Fully controlled bridge feeding a purely resistive load How do the output voltage and output current waveshapes compare in a fully controlled bridge converter with a purely resistive load?

Introduction / Context:
For resistive (R) loads, voltage and current are in phase and proportional. In controlled rectifiers, this means the instantaneous current waveform directly tracks the instantaneous output voltage during conduction, which drives the answer to this question.

Given Data / Assumptions:

• Fully controlled single-phase bridge.
• Purely resistive load (no L or C).
• Ideal switches; ignore commutation overlap.

Concept / Approach:

With R load, Ohm’s law applies instantaneously: i(t) = v(t) / R. Therefore, whenever the converter outputs a particular v(t), the current has exactly the same shape scaled by 1/R. Changes in firing angle α clip or shift voltage segments, and the current follows identically during conduction intervals.

Step-by-Step Solution:

Verification / Alternative check:

Examining waveforms for various α values shows identical shapes where conduction occurs. At α ≥ 90°, the average becomes negative, yet i(t) continues to mirror v(t) during the fired periods.

Why Other Options Are Wrong:

Claims of dissimilarity arise when inductance is present; with pure R, similarity always holds. The special case α = 0° is not required for similarity; it applies for any α.

Common Pitfalls:

Confusing R loads with RL loads; in RL, current lags and waveforms are not similar, especially with freewheeling intervals.

Final Answer:

They are always similar in shape