Separately excited DC motor fed via a single-phase semi-converter (constant field): If speed is 600 rpm at firing angle α = 0°, estimate the speed when α = 90° (ignore drop and assume speed ∝ average armature voltage).

Introduction / Context:
Phase-controlled rectifiers set the average DC voltage applied to a DC motor armature. With constant field, speed is approximately proportional to average armature voltage. A semi-converter has a specific Vdc–α relationship that lets us estimate speed changes as the firing angle increases.

Given Data / Assumptions:

• Semi-converter (half-controlled bridge) on single-phase AC.
• Constant field flux; neglect armature reaction and losses.
• Speed ∝ average armature voltage Vdc.
• Speed at α = 0° is 600 rpm.

Concept / Approach:

For a single-phase semi-converter: Vdc = (Vm/π) * (1 + cos α), where Vm is the peak of the AC line-to-neutral voltage presented to the converter. Hence Vdc at α = 90° is half of Vdc at α = 0° because cos 90° = 0.

Step-by-Step Solution:

Verification / Alternative check:

The linear proportion holds well when armature current ripple and drops are small relative to applied voltage. Practical speeds may differ slightly due to commutation overlap and IR drop, but 300 rpm is the correct first-order estimate.

Why Other Options Are Wrong:

• 150 rpm or 75 rpm imply larger than predicted reduction.
• Zero speed would require α near 180° and torque loss, not 90°.

Common Pitfalls:

Using the full-converter formula Vdc = (2Vm/π) * cos α instead of the semi-converter expression; mixing up degrees and radians for cos.

Final Answer:

about 300 rpm