GATE 2026 EE Q.20: solution and answer

GATE 2026 · EEQ.20 · 1 mark · MCQ

A certain application requires power at a frequency of 16.67 Hz, while the available grid frequency is 50 Hz. A 3-phase synchronous motor connected to the 50 Hz, 3-phase grid, driving a synchronous generator, is used for this application.

Which one of the following combinations is a suitable choice for the number of poles in the motor and generator, respectively?

(A)
2, 6
(B)
6, 2
(C)
4, 6
(D)
6, 4

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Answer and solution

ANSWER

The correct answer is option (B): 6 poles for the motor, 2 poles for the generator.

Official answer: BSource: Official answer key, GATE 2026 EE (IIT Guwahati)

Since the motor and generator are mechanically coupled on the same shaft, both rotate at the same synchronous speed $N$ (in rpm).
For the motor on the 50 Hz grid: $N = \frac{120 f _{m o t or}}{P _{m o t or}} = \frac{120 \times 50}{P _{m o t or}}$ .
For the generator producing 16.67 Hz at the same speed: $f_{g e n} = \frac{P _{g e n} N}{120}$ .
Eliminating $N$ : $\frac{f _{g e n}}{f _{m o t or}} = \frac{P _{g e n}}{P _{m o t or}}$ , so $\frac{P _{g e n}}{P _{m o t or}} = \frac{16.67}{50} = \frac{1}{3}$ , i.e. $P_{m o t or} = 3 P_{g e n}$ .
Check option (B): $P_{m o t or} = 6, P_{g e n} = 2$ . Then $N = \frac{120 \times 50}{6} = 1000 rpm$ , and $f_{g e n} = \frac{2 \times 1000}{120} = 16.67 Hz$ — matches exactly.
No other option satisfies $P_{m o t or} = 3 P_{g e n}$ : (A) 2, 6 needs $3 \times 6 = 18 \neq = 2$ ; (C) 4, 6 needs $3 \times 6 = 18 \neq = 4$ ; (D) 6, 4 needs $3 \times 4 = 12 \neq = 6$ .
Hence the correct combination is 6 poles for the motor and 2 poles for the generator — option (B).

KEY IDEA

A synchronous motor and a synchronous generator on the same mechanical shaft run at the same speed $N$ ; each machine's own pole count then converts that common $N$ to its own electrical frequency via $N = \frac{120 f}{P}$ .

WATCH OUT FOR

Assuming the frequency ratio equals the pole ratio directly in the same sense (i.e. picking $P_{m o t or} : P_{g e n} = f_{m o t or} : f_{g e n}$ ) instead of inversely — since speed is fixed, a lower output frequency needs FEWER poles on the generator, not more, so $P_{g e n} / P_{m o t or} = f_{g e n} / f_{m o t or}$ .

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GATE 2026 EE Q.20: solution and answer

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