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Double revolving field theory

Short answer

A pulsating field equals two half-magnitude fields rotating oppositely at synchronous speed. At standstill, forward and backward torques cancel, net torque zero; given initial rotation, forward torque dominates and motor runs. Thus not self-starting but works once started.

Using double revolving field theory, explain the working of a single phase induction motor (with a neat diagram).

A single-phase winding carrying a.c. produces a pulsating (alternating) field, φ = φ_m cos ωt, along a fixed axis. It does not rotate. The double revolving field theory says that such a pulsating field is equal to two rotating fields of half the magnitude, turning in opposite directions at synchronous speed.

Resolution of the pulsating flux

ϕmcos⁡ωt=ϕm2ejωt+ϕm2e−jωt\phi_m\cos\omega t = \frac{\phi_m}{2}e^{j\omega t} + \frac{\phi_m}{2}e^{-j\omega t}

i.e. the sum of two vectors, each of length φ_m/2:

  • forward field φ_f = φ_m/2, rotating at +N_s (say anticlockwise)
  • backward field φ_b = φ_m/2, rotating at −N_s (clockwise)
 ωt = 0         ωt = 90°             ωt = 180°
 ff and fb      ff and fb point      ff and fb both
 both point     opposite ways, at    point the other
 along +axis:   90° to the axis:     way along the axis:
 sum = φm       sum = 0              sum = -φm

 ff = fb = φm/2, turning in opposite directions;
 resultant = φm cos ωt along the fixed axis

The components at right angles to the axis cancel at every instant, and the components along the axis add to give φ_m cos ωt.

Working of the motor

Let the rotor turn in the forward direction at speed N. The slip with respect to the forward field is s_f = (N_s − N)/N_s = s. The slip with respect to the backward field is

s_b = (N_s + N)/N_s = 2 − s

Each field induces rotor currents and develops its own torque, T_f (forward) and T_b (backward). The net torque is T = T_f − T_b.

ConditionSlip f / bResult
Standstill (s = 1)1 / 1T_f = T_b, so net torque is zero
Rotor started forwards / 2 − sT_f > T_b, net forward torque
Rotor near N_s (s → 0)~0 / ~2T_f large, T_b small
 Torque
  |
  |                                              ffffff
  |                                         fffff      f
  |                                   ffffff    ********f
  |                        fffffffffff      ****        *f
  | fffffffffffffffffffffff             ****             *f
  |f                              ******
--+b*------------------------*****------------------------*f> N
  |                    ******                              b
  | b*             ****             bbbbbbbbbbbbbbbbbbbbbbb
  |  b*        ****      bbbbbbbbbbb
  |   b********    bbbbbb
  |    b      bbbbb
  |     bbbbbb
  |
  -Ns                          0                          +Ns

 f = Tf (forward field), b = Tb (backward field, acting in reverse)
 * = net torque T = Tf - Tb: zero at N = 0, positive for forward
     running, and zero again just below +Ns

The net torque is zero at standstill, so the single-phase induction motor has no starting torque. If the rotor is given a push in either direction, it keeps rotating in that direction, since the torque of the field rotating that way (T_f or T_b) becomes larger than the other. Near synchronous speed the backward torque is small, because the slip 2 − s for the backward field is near 2: the rotor current it causes is at nearly twice the supply frequency, lags heavily and gives little torque. The net torque falls to zero at a speed slightly below N_s, so the motor never reaches N_s.

✓ Answer checked by Claude Opus 5.5 (high effort)

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