Induction Motors

The workhorse of industry: rugged, cheap three-phase motors that run by 'slipping' behind a rotating magnetic field. Learn synchronous speed, slip, the torque–slip curve, power flow, starting methods and VFD control.

About 70 % of industrial electrical energy goes into induction motors driving pumps, fans, compressors and conveyors. Three-phase currents in the stator create a magnetic field that rotates at synchronous speed. That field induces currents in the rotor — there's no electrical connection to it — and those currents produce torque.

The rotor always turns a little slower than the field. That difference, called slip, is what makes induction possible and what the whole theory revolves around.

An analogy

Imagine a magnet swept around a copper disc: the disc gets dragged along, but it can never quite catch the magnet — if it did, there'd be no relative motion, no induced current and no pull. The rotor is that disc.

The rotating magnetic field

Three stator windings 120° apart, fed with three currents 120° apart in time, produce a magnetic field of constant strength that rotates at N_s = 120f/P. A 4-pole motor on 50 Hz: 1500 rpm. Swapping any two supply phases reverses the rotation.

Squirrel-cage vs wound rotor

A squirrel-cage rotor is aluminium or copper bars shorted by end rings — almost indestructible and maintenance-free. A wound rotor brings its windings out through slip rings so external resistance can raise starting torque; it is used for very heavy starts.

Slip and the torque–slip curve

At no load, slip is tiny (<1 %). As load increases, slip increases, inducing more rotor current and more torque. Near synchronous speed torque rises almost linearly with slip (the stable operating region); beyond the breakdown torque it falls and the motor stalls. Typical full-load slip: 2–5 %.

Power flow

Input → stator losses → air-gap power P_ag → rotor copper loss (s·P_ag) → mechanical power ((1 − s)·P_ag) → friction/windage → shaft output. Running at high slip is inherently inefficient, since a fraction s of the air-gap power becomes rotor heat.

Starting methods

A direct-on-line (DOL) start draws 6–8× rated current, which can dip the supply voltage.

• Star–delta: 1/3 current and 1/3 torque; for light-starting loads.
• Autotransformer / soft starter: reduced voltage, adjustable ramp.
• Variable-frequency drive (VFD): full torque at rated current from zero speed, plus speed control and energy savings.

Variable-frequency drives and V/f control

Changing the supply frequency changes N_s, so speed can be controlled. Keeping V/f constant holds flux (and torque capability) constant below base speed. Vector (field-oriented) control goes further, controlling flux and torque independently with servo-like dynamic performance. For pumps and fans, power ∝ speed³, so running 20 % slower saves ~50 % energy.

Formulas

• Synchronous speed: N_s = 120 · f / P
• Slip: s = (N_s − N_r) / N_s
• Rotor frequency: f_r = s · f
• Power split: P_rotor-cu = s·P_ag, P_mech = (1 − s)·P_ag
• Slip at max torque: s_Tmax = R₂ / X₂

Worked example: Speed, slip and rotor frequency

Given: 4-pole motor; f = 50 Hz; Full-load speed = 1440 rpm

Find: Synchronous speed, slip and rotor current frequency.

• N_s = 120 × 50 / 4 = 1500 rpm
• s = (1500 − 1440)/1500 = 0.04 = 4 %
• f_r = s·f = 0.04 × 50 = 2 Hz

Answer: 1500 rpm, 4 % slip, 2 Hz rotor currents

Worked example: Rotor power split

Given: Air-gap power P_ag = 10 kW; s = 0.04; Friction & windage = 150 W

Find: Rotor copper loss and shaft output.

• P_rotor-cu = s·P_ag = 0.04 × 10 000 = 400 W
• P_mech = (1 − s)·P_ag = 9600 W
• P_shaft = 9600 − 150 = 9450 W

Answer: 400 W rotor loss, 9.45 kW at the shaft

Common mistakes

• Wrong: Expecting an induction motor to run at synchronous speed. — Right: With zero slip there's no induced rotor current and no torque; it always runs slightly slower.
• Wrong: Using star–delta starting on a high-inertia or loaded conveyor. — Right: Star–delta gives only 1/3 starting torque; heavy loads may fail to accelerate — use a VFD or soft starter sized for the load.

Applications

• Pumps, fans and compressors: The bulk of industrial load; fitting VFDs to them is one of the most cost-effective energy-efficiency measures in industry.
• Conveyors and machine tools: Squirrel-cage motors tolerate dust, vibration and frequent starts with almost no maintenance.
• Electric vehicles: Tesla's early cars used induction motors: no rare-earth magnets and easy field weakening at high speed.

Key takeaways

• N_s = 120f/P; the rotor always slips behind it.
• Rotor copper loss = s·P_ag, so low slip means high efficiency.
• DOL draws 6–8× FLC; star–delta, soft starters and VFDs reduce it.
• VFDs control speed; pumps and fans save energy as speed³.

Going further

• Equivalent circuit: Per phase, an induction motor is a transformer with a secondary load of R₂/s. Splitting R₂/s = R₂ + R₂(1 − s)/s separates rotor copper loss from mechanical power, which is why the model and the power-flow equations match exactly.
• Efficiency classes: IEC 60034-30 defines IE1–IE5. Regulations in many countries mandate IE3 or better for new motors because motor losses are such a large share of industrial energy use.
• Sensorless vector control: Modern drives estimate rotor flux and speed from measured voltages and currents using observers, giving high dynamic performance without a shaft encoder.