Integral-cycle (on–off) control of an AC regulator with a resistive load If the duty cycle is a (0 < a ≤ 1), what is the overall input power factor seen by the supply?

Introduction / Context:Integral-cycle control (burst firing) turns whole cycles of the AC waveform on and off to control average power. With a purely resistive load, current and voltage are in phase during each “on” cycle, yet the waveform is discontinuous over long intervals, affecting the overall power factor seen by the supply.

Given Data / Assumptions:

• Resistive load (displacement factor during “on” cycles = 1).
• Duty cycle a is the fraction of time the regulator delivers full cycles to the load.
• “On” cycles are ideal (no device drops); “off” cycles draw zero current.

Concept / Approach:

Overall power factor PF is real power divided by apparent power over the averaging interval. Real power scales directly with the duty cycle a (since only a portion of cycles deliver power). Apparent power scales with the overall RMS current, which is reduced by √a because current flows only during an a-fraction of time. Combining these yields PF = √a for a resistive load under integral-cycle control.

Step-by-Step Solution:

Verification / Alternative check:

Check limits: if a = 1 (always on), PF = 1 (unity). If a = 0.25, PF = 0.5, matching the intuitive reduction due to burst operation.

Why Other Options Are Wrong:

PF = a or a^2 overestimates the reduction; PF = 1 ignores the discontinuous nature of current and is only valid for a = 1.

Common Pitfalls:

Confusing displacement factor (which remains 1 in “on” cycles) with overall power factor; forgetting to compute RMS over the entire time window including “off” intervals.

Final Answer:

√a