Single-phase phase-controlled heater with firing angle α = 90° What fraction of maximum heating (power) is obtained at α = 90° for a purely resistive load?

Introduction / Context:Single-phase AC regulators control RMS voltage by delaying conduction each half-cycle. For a resistive heater, average power is proportional to the square of the RMS output voltage delivered to the element.

Given Data / Assumptions:

• Purely resistive heater load.
• Firing angle α = 90° (π/2 radians).
• Supply is sinusoidal with RMS value V_rms.

Concept / Approach:The RMS output voltage of a single-phase controller with R load is given by a standard expression involving α. The normalized power P/P_max equals (V_out,rms/V_rms)^2.

Step-by-Step Solution:For R load, V_out,rms^2/V_rms^2 = (1/π) * (π − α) + (1/2π) * sin(2α).Insert α = π/2 → V_out,rms^2/V_rms^2 = (1/π)*(π − π/2) + (1/2π)*sin(π) = 1/2 + 0 = 0.5.Therefore power fraction P/P_max = 0.5 → 50% of maximum heating.

Verification / Alternative check:At α = 0°, P/P_max = 1; at α = 90°, symmetry implies equal on/off areas → 50% power; at α → 180°, power tends to 0.

Why Other Options Are Wrong:

• 25% and 75% contradict the analytical result 0.5.
• “None of the above” is incorrect because 50% is exact.

Common Pitfalls:

• Confusing voltage fraction (≈0.707) with power fraction (square → 0.5).

Final Answer:50% of maximum