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?
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A84% of maximum
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B60% of maximum
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C40% of maximum
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D16% of maximum
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E20% of maximum
Answer
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