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Double field revolving theory

Short answer

A pulsating field equals two constant-magnitude fields rotating oppositely at synchronous speed. At standstill torques cancel, so starting torque is zero. When running, forward torque dominates, giving net torque; motor not self-starting but runs if started.

State and explain the double field revolving theory of single phase induction motor with detailed diagram (and expressions).

Double field revolving theory states that a pulsating (alternating) magnetic field of maximum value ϕm\phi_m can be replaced by two fields, each of constant magnitude ϕm/2\phi_m/2, rotating in opposite directions at synchronous speed Ns=120f/PN_s = 120f/P.

Pulsating field resolved

The single-phase stator winding produces a flux that only alternates along its axis:

ϕ=ϕmcos⁡ωt\phi = \phi_m \cos\omega t

This is equal to the sum of two rotating phasors ϕf\phi_f (forward) and ϕb\phi_b (backward), each ϕm/2\phi_m/2, turning at ω\omega in opposite directions:

ϕ=ϕm2ejωt+ϕm2e−jωt=ϕmcos⁡ωt\phi = \frac{\phi_m}{2}e^{j\omega t} + \frac{\phi_m}{2}e^{-j\omega t} = \phi_m\cos\omega t
 wt = 0      wt = 90 deg     wt = 180 deg
  ^ ^          <-- -->          v v
  | |  phi_f  phi_b               | |
  sum = phi_m  sum = 0       sum = -phi_m
 (both along  (cancel)       (both reversed)
  the axis)
 phi_f turns anticlockwise, phi_b clockwise

At every instant the vertical components add and the horizontal components cancel, so the resultant is the pulsating field.

Torque at standstill - not self-starting

  • At standstill each rotating field induces equal rotor currents (slip = 1 for both), so the forward torque TfT_f and backward torque TbT_b are equal and opposite.
  • Net starting torque T=Tf−Tb=0T = T_f - T_b = 0. Hence a single-phase induction motor is not self-starting.

Torque when running

If the rotor turns at speed NN in the forward direction:

sf=Ns−NNs=s,sb=Ns+NNs=2−ss_f = \frac{N_s - N}{N_s} = s, \qquad s_b = \frac{N_s + N}{N_s} = 2 - s

Using the induction motor torque expression for each field:

Tf=KsR2R22+(sX2)2,Tb=−K(2−s)R2R22+((2−s)X2)2T_f = \frac{K s R_2}{R_2^2 + (sX_2)^2}, \qquad T_b = -\frac{K (2-s) R_2}{R_2^2 + \left((2-s)X_2\right)^2} T=Tf+TbT = T_f + T_b
  • Near normal speed ss is small, 2−s≈22 - s \approx 2: the backward rotor current sees a large frequency (2−s)f(2-s)f and high reactance, so TbT_b is small and Tf≫∣Tb∣T_f \gg |T_b|. The motor keeps running in whichever direction it was started.
        T
        |            _
        |         .-' \   forward run
        |      .-'     |  (Tf > Tb)
 -Ns ---+----.'--------+---> speed
     |  |  .' 0        Ns
      \_|.'
 reverse run      net T = 0 at N = 0
 (mirror image)

The resultant torque-speed curve passes through zero at standstill and is symmetrical: positive in the forward direction, negative in the reverse direction.

Equivalent circuit (based on DFRT)

The rotor is split into two halves, one for each field: forward half with 0.5R2′/s0.5R_2'/s and 0.5X2′0.5X_2', backward half with 0.5R2′/(2−s)0.5R_2'/(2-s) and 0.5X2′0.5X_2', each with 0.5Xm0.5X_m in parallel, in series with the stator R1+jX1R_1 + jX_1.

 o--R1--X1--+--0.5X2'--0.5R2'/s ------+
 V          |  (0.5Xm parallel)  forward
            +--0.5X2'--0.5R2'/(2-s)---+
            |  (0.5Xm parallel)  backward
 o----------+-------------------------+

Hence, an auxiliary means (split phase, capacitor, shaded pole) is needed only to start the motor; once running, the forward field dominates.

✓ Answer checked by Claude Sonnet 5.5

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