Difficulty: Medium
Correct Answer: α < ωt < π and π < ωt < 2π
Explanation:
Introduction / Context:This question examines current continuity and commutation in a half-wave controlled rectifier feeding an R–L load with a free-wheeling diode (FWD). Understanding when the SCR and the diode conduct is essential for predicting output voltage and current waveforms.
Given Data / Assumptions:
Concept / Approach:During the positive half-cycle, after firing at ωt = α, the SCR conducts until the source crosses zero at ωt = π. When the source becomes negative (ωt > π), the inductor tries to keep current flowing; the FWD provides a path, clamping load voltage near zero and preventing negative current through the SCR. Thus, the diode conducts from π to 2π (until the source becomes positive again and the next firing occurs).
Step-by-Step Solution:
SCR interval: from ωt = α (trigger) to ωt = π (natural commutation at zero crossing).FWD interval: from ωt = π to ωt = 2π (freewheeling during negative half-cycle).Therefore: SCR → α < ωt < π; FWD → π < ωt < 2π.Verification / Alternative check:
Waveforms show load voltage equals source during SCR conduction and near zero during freewheeling; load current is continuous if L is adequate.Why Other Options Are Wrong:
Extending beyond 2π or trimming by α in the negative half-cycle is inconsistent with freewheeling action.Common Pitfalls:
Forgetting that the FWD takes over at π; assuming SCR continues into negative half-cycle (it does not).Final Answer:
α < ωt < π and π < ωt < 2π
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