In a single-phase, single-winding induction motor, let Ns be the synchronous speed and N be the rotor speed. What will be the rotor slip with respect to the backward rotating field?
In a single-phase, single-winding induction motor, let Ns be the synchronous speed and N be the rotor speed. What will be the rotor slip with respect to the backward rotating field? Correct Answer <span class="math-tex">\(\dfrac{N_s+N}{N_s}\)</span>
Concept of slip:
- The speed of the rotor structure with respect to stator structure is Nr
- With the rotor running at Nr, the relative speed of the stator rotating magnetic field with respect to the rotor conductors is Ns - Nr in the direction of Ns
- This relative speed is referred to as slip speed.
Where Ns is the synchronous speed of the machine
Ns = 120 f / p
Where,
f is the frequency of the machine
p is the number of poles of the machine
Slip speed = Ns - Nr
Per unit slip or slip s = (Ns - Nr) / Ns
Range of slip (s) is 0 < s < 1
If the rotor is made to revolve in a direction opposite to the rotating magnetic field:
The relative speed between the rotor winding and the rotating magnetic flux becomes Ns + Nr
Backward slip sb = (Ns + Nr) / Ns = 2 - s
As slip varies from 0 < s < 1, backward slip varies from 2 < s < 1
∴ The range of slip variation for this mode is 2 > s > 1
