Induction motor with constant V/f control – starting torque at reduced frequency A 3-phase induction motor is operated with constant V/f control. Its starting torque at 50 Hz is 100 N·m. What happens to the starting torque if the supply frequency is reduced to 25 Hz while maintaining the same V/f ratio?

Introduction / Context:
With variable-frequency drives, constant V/f control keeps air-gap flux approximately constant. The starting (slip = 1) torque, however, depends not only on flux but also on the relative magnitudes of rotor resistance and standstill reactance, which scale with frequency. Hence, predicting starting torque at reduced frequency requires considering both V and f effects.

Given Data / Assumptions:

• V/f kept constant from 50 Hz to 25 Hz.
• Starting torque at 50 Hz = 100 N·m.
• Rotor parameters not specified (R2, X2 at standstill).

Concept / Approach:

At start (s = 1): T_start ∝ V^2 * R2 / (R2^2 + X2^2). With constant V/f, V ∝ f and X2 ∝ f, but R2 is approximately constant. Reducing f lowers both V and X2 proportionally; the ratio R2/X2 changes, potentially moving operation closer to maximum torque condition. Depending on whether X2 dominated at 50 Hz or R2 dominated at 25 Hz, starting torque can increase, decrease, or stay close to the original value.

Step-by-Step Solution:

Verification / Alternative check:

Many drives texts note that with constant V/f, breakdown torque remains roughly constant, while starting torque can vary with frequency according to rotor circuit parameters and stator/rotor leakage reactances.

Why Other Options Are Wrong:

• Always more / always less: ignores parameter dependence.
• “Very small”: not correct under constant V/f, which preserves flux.

Common Pitfalls:

Assuming T_start scales simply with V^2; overlooking the frequency dependence of X2 and the resulting change of R2/X2 at start.

Final Answer:

may be more or less than 100 N·m depending on motor parameters