Correct Answer - Option 4 : 1440 rpm
The synchronous speed is given by
\({N_s} = \frac{{120f}}{P}\)
f is the frequency in Hz or C/s
P is the number of poles
The induction motor rotates at a speed (Nr) close to but less than the synchronous speed.
Slip of an induction motor is given by,
\(s = \frac{{{N_s} - {N_r}}}{{{N_s}}}\)
⇒ Nr = (1 – s) Ns
Where Ns is the synchronous speed
And Nr is the rotor speed
The frequency of rotor current = sf
Application:
Given,
P = 4
f = 50 Hz
s = 4% = 0.04
∴ \({N_s} = \frac{{120f}}{P}=\frac{120× 50}{4}= 1500\ rpm\)
From above concept,
Nr = (1 – s) Ns = (1 - 0.04) × 1500 = 1440 rpm
