Difficulty: Medium
Correct Answer: Less than 12.5 A
Explanation:
Introduction / Context:In a single-phase half-wave controlled rectifier (SCR triggered only on positive half-cycles), the firing angle α determines the conduction interval within each positive half-cycle. Increasing α reduces the conduction width, decreasing both average and RMS currents. This question asks for a qualitative comparison at α = 80° versus α = 160°.
Given Data / Assumptions:
Concept / Approach:
RMS current depends on the square of instantaneous current over the conduction interval. As α increases, the conduction window (α to π) narrows. Going from α = 80° (100° conduction) to α = 160° (20° conduction) dramatically reduces the area under i^2(t), so the RMS value falls much more than by a simple factor of 2; hence it will be less than 12.5 A (half of the original), not merely less than 25 A.
Step-by-Step Solution:
For half-wave resistive load: I_rms^2 ∝ ∫_α^π sin^2 θ dθ.Compute trend: ∫_80°^180° sin^2 θ dθ versus ∫_160°^180° sin^2 θ dθ → the latter is much smaller because the integration range is only 20° near the zero crossing.Therefore I_rms(160°) ≪ I_rms(80°), clearly below 12.5 A.Verification / Alternative check:
Limiting behavior: as α → 180°, conduction interval tends to zero, and RMS current tends to zero; thus increasing α from 80° to 160° must decrease RMS current below half the former value.
Why Other Options Are Wrong:
25 A assumes no change with α; 12.5 A assumes linear halving with α, which is incorrect because RMS depends on the integral of sin^2 over the conduction window; “Less than 25 A” is true but not as informative as “Less than 12.5 A”.
Common Pitfalls:
Equating proportional change in α with proportional change in RMS; confusing average with RMS values; ignoring that conduction near π contributes very little to RMS due to low instantaneous values.
Final Answer:
Less than 12.5 A
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