About two-thirds of all industrial electrical energy is consumed by motors, and most of them are three-phase induction motors — in pumps, fans, compressors, conveyors and mills. That is why the induction motor is, after the transformer, the most examined machine in PGCB, BPDB, BREB/PBS, DPDC, DESCO, NWPGCL, BCS (EEE) and every power-sector viva. This guide explains the rotating magnetic field with an animation, then covers slip, torque, power flow, starting and speed control with solved problems and MCQs.
- Three-phase currents in a three-phase stator winding produce a rotating magnetic field (RMF) of constant magnitude (1.5 Φm) at synchronous speed Ns = 120f/P.
- The RMF induces current in the rotor; the interaction produces torque. The rotor always runs slower than Ns → asynchronous motor.
- Slip s = (Ns − N)/Ns; typically 2–6 % at full load. Rotor frequency fr = s f.
- Air-gap power : rotor copper loss : mechanical power = 1 : s : (1 − s).
- Maximum torque occurs at s = R₂/X₂ and is independent of rotor resistance; torque ∝ V².
1. Construction
An induction motor has two main parts separated by a small air gap (0.4–4 mm — kept small to reduce magnetising current and improve power factor):
- Stator: a hollow cylinder of insulated silicon-steel laminations with slots carrying a three-phase distributed winding (star or delta), wound for P poles.
- Rotor: a laminated cylinder on the shaft, of one of two types:
| Feature | Squirrel-cage rotor | Slip-ring (wound) rotor |
|---|---|---|
| Winding | Copper/aluminium bars short-circuited by end rings | Three-phase winding, star-connected, ends brought to slip rings |
| External resistance | Not possible | Can be added through brushes for starting / speed control |
| Starting torque | Low to moderate | High (with rotor resistance) |
| Cost & maintenance | Cheap, rugged, almost maintenance-free | Costlier; brushes and slip rings need maintenance |
| Use | ≈ 90 % of motors: pumps, fans, compressors | Cranes, hoists, lifts, mills needing high starting torque |
Rotor bars are usually skewed to reduce magnetic hum and prevent cogging (crawling) caused by slot harmonics.
2. The Rotating Magnetic Field (Animation)
The three stator phase windings are placed 120° apart in space, and they carry currents that are 120° apart in time. Each winding produces a pulsating flux along its own axis. Their vector sum, however, is a flux of constant magnitude 1.5 Φm that rotates at synchronous speed.
| Poles (P) | 2 | 4 | 6 | 8 | 10 |
|---|---|---|---|---|---|
| Ns at 50 Hz (rpm) | 3000 | 1500 | 1000 | 750 | 600 |
Reversing direction: interchange any two of the three supply lines — the phase sequence and hence the direction of the RMF reverses.
3. Working Principle
- The RMF sweeps past the stationary rotor conductors at Ns.
- By Faraday’s law, an EMF is induced in the rotor conductors; because they are short-circuited, rotor currents flow.
- Current-carrying conductors in a magnetic field experience a force (F = BIL). By Lenz’s law, the force tends to reduce the relative motion — so the rotor turns in the same direction as the field.
- The rotor can never reach Ns: at Ns there would be no relative motion, no induced EMF, no current and no torque. Hence it is an asynchronous machine — like a transformer with a rotating secondary.
4. Slip and Rotor Frequency
s = (Ns − N) / Ns, N = Ns(1 − s), fr = s f- At standstill (start): N = 0, s = 1, fr = f.
- At no load: s ≈ 0.5–1 %. At full load: s ≈ 2–6 % (larger motors have lower slip).
- Rotor EMF and reactance at slip s: E2s = s E2, X2s = s X2.
- s > 1 → braking (plugging); s < 0 (N > Ns) → the machine works as an induction generator (used in wind turbines).
5. Torque Equation and Torque-Slip Curve
T = (3 / ωs) · s E2² R2 / (R2² + (s X2)²) ⇒ T ∝ s V² R2 / (R2² + s² X2²)- Maximum (breakdown / pull-out) torque occurs when R2 = s X2, i.e. sm = R2/X2; Tmax = (3/ωs) E2²/(2X2) — independent of R2.
- Starting torque (s = 1) is maximum when R2 = X2 — that is why slip-ring motors add rotor resistance at start.
- Low slip (normal running): T ∝ s → nearly linear, stable region. High slip: T ∝ 1/s → unstable region.
- T ∝ V²: a 10 % drop in supply voltage reduces torque by about 19 %.
6. Power Flow and Losses
Stator input → (minus stator copper and core loss) → air-gap power Pg → splits into rotor copper loss and gross mechanical power → (minus friction and windage) → shaft output.
Pg : Pcu2 : Pm = 1 : s : (1 − s)So rotor copper loss = s × air-gap power — a motor running at high slip is inefficient. Gross torque T = Pg/ωs = Pm/ω.
7. Starting Methods
At start s = 1, so the rotor behaves like a short-circuited transformer secondary: a direct-on-line motor draws 5–7 times full-load current at a low power factor. Starters limit this current:
| Method | How it works | Starting current / torque | Used for |
|---|---|---|---|
| Direct-on-line (DOL) | Full voltage applied directly | Full (5–7 × IFL), full torque | Small motors (a few kW) |
| Star-delta | Start in star, run in delta | 1/3 of DOL current and torque | Medium motors designed for delta running |
| Auto-transformer | Reduced voltage x·V at start | Line current and torque ∝ x² | Large squirrel-cage motors |
| Rotor resistance | External resistance in rotor (slip-ring) | Low current, high starting torque | Cranes, hoists, compressors |
| Soft starter | Thyristors ramp voltage smoothly | Adjustable | Pumps, conveyors |
| VFD | Starts at low frequency and voltage | ≈ rated current, full torque | Process control, energy saving |
8. Speed Control
- Variable-frequency drive (V/f control) — the most common modern method; keeping V/f constant keeps flux constant.
- Pole changing — gives two or more fixed speeds (e.g. 1500/750 rpm).
- Rotor resistance control (slip-ring) — simple but wasteful (losses = s × Pg).
- Stator voltage control — for fans and pumps; limited range.
- Slip-power recovery (Kramer, Scherbius) — large slip-ring drives.
9. Solved Problems
Problem 1. A 4-pole, 50 Hz induction motor runs at 1440 rpm. Find slip and rotor frequency.
- Ns = 120 × 50 / 4 = 1500 rpm
- s = (1500 − 1440)/1500 = 0.04 (4 %)
- fr = 0.04 × 50 = 2 Hz
Problem 2. The rotor input (air-gap power) of the same motor is 20 kW. Find rotor copper loss, gross mechanical power and gross torque.
- Pcu2 = s Pg = 0.04 × 20 = 0.8 kW
- Pm = (1 − s) Pg = 19.2 kW
- T = Pg/ωs = 20 000 / (2π × 1500/60) = 20 000/157.08 ≈ 127.3 N·m
Problem 3. A motor has R2 = 0.2 Ω and X2 = 1 Ω per phase (referred to standstill). Find the slip at maximum torque and the speed if Ns = 1000 rpm.
sm = R2/X2 = 0.2; N = 1000 × (1 − 0.2) = 800 rpm.
10. One-Line Exam Facts
- Induction motor = rotating transformer; it is singly excited (only stator connected to supply).
- Power factor is low at no load (≈ 0.1–0.3) and about 0.8–0.9 at full load.
- A single-phase induction motor is not self-starting — its field pulsates (double-revolving-field theory); a capacitor or split-phase winding is added.
- If one phase fails while running (single phasing), the motor keeps running but overheats; at start it will not start.
- Crawling: running at about 1/7 of Ns due to 7th space harmonic. Cogging: refusal to start when rotor and stator slots are equal.
- Maximum torque ∝ 1/X2; full-load torque is typically 40–50 % of maximum torque.
11. Practice MCQs (click to see answer)
1. Synchronous speed of a 6-pole, 50 Hz induction motor is: (a) 3000 rpm (b) 1500 rpm (c) 1000 rpm (d) 750 rpm
2. The magnitude of the rotating magnetic field in a 3-phase motor is: (a) Φm (b) 1.5 Φm (c) 2 Φm (d) 3 Φm
3. Slip of an induction motor at standstill is: (a) 0 (b) 0.5 (c) 1 (d) Infinity
4. Rotor copper loss equals: (a) s × rotor input (b) (1 − s) × rotor input (c) rotor input / s (d) s² × rotor input
5. Direction of rotation of a 3-phase induction motor can be reversed by: (a) Reversing the rotor (b) Interchanging any two supply lines (c) Changing frequency (d) Adding a capacitor
6. Maximum torque of an induction motor is: (a) Proportional to R2 (b) Independent of R2 (c) Inversely proportional to R2 (d) Zero
7. Star-delta starter reduces starting current to: (a) 1/√3 (b) 1/3 (c) 1/2 (d) 1/9 of DOL value
8. Torque of an induction motor is proportional to: (a) V (b) V² (c) 1/V (d) √V
9. A 4-pole, 50 Hz motor runs at 1470 rpm. Rotor frequency is: (a) 1 Hz (b) 2 Hz (c) 0.5 Hz (d) 50 Hz
10. Slip-ring induction motors are preferred where: (a) Speed must be constant (b) High starting torque is needed (c) Cost must be minimum (d) Load is light
11. When an induction machine runs above synchronous speed, it acts as a: (a) Motor (b) Brake (c) Generator (d) Transformer
12. A single-phase induction motor is not self-starting because: (a) Its field is pulsating, not rotating (b) It has no rotor (c) Slip is zero (d) It has high resistance
12. বাংলায় দ্রুত রিভিশন
- তিন-ফেজ স্টেটর কারেন্ট একটি ঘূর্ণায়মান চৌম্বক ক্ষেত্র তৈরি করে, যার মান স্থির (1.5 Φm) এবং গতি Ns = 120f/P।
- রোটরে EMF আবিষ্ট হয়, কারেন্ট চলে এবং টর্ক তৈরি হয়; রোটর কখনো Ns-এ পৌঁছায় না — তাই একে অ্যাসিনক্রোনাস মোটর বলে।
- স্লিপ s = (Ns − N)/Ns; পূর্ণ লোডে ২–৬%; রোটর কম্পাঙ্ক = s·f।
- এয়ার-গ্যাপ পাওয়ার : রোটর কপার লস : যান্ত্রিক শক্তি = 1 : s : (1 − s)।
- সর্বোচ্চ টর্ক যখন s = R₂/X₂; এটি R₂-এর উপর নির্ভর করে না; টর্ক ∝ V²।
- যেকোনো দুটি লাইন অদলবদল করলে ঘূর্ণনের দিক বদলে যায়। স্টার-ডেল্টা স্টার্টার শুরুর কারেন্ট ১/৩ করে।
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References: B. L. Theraja, Electrical Technology Vol. II; P. S. Bimbhra, Electrical Machinery; Chapman, Electric Machinery Fundamentals. Images and animations: original work by Totalecer.
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