--- title: "DFRT explanation of no starting torque" subject: "Electric Machines II" chapter: 3 key: "semester_5/electric_machines_ii:dfrt-not-self-starting" source: https://www.bctengineer.com/q/semester_5/electric_machines_ii/dfrt-not-self-starting site: BCT Engineering --- # DFRT explanation of no starting torque Electric Machines II · Chapter 3: Fractional Kilowatt Motors - Asked in: 2070 Asar, 8 marks - 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 Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/electric_machines_ii/dfrt-not-self-starting - Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machines_ii/chapter-3 ## Question Starting from double field revolving theory, explain why single phase induction motors are not self starting? ## Short answer The pulsating field of a single-phase winding resolves into two equal counter-rotating fields. At standstill both have slip=1, producing equal and opposite torques, so net starting torque is zero. The motor merely vibrates until a unidirectional initial rotation is given. ## Answer According to **double field revolving theory**, the pulsating field of a single-phase winding is equivalent to two equal fields, each of half the maximum value, rotating in opposite directions at synchronous speed. At standstill these two fields produce equal and opposite torques, so the net starting torque is zero and the motor is **not self-starting**. #### Resolution of the pulsating field The flux of a single-phase winding at space angle $\theta$ is $$ \begin{aligned} \phi(\theta, t) &= \phi_m \cos\omega t \cos\theta \\ &= \frac{\phi_m}{2}\cos(\theta - \omega t) + \frac{\phi_m}{2}\cos(\theta + \omega t) \end{aligned} $$ - First term: field of amplitude $\phi_m/2$ rotating **forward** (anticlockwise) at $\omega$. - Second term: equal field rotating **backward** (clockwise) at $\omega$. ```text t = 0 t = T/8 t = T/4 phi_f phi_b phi_f phi_b phi_f phi_b ^ ^ \ / <-- --> | | \ / resultant resultant resultant = phi_m = 0.707phi_m = 0 (vertical) (vertical) (sum cancels) ``` The two vectors always add along the winding axis, so the resultant only pulsates between $+\phi_m$ and $-\phi_m$. #### Slips with respect to the two fields If the rotor runs at $N$ in the forward direction: $$ s_f = \frac{N_s - N}{N_s} = s, \qquad s_b = \frac{N_s + N}{N_s} = 2 - s $$ #### Torques Each field acts like the field of a three-phase motor and produces a torque–slip curve: - Forward torque $T_f$ (positive) by the forward field - Backward torque $T_b$ (negative) by the backward field - Net torque $T = T_f - T_b$ ```text T | Tf | /\ | ____/ \ Tnet (solid) | / \ \ --+-/------+---\------> N -Ns \ / 0 +Ns | \_/ Tb | s: 2 1 0 ``` #### Why not self-starting - At standstill $N = 0$: $s_f = s_b = 1$. - Both fields cut the rotor at the same speed, induce equal rotor emfs and currents, and produce **equal and opposite torques**: $T_f = T_b$. - Net starting torque $T_{st} = T_f - T_b = 0$. The rotor just vibrates and hums. #### Behaviour once started If the rotor is turned forward by some means, $s_f < 1$ and $s_b > 1$. The forward torque increases and the backward torque falls (the backward rotor current is at nearly $2f$ and largely reactive), so $T_f > T_b$ and the motor accelerates to near synchronous speed. It can run equally well in either direction, depending on the initial push. In the equivalent-circuit form, at $s = 1$, $Z_F = Z_B$ and $P_{gF} = P_{gB}$, giving $T = (P_{gF} - P_{gB})/\omega_s = 0$. #### Remedy Make the motor temporarily two-phase: add an auxiliary winding displaced $90^\circ$ in space carrying a current displaced in time (split-phase, capacitor-start, PSC, CSCR) or use shading coils. This produces a true rotating field and hence starting torque. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The flux phi_m cos(wt) cos(theta) equals: A. Two fields phi_m/2 rotating in opposite directions B. One field phi_m rotating forward C. Two fields phi_m in same direction D. Four fields of phi_m/4 Answer: A. Trigonometric identity splits it into (phi_m/2)cos(theta-wt) + (phi_m/2)cos(theta+wt). (Key point: Resolution of field) 2. At N = 0 the slips s_f and s_b are: A. 0 and 2 B. 1 and 1 C. 1 and 0 D. 2 and 2 Answer: B. s_f = s_b = 1 at standstill. (Key point: Slips at standstill) 3. The net starting torque is zero because: A. T_f = T_b and they act in opposite directions B. The rotor has no flux C. T_f is zero D. T_b is greater Answer: A. Equal rotor emfs and currents give equal and opposite torques. (Key point: Net starting torque) 4. If the rotor is turned forward, then: A. s_f < 1 and s_b > 1, so T_f > T_b and it accelerates B. Torque stays zero C. Backward torque exceeds forward D. The motor stops Answer: A. Forward torque rises, backward torque falls, so the motor keeps running in the direction it was pushed. 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