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Slip and the Torque-Slip Curve

Walk the curve from standstill to synchronism and find that the useful part is a narrow sliver at the end.

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An induction motor's rotor must lag the rotating field for any current to be induced, so slip is the mechanism rather than a defect — and the torque-slip curve rises from a poor starting torque to a breakdown peak before falling to zero exactly at synchronous speed.

Slip

s = (N_s − N)/N_s — 1 at standstill, 0 at synchronous speed. At s = 0 there is no relative motion, so no flux is cut, no current is induced and no torque is produced.

The machine therefore cannot reach synchronous speed, even with zero friction. Full-load slip is typically 2–5%.

The curve, walked from standstill

  1. 1At standstill (s = 1) the rotor sees full supply frequency, so its reactance sX₂ is large. Current is 6–7 times rated but badly lagging, giving only about 1.5 times rated torque.
  2. 2As speed rises, rotor frequency and reactance fall, current comes into phase, and torque rises.
  3. 3At the breakdown point (R₂ = sX₂) torque peaks at typically 2–3 times full load. Load it past this and it stalls.
  4. 4Past the peak the curve is steep and nearly straight — the normal operating region, at 2–5% slip.
  5. 5At s = 0 the torque is exactly zero, so it settles just short of synchronism.

The machine is self-regulating with no controller: more load raises the slip, which raises the torque. And the entire useful range is the narrow sliver between the peak and s = 0 — everything to the left exists only during starting.

Rotor resistance

Increasing rotor resistance moves the peak toward standstill without changing its magnitude — only its position depends on R₂.

A wound-rotor machine exploits this: add external resistance through slip rings to start at maximum torque, then progressively short it out for efficient running. High resistance is excellent at starting and wasteful once running, which is precisely why it is made switchable.

Starting a cage motor

MethodCurrentTorque
Direct-on-line6–7× ratedFull
Star-delta1/31/3 — reduced by the square
AutotransformerSet by tapReduced by the square of the tap
Variable-frequency driveNear ratedFull torque from zero speed

Reducing voltage cuts torque by the square, which is why star-delta is so weak. A VFD reduces frequency alongside voltage to hold V/f — and therefore flux — constant, which is why it displaced every other method once power electronics became cheap.

The numbers you will be asked for

Slip

s = (N_s − N) / N_s

1 at standstill, 0 at synchronism.

Rotor frequency

f_r = s · f

Which is why rotor reactance falls as it speeds up.

Torque

T ∝ s·R₂ / (R₂² + (s·X₂)²)

The whole curve is in this expression.

Maximum torque

at s = R₂ / X₂

Position depends on R₂; magnitude does not.

Rotor copper loss

P_cu,rotor = s · P_airgap

So high slip means high rotor heating.

Advantages and disadvantages

Advantages

  • Self-starting and self-regulating, with no controller.
  • No brushes or slip rings in a cage machine, so almost no maintenance.
  • Rugged and sealed, so it suits hostile environments.
  • Rotor resistance can shift the peak where a wound rotor is available.

Disadvantages

  • Starting current is 6–7 times rated for modest torque.
  • Speed is nearly fixed without a variable-frequency drive.
  • Rotor loss is proportional to slip, so high-slip running overheats it.
  • It always draws magnetising current, so the power factor is poor at light load.

Watch it work

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question 1 / 4

One question at a time. Pick an answer to see why it is right or wrong, then move on — there is no score to keep and nothing is saved.

Why can an induction motor never reach synchronous speed, even with no friction?
Adding rotor resistance in a wound-rotor machine — what does it change?
Why is starting torque poor despite a starting current of six or seven times rated?
Star-delta starting reduces the starting current to a third. What happens to the torque?

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4 still unanswered — the dots above jump straight to them.

 

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