc ≈ 0.318, Vrms = 0.5, consistent with standard rectifier tables.
Why Other Options Are Wrong:2Vm/π or Vm/√2 pairings: These correspond to full-wave average or sinusoidal RMS, not half-wave rectified values.
Common Pitfalls:Confusing full-wave with half-wave results; mixing peak and RMS of the source instead of the rectified waveform.
Final Answer:Vdc = Vm/π, Vrms = Vm/2