Three-phase full-wave AC voltage regulator (star-connected resistive load) A 3-phase full-wave regulator (thyristor AC voltage controller) feeds a star-connected purely resistive load from a 400 V line-to-line, 50 Hz supply. If the firing angle is set to 160° (very close to natural commutation), what will be the approximate line-to-line RMS output voltage delivered to the load?

Introduction / Context:
Three-phase full-wave AC voltage regulators use phase-controlled thyristors to chop portions of the input sinusoid, thereby reducing the RMS output voltage to the load. Understanding how the firing angle near 180° limits the conduction interval helps predict the delivered voltage, especially for resistive star-connected loads.

Given Data / Assumptions:

• Three-phase controller, star-connected purely resistive load.
• Input line-to-line RMS voltage = 400 V, 50 Hz.
• Firing angle α = 160°.
• Ideal, symmetrical firing and no commutation overlap assumed for conceptual calculation.

Concept / Approach:

In a 3-phase controller, each phase pair conducts when its line-to-line voltage is positive and its corresponding thyristors have been fired. As the firing angle α increases, the effective conduction window shrinks. Beyond about 150°, conduction is highly discontinuous and the effective RMS output collapses toward zero for purely resistive loads because there is almost no time for significant current to flow before the line voltage changes polarity.

Step-by-Step Solution:

Verification / Alternative check:

Rules of thumb and waveforms of 3-phase AC regulators show that beyond α ≈ 150°, output becomes vanishingly small for R-loads; at α very near 180°, it is effectively zero.

Why Other Options Are Wrong:

400 V and about 250 V correspond to little or no control (α near 0°). About 100 V or about 20 V still suggest material conduction which is not the case at α = 160° for an R-load in a 3-phase scheme.

Common Pitfalls:

Confusing single-phase behavior with three-phase conduction windows; forgetting that with resistive loads, late firing severely truncates current flow. Also, neglecting that RMS values depend on the square of the voltage over the conduction interval, which becomes tiny here.

Final Answer:

zero