Physics • Year 12 • Module 6: Electromagnetism • Lesson 16
AC Induction Motors and Generators
Apply the synchronous speed formula, interpret motor performance data, compare motor types, and predict outcomes in generator scenarios.
1. Synchronous speed calculations — motor and generator scenarios
Use the formula ns = 120f / p for all calculations. Show full working. 8 marks
1.1 A 6-pole induction motor is connected to a 50 Hz supply. Calculate the synchronous speed of the stator field in RPM. (1 mark)
1.2 The rotor of this 6-pole motor turns at 970 RPM under full load. Calculate (a) the slip speed and (b) the percentage slip. (3 marks)
1.3 A hydroelectric power station generator in the Snowy Mountains must produce 50 Hz electricity and has 12 poles. At what rotational speed (RPM) must the turbine spin? (1 mark)
1.4 A large thermal power station uses 2-pole generators. If the turbines are running at 2850 RPM instead of the design speed, calculate the frequency of the AC output. Explain why maintaining exactly 50 Hz is important for the national electricity grid. (3 marks)
2. Interpret motor performance data
A factory engineer tests three induction motors (M1, M2, M3) connected to a 50 Hz supply. The table records the motor specifications and measured rotor speeds. 7 marks
| Motor | Number of poles | Synchronous speed (RPM) | Measured rotor speed (RPM) | Slip speed (RPM) | % Slip |
|---|---|---|---|---|---|
| M1 | 2 | 2910 | |||
| M2 | 4 | 1440 | |||
| M3 | 8 | 720 |
2.1 Complete all blank cells in the table above. (6 marks)
2.2 The engineer observes that all three motors have a similar percentage slip even though their synchronous speeds are very different. Explain why this is expected based on the operating principle of induction motors. (1 mark)
3. Compare: induction motor vs power station generator
Complete the two-column table below. For each feature, write a concise description that contrasts the two devices. 10 marks (1 per cell)
| Feature | AC Induction Motor | Power Station Generator |
|---|---|---|
| Energy conversion | ||
| Part that rotates | ||
| How the rotating field / flux is produced | ||
| Role of slip | ||
| Are brushes required? |
4. Predict and justify — load change in an industrial motor
A 4-pole induction motor operating at 50 Hz runs at 1455 RPM when driving a conveyor belt. A heavy load is suddenly placed on the belt. 5 marks
4.1 Calculate the percentage slip before the load is applied. (2 marks)
4.2 Predict what happens to (a) the rotor speed, (b) the percentage slip, and (c) the current drawn from the supply when the heavy load is applied. Justify each prediction using the operating principles of the induction motor. (3 marks)
Q1.1 — Synchronous speed of 6-pole motor
ns = 120 × 50 / 6 = 1000 RPM. (1 mark)
Q1.2 — Slip speed and percentage slip
(a) Slip speed = 1000 − 970 = 30 RPM. (1 mark)
(b) % slip = 30 / 1000 × 100 = 3.0%. (1 mark for correct formula and substitution; 1 mark for correct answer)
Q1.3 — Snowy Mountains generator speed
ns = 120 × 50 / 12 = 500 RPM. (1 mark) Note: generators run at synchronous speed (no slip).
Q1.4 — Off-speed generator frequency and grid importance
f = ns × p / 120 = 2850 × 2 / 120 = 47.5 Hz. (1 mark for formula rearrangement; 1 mark for correct value)
Grid frequency must be maintained at exactly 50 Hz because all connected AC equipment (motors, transformers, clocks, synchronised generators) is designed for 50 Hz. A deviation causes synchronous generators to fall out of step, motors to run at incorrect speeds, and protection relays to trip; sustained frequency deviation can cause grid-wide blackouts. (1 mark — any valid consequence)
Q2.1 — Motor performance table
M1: ns = 3000 RPM; slip speed = 3000 − 2910 = 90 RPM; % slip = 90/3000 × 100 = 3.0%.
M2: ns = 1500 RPM; slip speed = 1500 − 1440 = 60 RPM; % slip = 60/1500 × 100 = 4.0%.
M3: ns = 750 RPM; slip speed = 750 − 720 = 30 RPM; % slip = 30/750 × 100 = 4.0%.
Award 2 marks per motor: 1 for synchronous speed, 1 for both slip values correct.
Q2.2 — Why similar % slip across motors
Percentage slip reflects the amount of relative motion between the stator field and rotor that is needed to maintain the torque required by the load. For motors of similar design driving similar fractional loads, the torque-producing mechanism (electromagnetic induction) requires the same fraction of relative motion regardless of absolute speed. Typical well-designed induction motors are built to operate at 2–5% slip at rated load. (Accept any response that links slip % to the load torque requirement and the induction mechanism.)
Q3 — Comparison table
Energy conversion: Motor: electrical → mechanical. Generator: mechanical → electrical.
Part that rotates: Motor: rotor (squirrel cage). Generator: rotor (DC-excited electromagnet).
How rotating field/flux is produced: Motor: stator windings carry three-phase AC, creating a rotating field. Generator: rotor carries DC-excited coil that rotates mechanically, sweeping flux through the stationary stator windings.
Role of slip: Motor: slip (2–5%) is essential — without it, no flux change in rotor, no induced current, no torque. Generator: no slip; generator runs at exactly synchronous speed to produce the correct output frequency.
Brushes required: Motor: No. Generator: Yes, to supply DC excitation current to the rotating rotor winding.
Q4.1 — % slip before load
ns = 120 × 50 / 4 = 1500 RPM. Slip speed = 1500 − 1455 = 45 RPM. % slip = 45 / 1500 × 100 = 3.0%. (1 mark synchronous speed and slip speed; 1 mark correct % slip)
Q4.2 — Effect of heavy load
(a) Rotor speed decreases: the extra mechanical load resists rotation and slows the rotor. (1 mark)
(b) Percentage slip increases: the rotor falls further behind the stator field, increasing relative motion and therefore flux change in the rotor bars. (1 mark)
(c) Current drawn increases: greater flux change induces larger currents in the rotor, which in turn require higher currents in the stator windings to maintain the rotating field; the motor draws more power from the supply to meet the increased mechanical demand. (1 mark)