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Question

In a three phase, 4 pole 50 Hz synchronous machine, if the frequency, pole number and load torque will be halved. The motor speed will be :

This question was previously asked in
CUET PG 2026 Agri-Business Management Question Paper (25-Mar-2026) (Shift 2)
The correct answer is
1500 rpm

The synchronous speed of an AC motor is determined by the supply frequency and the number of poles in the machine. The formula for synchronous speed ($N_s$) is:

$N_s = \frac{120 \times f}{P}$

Where:

  • $N_s$ = Synchronous speed in revolutions per minute (rpm)
  • $f$ = Supply frequency in Hertz (Hz)
  • $P$ = Number of poles

Synchronous Speed Calculation: Initial State

Given initial parameters:

  • Frequency ($f_1$) = 50 Hz
  • Number of poles ($P_1$) = 4

Calculate the initial synchronous speed ($N_{s1}$):

$N_{s1} = \frac{120 \times 50}{4} = \frac{6000}{4} = 1500 \text{ rpm}$

Synchronous Speed Calculation: Changed Parameters

The problem states that the frequency and pole number are halved. The load torque change does not affect the synchronous speed itself.

New parameters:

  • New frequency ($f_2$) = $\frac{f_1}{2} = \frac{50}{2} = 25 \text{ Hz}$
  • New number of poles ($P_2$) = $\frac{P_1}{2} = \frac{4}{2} = 2 \text{ poles}$

Calculate the new synchronous speed ($N_{s2}$):

$N_{s2} = \frac{120 \times f_2}{P_2} = \frac{120 \times 25}{2} = \frac{3000}{2} = 1500 \text{ rpm}$

Motor Speed Conclusion

After halving the frequency and the number of poles, the new synchronous speed of the motor remains 1500 rpm. For a synchronous motor, the operating speed is the synchronous speed.

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