Difficulty: Easy
Correct Answer: similar only if firing angle α = 0°
Explanation:
Introduction / Context:
Power-factor and harmonic performance in controlled rectifiers are tied to how closely input current follows the input voltage. With resistive loads, conduction intervals depend on firing angle.
Given Data / Assumptions:
Concept / Approach:
With an R load, current is proportional to instantaneous output voltage during conduction. If α = 0°, conduction occurs during the positive half-cycle similar to a diode rectifier, giving a current waveform that mirrors the voltage half-cycle. For α > 0°, conduction begins later, so current is phase-delayed and truncated, no longer matching the voltage shape.
Step-by-Step Solution:
Set α = 0° → output equals the positive half-cycle of input → current is a half-sine in phase.Increase α → conduction shifts and shortens; waveform differs from the input voltage.Hence, similarity occurs only at α = 0°.
Verification / Alternative check:
Drawing input current for α = 0° shows an in-phase half-sine. For α = 30° or 60°, the current duration and phase shift clearly differ.
Why Other Options Are Wrong:
Always/Never: Too absolute; α determines similarity.
High α: Produces shorter, delayed conduction—not similar.
Common Pitfalls:
Assuming R-load guarantees shape similarity; the firing angle changes conduction timing.
Final Answer:
similar only if firing angle α = 0°
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