A 6-pole, 50-Hz, single-phase induction motor is running at 950 rpm. The backward slip of the motor is:

Concept:

Double field revolving theory:

• According to the double field revolving theory, we can resolve any alternating quantity into two components.
• Each component has a magnitude equal to half of the maximum magnitude of the alternating quantity, and both these components rotate in the opposite direction to each other.

For example,

• A flux, φ can be resolved into two components \(\frac{{{\phi _m}}}{2}\;and - \frac{{{\phi _m}}}{2}\)
• Each of these components rotates in the opposite direction i.e. if one \(\frac{{{\phi _m}}}{2}\) is rotating in a clockwise direction then the other \(\frac{{{\phi _m}}}{2}\) rotates in an anticlockwise direction.
• In a single-phase induction motor, let us call these two components of flux as forwarding component of flux \({\phi _f}\) and the backward component of flux \({\phi _b}\).
• The resultant of these two components of flux at any instant of time gives the value of instantaneous stator flux at that particular instant.
  \({\phi _r} = {\phi _f} + {\phi _b}\)

The forward flux has a slip of s and the backward flux has a slip of 2 - s.

sf = s, and sb = 2 - s

\(s = \frac{{N_s - N_r}}{{N_s}} \)

\({N_s} = \frac{{120 \times f}}{P}\)

Where, Ns = Synchronous speed in rpm

Nr = Rotor speed in rpm

f = Supply frequency in Hz

P = Numeber of poles

Calculation:

Number of poles (P) = 6

Synchronous speed \({N_s} = \frac{{120 \times 50}}{6} = 1000\;rpm\)

Rotor speed (Nr) = 950 rpm

Slip, \(s = \frac{{1000 - 950}}{{1000}} = 0.05\)

Forward slip (sf) = 0.05

Backward slip (sb) = 2 – 0.05 = 1.95

= \(\frac{39}{20}\)

The correct answer is option "3"