# What would be the backward field slip of a single phase induction motor running at N r.p.m, when it is given that is synchronous speed is Ns and slip

170 views
in General
closed
What would be the backward field slip of a single phase induction motor running at N r.p.m, when it is given that is synchronous speed is Ns and slip with respect to forward field is s?
1. s
2. -s
3. 1 - s
4. 2 - s

by (30.0k points)
selected

Correct Answer - Option 4 : 2 - s

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 = Number of poles