Power in a single-phase half-wave controlled rectifier (R load) A single-phase half-wave converter feeds a purely resistive load. Let Pdc be the DC output power computed as Vdc * Idc, and Pac be the total real power actually delivered to the load (including ripple components). Which statement is correct for any firing angle?

Introduction / Context:
In rectifiers, learners often equate the product Vdc * Idc with the “power in the load.” However, when the load voltage and current are not pure DC but contain ripple, the real (average) power dissipated in a resistive load includes contributions from both the DC component and the AC ripple components.

Given Data / Assumptions:

• Converter: single-phase half-wave controlled rectifier.
• Load: purely resistive (R), so instantaneous power is v(t) * i(t) and is always non-negative when current flows.
• Definitions: Pdc = Vdc * Idc; Pac = average of v(t) * i(t) over one cycle.

Concept / Approach:
For any periodic waveform, the average power in a resistor equals the sum of the powers due to the DC component and each AC harmonic. The DC product Vdc * Idc accounts only for the DC component. Because there is ripple, additional positive power results from the AC components, thus Pac exceeds Pdc.

Step-by-Step Solution:
Express v(t) and i(t) as DC + AC ripple.Real power in R: P = average{v(t) * i(t)} = Vdc * Idc + average{v_ac(t) * i_ac(t)}.Since i(t) in R is in phase with v(t), the ripple term contributes a positive average.Therefore, Pac = Pdc + positive ripple power ⇒ Pdc < Pac.

Verification / Alternative check:
Fourier analysis shows that for a resistive load, each harmonic contributes I_n² * R / 2 to the average power (when voltage and current contain that harmonic), hence total real power surpasses the DC-only estimate.

Why Other Options Are Wrong:
Pdc = Pac: Only true for pure DC with zero ripple, which never occurs in half-wave control.
Conditional statements with angle: Even at α = 0°, half-wave output still has significant ripple; equality never holds.

Common Pitfalls:
Ignoring ripple contribution to power; confusing apparent and real power; assuming R load eliminates ripple power—which it does not.

Final Answer:
Pdc < Pac for all firing angles