Highly inductive load with single-phase full control For the single-phase full-controlled bridge shown (two devices conducting per interval) supplying a highly inductive load, each thyristor is fired at angle α. What are the conduction intervals of Th1 and Th2, respectively, over one electrical cycle?

Introduction / Context:
With a highly inductive (large L) load, current is nearly constant and cannot change direction instantaneously. In a full-controlled single-phase bridge, each thyristor therefore conducts for more than a half-cycle once triggered, overlapping into the next half-cycle by approximately the firing angle. Determining conduction intervals is a staple skill in rectifier analysis.

Given Data / Assumptions:

• Single-phase full bridge, firing angle α for each device pair.
• Load highly inductive → continuous current; commutation at device firing instants.
• Ideal devices, negligible overlap due to source inductance (focus on load inductance effect).

Concept / Approach:

For continuous current, when a new pair is fired at angle α, current transfers from the outgoing pair to the incoming pair; the outgoing thyristor pair then turns off. Each thyristor thus conducts for a duration of π radians. Counting from its own firing instant, the conduction ends one half-cycle later at α + π. Therefore, the first device pair conducts from α to α + π; the complementary pair conducts from π + α to 2π + α, and so on.

Step-by-Step Solution:

Verification / Alternative check:

Sketching the current flow for a large L load shows rectangle-like current with phase-shifted commutations at every α beyond the standard π boundaries, matching the intervals above.

Why Other Options Are Wrong:

• 0 to π / π to 2π (option a) ignore the firing delay α.
• α to π; π + α to 2π (option b) truncate conduction to less than π radians.
• α to π − α; π + α to 2π − α (option d) are characteristic of discontinuous or resistive cases, not highly inductive continuous current.

Common Pitfalls:

Confusing effects of load inductance (extends conduction) with source inductance (causes overlap but not the same interval extension). For “highly inductive” analyses we assume continuous current transfer at firing instants.

Final Answer:

α to π + α and π + α to 2π + α