GATE 2020 EE Q.17: solution and answer

GATE 2020 · EEQ.17 · 1 mark · NAT

A three-phase, 50 Hz, 4-pole induction motor runs at no-load with a slip of 1 %. With full load, the slip increases to 5 %. The % speed regulation of the motor (rounded off to 2 decimal places) is ____________.

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Answer and solution

ANSWER

The % speed regulation of the motor is 4.21%.

Official answer: 4.10 to 4.40Source: Official answer key, GATE 2020 EE (IIT Delhi)

Synchronous speed: $N_{s} = \frac{120 f}{P} = \frac{120 \times 50}{4} = 1500 rpm$ , fixed by the 50 Hz supply and 4 poles — it does not change with load.
Rotor speed from slip: $N = N_{s} (1 - s)$ .
No-load speed: $N_{N L} = N_{s} (1 - 0.01) = 0.99 N_{s}$ .
Full-load speed: $N_{F L} = N_{s} (1 - 0.05) = 0.95 N_{s}$ .
% speed regulation, referenced to the full-load (rated) speed, exactly as voltage regulation is referenced to full-load voltage: $% Speed Regulation = \frac{N _{N L} - N _{F L}}{N _{F L}} \times 100$
Substitute: $= \frac{0.99 N _{s} - 0.95 N _{s}}{0.95 N _{s}} \times 100 = \frac{0.04}{0.95} \times 100$
Evaluate: $= 4.2105 \dots % \approx 4.21%$

KEY IDEA

Synchronous speed $N_{s} = 120 f / P$ is fixed by supply frequency and pole count and does not change with load; only the slip changes between no-load and full-load. Speed regulation is defined the same way as voltage regulation — as a percentage change measured relative to the full-load (rated operating) value, i.e. $(N_{N L} - N_{F L}) / N_{F L} \times 100$ .

WATCH OUT FOR

Using the synchronous speed $N_{s}$ in the denominator instead of the full-load speed $N_{F L}$ — that gives $0.04/1 \times 100 = 4.00%$ , which is outside the accepted key range because it silently redefines "regulation" against the wrong reference speed.

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GATE 2020 EE Q.17: solution and answer

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