Half-wave vs full-wave controlled rectifiers – efficiency and ripple (Assertion-Reason) Assertion (A): A half-wave controlled rectifier has poorer efficiency and a higher ripple factor than a full-wave controlled rectifier. Reason (R): Adding a freewheeling diode to a half-wave controlled rectifier improves the load current waveform and the circuit power factor.

Introduction / Context:Converter performance is judged by efficiency, ripple, and power factor. Half-wave circuits utilize only one half cycle, leading to inferior performance relative to full-wave circuits that exploit both half cycles.

Given Data / Assumptions:

• Comparison at similar load and firing conditions.
• Freewheeling diode considered where applicable for inductive loads.

Concept / Approach:

Full-wave converters process both half cycles, delivering higher average output power for the same peak values and significantly lower ripple. Half-wave converters inherently waste half the source cycle and impose larger DC offsets and ripple on the load.

Step-by-Step Solution:

Verification / Alternative check:

Waveform sketches show continuous load current with a freewheel path, yet average power remains lower than full-wave because negative half cycles are still not rectified to the load.

Why Other Options Are Wrong:

• A: Implies correct causal link, which is absent.
• C/D: Would deny a correct statement (either A or R), which they are not.

Common Pitfalls:

Confusing “improvement” within a half-wave circuit with “equivalence” to a full-wave circuit. Even with freewheeling, half-wave utilization remains inferior.

Final Answer:

Both A and R are correct but R is not correct explanation of A