Question

In a single-phase, single-winding induction motor, let N_s be the synchronous speed and N be the rotor speed. What will be the rotor slip with respect to the backward rotating field?
1. \(\dfrac{N_s+N}{N}\)
2. \(\dfrac{N_s-N}{N_s}\)
3. \(\dfrac{N_s+N}{N_s}\)
4. \(\dfrac{N_s-N}{N}\)

Answer 1

Correct Answer - Option 3 : \(\dfrac{N_s+N}{N_s}\)

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