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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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

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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