Integral-cycle (on–off) control timing For integral-cycle control of a power load, the on–off control period should be compared to the mechanical time constant of the load in which way?

Introduction / Context:
Integral-cycle control (burst firing) applies whole cycles of AC to a load and then skips cycles. For mechanical systems (e.g., heaters with thermal inertia, motors with inertia), choosing the modulation period relative to the system’s time constant affects ripple and performance.

Given Data / Assumptions:

• Power delivered in bursts of complete cycles (on) and off intervals.
• Load exhibits a dominant mechanical or thermal time constant τ_m.
• Goal: smooth average response without large oscillations.

Concept / Approach:
If the on–off period is much shorter than τ_m, the load averages the power, producing a smooth output (e.g., stable temperature or speed). If the period is longer than τ_m, the response will swing noticeably between on and off, creating unacceptable ripple.

Step-by-Step Solution:
Define desired behavior: minimal ripple in the controlled variable.Choose control period T_ctrl such that T_ctrl ≪ τ_m.Therefore the correct relation is that the period should be less than the mechanical time constant.

Verification / Alternative check:
Classical control guidance: sample or modulation intervals should be small compared with the plant time constant to approximate continuous control.

Why Other Options Are Wrong:

• “Greater than”: increases ripple and oscillations.
• “Equal”: offers no general guarantee of smoothness.
• “Unrelated”: incorrect; τ_m directly informs suitable burst periods.

Common Pitfalls:

• Setting very long bursts for convenience, causing thermal or speed cycling.
• Ignoring differences between electrical and mechanical/thermal time constants.

Final Answer:
It should be less than the mechanical time constant