--- title: "Double revolving field theory" subject: "Electrical Machine" chapter: 7 key: "semester_4/electrical_machine:double-field-revolving-theory" source: https://www.bctengineer.com/q/semester_4/electrical_machine/double-field-revolving-theory site: BCT Engineering --- # Double revolving field theory Electrical Machine · Chapter 7: Fractional Kilowatt Motors - Asked in: 2074 Bhadra, 6 marks; 2072 Asoj; 2070 Magh, 8 marks (asked 3 times) - Video: [double revolving field theory of single phase induction motor | double revolving theory](https://www.youtube.com/watch?v=pXmPCSc9di0) — Mind of Engineer - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/electrical_machine/double-field-revolving-theory - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machine/chapter-7 ## Question Using double revolving field theory, explain the working of a single phase induction motor (with a neat diagram). ## 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. ## Answer 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 $$\phi_m\cos\omega t = \frac{\phi_m}{2}e^{j\omega t} + \frac{\phi_m}{2}e^{-j\omega t}$$ 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) ```text 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. | Condition | Slip f / b | Result | |---|---|---| | Standstill (s = 1) | 1 / 1 | T_f = T_b, so net torque is zero | | Rotor started forward | s / 2 − s | T_f > T_b, net forward torque | | Rotor near N_s (s → 0) | ~0 / ~2 | T_f large, T_b small | ```text 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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. According to the double revolving field theory, a pulsating field phi_m cos(wt) is equivalent to: A. one rotating field of magnitude phi_m at N_s B. two fields of magnitude phi_m/2 rotating in the same direction at N_s C. two fields of magnitude phi_m rotating in opposite directions at N_s D. two fields of magnitude phi_m/2 rotating in opposite directions at N_s Answer: D. The pulsating flux is the sum of a forward and a backward field, each phi_m/2, turning at synchronous speed in opposite directions. (Key point: Pulsating field resolution) 2. If the rotor runs forward with slip s, what is the slip with respect to the backward field? A. s B. 1 - s C. 2 - s D. s - 2 Answer: C. The backward field moves opposite to the rotor, so s_b = (N_s + N)/N_s = 2 - s. (Key point: Slip for the backward field) 3. Why is the net torque zero at standstill? A. Both fields have slip 1, so T_f = T_b and the two torques cancel B. The backward field is zero at standstill C. The rotor has no resistance at standstill D. The forward field cancels the supply flux Answer: A. At N = 0 both slips are 1, the fields are equal, so T_f = T_b and T = T_f - T_b = 0. (Key point: Zero starting torque) 4. What happens when the rotor is pushed in one direction? A. It stops again because the net torque is still zero B. It keeps rotating in that direction because the torque of the field turning that way becomes larger C. It reverses because the backward field is stronger D. It reaches synchronous speed exactly Answer: B. Once the rotor moves, the slips differ and the field rotating the same way gives more torque. The net torque falls to zero slightly below N_s, so the motor never reaches N_s. 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