--- title: "Operating principle of 3-phase induction motor" subject: "Electrical Machines I" chapter: 5 key: "semester_4/electrical_machines_i:induction-motor-operating-principle" source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/induction-motor-operating-principle site: BCT Engineering --- # Operating principle of 3-phase induction motor Electrical Machines I · Chapter 5: Three-Phase Induction Machines - Asked in: 2070 Bhadra, 8 marks - Video: [How Electric Motors Work - 3 phase AC induction motors ac motor](https://www.youtube.com/watch?v=59HBoIXzX_c) — The Engineering Mindset - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/electrical_machines_i/induction-motor-operating-principle - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-5 ## Question Explain the operating principle of 3 phase induction motor in detail. ## Short answer A balanced 3‑phase supply produces a rotating magnetic field at synchronous speed Ns = 120f/P. This field induces rotor currents; their interaction creates torque, turning the rotor slightly below Ns with slip s = (Ns−N)/Ns. Torque demands more slip, increasing rotor current. Swapping two lines reverses rotation. ## Answer A 3-phase induction motor works on the principle of **electromagnetic induction**: the rotating magnetic field set up by the stator induces current in the rotor, and the force between this current and the field turns the rotor. It is like a transformer with a rotating secondary. #### 1. Production of the rotating magnetic field A balanced 3-phase supply is fed to three stator windings placed 120° apart in space. Their currents are 120° apart in time: $$ \phi_R = \phi_m\sin\omega t,\quad \phi_Y = \phi_m\sin(\omega t - 120^\circ),\quad \phi_B = \phi_m\sin(\omega t - 240^\circ) $$ Adding these at any instant gives a resultant flux of constant magnitude $\phi_r = 1.5\phi_m$ whose axis rotates uniformly. For example, at $\omega t = 0^\circ, 60^\circ, 120^\circ$ the resultant has the same size but has turned by 60°, 120° of electrical angle. The field rotates at **synchronous speed** $$ N_s = \frac{120 f}{P} $$ #### 2. Induced emf and current in the rotor At start the rotor is stationary. The RMF cuts the rotor conductors at speed $N_s$ and induces an emf (Faraday's law). Since the rotor conductors are short-circuited by end rings (cage) or through slip rings and resistors (wound rotor), a rotor current flows. #### 3. Production of torque The current-carrying rotor conductors lie in the stator field and experience a force $F = BIl$. The forces on all conductors add up to a torque. By **Lenz's law** the rotor current opposes its cause — the relative motion between field and rotor — so the rotor turns **in the same direction as the field**, trying to catch it. ```text Stator RMF at Ns --> --------------------- x x x rotor conductors (current) --------------------- Force on rotor --> rotor speed N < Ns ``` #### 4. Slip The rotor can never reach $N_s$: at $N_s$ there would be no relative motion, no induced emf, no current and no torque. It runs at $N$ slightly less than $N_s$: $$ s = \frac{N_s - N}{N_s}, \qquad N = N_s(1 - s) $$ Typical slip: 0.5% at no load, 2–5% at full load. #### 5. Rotor frequency and emf The rotor sees the field at the relative speed $N_s - N = sN_s$, so $$ f_2 = s f, \qquad E_{2r} = sE_2, \qquad X_{2r} = sX_2 $$ #### 6. Effect of load When load increases, the rotor slows slightly, slip increases, rotor emf and current increase, and torque increases until it matches the load. Thus the motor runs at almost constant speed. Example: 6-pole, 50 Hz motor: $N_s = 1000$ rpm; at 4% slip, $N = 960$ rpm and $f_2 = 2$ Hz. Reversal: interchanging any two supply terminals reverses the RMF and so the direction of rotation. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The three stator fluxes of a balanced 3-phase winding add to a resultant that is: A. Constant at 1.5 phi_m and rotating at Ns = 120f/P B. Zero at all instants C. Equal to 3 phi_m and fixed in space D. Pulsating at supply frequency without rotating Answer: A. The vector sum has constant magnitude 1.5 phi_m and its axis rotates uniformly. (Key point: Resultant flux) 2. Why does the rotor turn in the same direction as the field? A. Because the rotor is magnetised permanently B. By Lenz's law, the rotor current opposes the relative motion, so the rotor tries to catch the field C. Because the stator current reverses D. Because friction drags it along Answer: B. The induced current opposes its cause, the relative speed, so the torque makes the rotor follow the field. (Key point: Direction of rotation) 3. Why can the rotor never reach synchronous speed? A. The supply frequency prevents it B. Because of the bearing friction only C. At Ns there is no relative motion, so no emf, current or torque D. Because the stator would overheat Answer: C. With zero relative speed the rotor conductors cut no flux, so torque vanishes and the rotor settles slightly below Ns. (Key point: Why slip exists) 4. What happens when the load on the motor increases? A. Speed rises sharply B. Slip decreases and torque falls C. The field stops rotating D. Rotor slows slightly, slip, rotor emf and current rise, and torque rises to match the load Answer: D. Increased slip gives larger rotor emf and current, so the motor reaches a new balance at almost the same speed. (Key point: Effect of load) ## Related in Electrical Machines I - [Construction and working of 3-phase induction motor](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Ainduction-motor-construction-working) — Electrical Machines I, chapter 5 - [Transformer operating principle and emf equation](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Atransformer-principle-emf-equation) — Electrical Machines I, chapter 2, asked 6× - [Induction motor as generator in grid-connected mode](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Ainduction-generator-grid-connected) — Electrical Machines I, chapter 5, asked 5× - [Induction motor torque and condition for maximum torque](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Aim-torque-derive-max-condition) — Electrical Machines I, chapter 5, asked 2× - [Rotor copper loss and efficiency of 35 kW induction motor](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Aim-4pole-1440rpm-35kw-losses) — Electrical Machines I, chapter 5 - [Induction motor working and torque equation](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Aim-working-torque-equation) — Electrical Machines I, chapter 5 ## Related in other subjects - [Capacitor start motor operating principle](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machines_ii%3Acapacitor-start-motor-principle) — Electric Machines II, chapter 3, asked 2× - [Split phase induction motor](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machine%3Asplit-phase-im) — Electrical Machine, chapter 7 - [Servo motor operating principle](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machine%3Aservo-motor-principle) — Electrical Machine, chapter 7, asked 2× - [Shaded pole motor operating principle](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machines_ii%3Ashaded-pole-motor) — Electric Machines II, chapter 3