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On–off (integral-cycle) control of an electric heater using thyristors: if the duty ratio α = 0.4, what fraction of maximum heating power is delivered?

Difficulty: Easy

Correct Answer: 40% of maximum

Explanation:


Introduction / Context:
On–off (burst or integral-cycle) control applies whole cycles of AC to a resistive heater, then skips cycles. Average power is proportional to the duty ratio over a control window. This is common in industrial heating due to low EMI compared to phase control.


Given Data / Assumptions:

  • Resistive load (heater), so P ∝ V_rms^2 and scales linearly with applied cycle fraction.
  • Duty ratio α = fraction of time (or cycles) power is applied within a fixed period.
  • Line voltage constant when applied.


Concept / Approach:
For integral-cycle control, during ON cycles the heater sees full voltage; during OFF cycles it sees zero. Over the control window, average power equals α * P_max, where P_max is power with continuous ON (α = 1). Thus heating fraction equals α expressed as a percentage.


Step-by-Step Solution:

Duty ratio α = 0.4.Average power P_avg = α * P_max = 0.4 * P_max.Hence heating fraction = 40% of maximum.


Verification / Alternative check:

Equivalent thermal response integrates power; for purely resistive loads, no reactive effects alter average energy per window.


Why Other Options Are Wrong:

16% would correspond to α = 0.16; 60% corresponds to α = 0.6; 84% to α = 0.84; 20% to α = 0.2.


Common Pitfalls:

Confusing phase-angle control (nonlinear relation) with integral-cycle control (linear in α).


Final Answer:

40% of maximum
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