Induction motor V/f control (constant flux) A 3-phase induction motor is controlled with constant V/f. At 50 Hz, its shaft speed is 1440 rpm. Estimate the expected speed at 25 Hz under similar load (assume slip percentage remains approximately the same).

Introduction / Context:
With constant V/f (volts per hertz) control, an induction motor roughly maintains constant air-gap flux. For the same torque condition, the slip percentage tends to remain similar, so the speed scales mainly with synchronous speed, which is proportional to frequency.

Given Data / Assumptions:

• Shaft speed at 50 Hz: 1440 rpm (typical of a 4-pole motor with Ns = 1500 rpm and about 4% slip).
• New frequency: 25 Hz (half of 50 Hz).
• Constant V/f, similar load → similar slip percentage.
• Neglect deep saturation and nonlinear effects.

Concept / Approach:

Synchronous speed Ns = 120 * f / P. Halving frequency halves Ns. If slip percentage s stays approximately the same, mechanical speed n ≈ (1 − s) * Ns at the new frequency, so it also halves approximately.

Step-by-Step Solution:

Verification / Alternative check:

Constant V/f keeps torque per ampere characteristics comparable; thus slip for the same load torque remains close, yielding half the original speed when frequency is halved.

Why Other Options Are Wrong:

• 1350 rpm corresponds to near 50 Hz operation, not 25 Hz.
• 360 rpm suggests ~50% of 720 rpm, which would require doubled slip percentage under the same torque—unlikely with proper V/f.
• “Either (b) or (c)” undermines the physics; (b) is the consistent result.

Common Pitfalls:

Confusing synchronous speed scaling with load-dependent slip; treating slip as a fixed rpm rather than a percentage under V/f control.

Final Answer:

about 720 rpm