A separately excited DC motor (constant field current) is fed from a single-phase fully controlled converter. At firing angle α = 0°, the speed is 500 rpm. Estimate the new speed when α = 45° (assume proportionality to average armature voltage).

Introduction / Context:In a DC drive with a fully controlled single-phase converter (bridge), the average armature voltage V_a,avg is proportional to cos α (continuous current assumption). With constant field, no-load speed is roughly proportional to armature voltage; under similar load, speed scales approximately with V_a,avg after accounting for small armature drop.

Given Data / Assumptions:

• Single-phase fully controlled converter feeding armature (separately excited motor).
• At α = 0°, speed n₀ = 500 rpm.
• At α = 45°, cos 45° ≈ 0.707.
• Assume speed ∝ V_a,avg ∝ cos α (armature drop neglected for estimation).

Concept / Approach:

For a single-phase full converter: V_dc = (2V_m/π) cos α. If all else is fixed, n ∝ V_dc. Hence, n(45°) ≈ 0.707 * n(0°).

Step-by-Step Solution:

Verification / Alternative check:

Including armature resistance would slightly lower speed, but result would still be close to ~340–350 rpm for typical small drops, validating the chosen option.

Why Other Options Are Wrong:

• 250 or 175 or 125 rpm: correspond to much larger α or additional drops not indicated.
• 500 rpm: would require α = 0° again.

Common Pitfalls:

• Using the half-wave formula V_avg ∝ (1 + cos α) instead of full-bridge cos α relation.
• Ignoring that constant field implies speed scales with armature voltage.

Final Answer:

about 350 rpm