Torque equation of three phase induction motor

The torque equation of a three-phase induction motor tells you how much shaft torque the motor produces at a given slip. It also explains why a slip-ring motor can start with a heavy load, and why a squirrel cage motor behaves differently. This article builds the equation step by step, then solves a worked example.

Table of Contents

• Basic torque relation
• Starting torque
• Torque under running condition
• Maximum (breakdown) torque
• Worked example
• Common mistakes
• Practice questions
• Frequently Asked Questions

Assumptions and symbols

We use per-phase values for the rotor and neglect the stator winding resistance and leakage reactance. With this assumption, the rotor emf is proportional to the supply voltage. All torques are in N·m.

• Φ = flux per stator pole (Wb)
• E₂ = rotor emf per phase at standstill (V)
• R₂ = rotor resistance per phase (Ω)
• X₂ = rotor reactance per phase at standstill (Ω). It equals 2πfL₂, where f is the supply frequency and L₂ is the rotor leakage inductance.
• s = slip = (N_s − N) / N_s
• N_s = synchronous speed in revolutions per second (rps) = (120f/P) / 60
• φ₂ = phase angle between rotor emf and rotor current

Note: The numbers in these symbols are an engineering convention used to distinguish between the two main parts of the induction motor:

• Subscript 1 refers to the Stator (the stationary input side).
• Subscript 2 refers to the Rotor (the rotating output side).

Basic torque relation

Torque comes from 'the rotor current acting on the rotating flux'. But only the part of the rotor current that is in phase with the rotor emf produces useful torque. That is why the rotor power factor appears in the equation.

T ∝ Φ I_r cosφ₂
T = k Φ I_r cosφ₂

Here I_r is the rotor current per phase under the operating condition (at standstill, I_r becomes I₂). Since the rotor emf is proportional to the flux (E₂ ∝ Φ):

T = k₁ E₂ I_r cosφ₂,   where k₁ = 3 / (2πN_s)

Why 3/(2πN_s)? The rotor input power is P₂ = 3E₂I_rcosφ₂. Torque is power divided by synchronous angular speed (ω_s = 2πN_s), so T = P₂/ω_s.

Therefore, T = (3E₂I_rcosφ₂) / (2πN_s))

Starting torque

Starting torque is the torque at the instant of starting, when the rotor is at rest (s = 1). Rotor frequency equals supply frequency, so the rotor reactance is the full X₂.

Z₂ = √(R₂² + X₂²)
I₂ = E₂ / √(R₂² + X₂²)
cosφ₂ = R₂ / √(R₂² + X₂²)

Substituting in T = k₁E₂I₂cosφ₂:

T_st = [3 / (2πN_s)] × E₂² R₂ / (R₂² + X₂²)

Compared with full-load torque, starting torque can be higher or lower. It depends on R₂. A cage motor has low R₂, so its starting power factor is poor and its starting torque is modest.

Condition for maximum starting torque

Keep the supply voltage constant. Then Φ, E₂ and X₂ stay constant, and only R₂ varies (possible in a slip-ring motor). Differentiating T_st with respect to R₂ and equating to zero gives:

R₂ = X₂
(so R₂² + X₂² = 2R₂²)

Maximum starting torque occurs when rotor resistance equals standstill rotor reactance. The rotor power factor is then cosφ₂ = 1/√2 ≈ 0.707 lagging.

Also Read: How Rotating Magnetic Field is Produced in Induction Motors

Torque under running condition

When the rotor turns at slip s, the rotating field cuts the rotor conductors at a slower relative speed. So the rotor frequency becomes sf, and:

• Rotor emf per phase: E_r = sE₂
• Rotor reactance per phase: sX₂
• Rotor resistance per phase: R₂ (unchanged)

I_r = sE₂ / √(R₂² + (sX₂)²)
cosφ₂ = R₂ / √(R₂² + (sX₂)²)

The rotor input power is P₂ = 3E₂I_rcosφ₂. Dividing by ω_s=2πN_s gives the running torque:

T = [3 / (2πN_s)] × s E₂² R₂ / (R₂² + (sX₂)²)

Three useful observations:

• Small slip (normal running): (sX₂)² is tiny compared with R₂², so T ∝ s. The torque-slip curve is nearly a straight line.
• Voltage: E₂ ∝ V, so T ∝ V² at a given slip.
• Large slip: R₂ becomes small next to sX₂, so T ∝ 1/s. Torque falls as slip increases.

Maximum (breakdown) torque

Differentiating T with respect to s and equating to zero shows that torque is maximum when rotor reactance equals rotor resistance:

sX₂ = R₂
s_m = R₂ / X₂ (slip at maximum torque)

Substituting s_m back into the torque equation:

T_max = 3E₂² / (4πN_sX₂)

Key points:

• T_max does not depend on R₂. Changing R₂ only shifts where (at what slip) the peak occurs.
• T_max ∝ V², and it falls as X₂ rises.
• If R₂ ≥ X₂, then s_m ≥ 1, so maximum torque is available at start.
• The ratio of starting to maximum torque is T_st/T_max = 2s_m / (1 + s_m²).
• From s = 0 to s_m the operation is stable. Beyond s_m, an increase in load slows the motor further and it can stall.

Worked example

A 4-pole, 50 Hz, three-phase slip-ring induction motor has a standstill rotor emf of 120 V per phase, R₂ = 0.1 Ω and X₂ = 1.0 Ω per phase. Find (a) the starting torque, (b) the torque at 4% slip, (c) the maximum torque and its slip, and (d) the external resistance per phase for maximum starting torque.

Step 1: Synchronous speed. N_s = 120 × 50 / 4 = 1500 rpm = 25 rps. So 2πN_s = 157.08 rad/s.

Common factor: 3E₂² / (2πN_s) = 3 × 120² / 157.08 = 275.0 (units chosen so torque is in N·m).

(a) Starting torque (s = 1):
T_st = 275.0 × 0.1 / (0.1² + 1.0²) = 275.0 × 0.0990 ≈ 27.2 N·m

(b) Torque at s = 0.04:
sX₂ = 0.04 Ω. Denominator = 0.1² + 0.04² = 0.0116.
T = 275.0 × (0.04 × 0.1) / 0.0116 ≈ 94.8 N·m

(c) Maximum torque:
s_m = R₂/X₂ = 0.1/1.0 = 0.1
T_max = 3 × 120² / (4π × 25 × 1.0) = 43200 / 314.16 ≈ 137.5 N·m

(d) Maximum starting torque: We need total R₂ = X₂ = 1.0 Ω. External resistance = 1.0 − 0.1 = 0.9 Ω per phase. Check: T_st = 275.0 × 1.0 / (1 + 1) = 137.5 N·m, which equals T_max.

Check: T_st/T_max = 27.2/137.5 = 0.198, and 2(0.1)/(1 + 0.01) = 0.198. They match.

What this shows: With only the rotor's own resistance, the starting torque (27.2 N·m) is much lower than the 4% slip torque (94.8 N·m). Adding external resistance raises the starting torque five-fold. This is the whole reason for the slip-ring motor.

Common mistakes

1. Forgetting the s in sX₂. Rotor reactance falls in proportion to slip because rotor frequency is sf.
2. Using N_s in rpm inside 2πN_s. Use rps, or write ω_s = 2πN_s/60 with N_s in rpm.
3. Mixing line and phase values. E₂ in the formulas is the per-phase value. The factor 3 accounts for three phases.
4. Saying maximum torque depends on rotor resistance. It does not. Only the slip at which it occurs changes.
5. Forgetting the V² dependence. A 10% voltage drop reduces torque by about 19%, not 10%.
6. Assuming maximum torque is always at start. That is true only when R₂ ≥ X₂.

Practice questions

1. A rotor has R₂ = 0.05 Ω and X₂ = 0.5 Ω per phase. At what slip is the torque maximum?
2. The supply voltage of an induction motor drops by 10%. By what percentage does the torque at a given slip change?
3. In a slip-ring motor, R₂ is doubled by adding external resistance. What happens to T_max and s_m?
4. State the condition for maximum starting torque.
5. Why is the starting torque of a normal cage motor low, even though the starting current is high?

Answers:

1. s_m = R₂/X₂ = 0.05/0.5 = 0.1.
2. T ∝ V², so new torque = 0.9² = 0.81 of the original. It falls by 19%.
3. T_max is unchanged. s_m doubles.
4. R₂ = X₂ (rotor resistance equals standstill rotor reactance).
5. Rotor resistance is low compared with reactance at standstill, so the rotor power factor is poor. Much of the current is reactive and produces little torque.

Summary

• Torque is proportional to flux, rotor current and rotor power factor: T = kΦI_rcosφ₂.
• Running torque: T = [3/(2πN_s)] × sE₂²R₂ / (R₂² + (sX₂)²).
• Starting torque is the same expression at s = 1. It is maximum when R₂ = X₂.
• Maximum torque occurs at s_m = R₂/X₂, and T_max = 3E₂²/(4πN_sX₂) is independent of R₂.
• Torque is proportional to V².

Also Read: Starting methods of three phase induction motors

FAQ on torque of three phase induction motors

What is the torque equation of a 3-phase induction motor?

T = [3/(2πN_s)] × sE₂²R₂ / (R₂² + (sX₂)²), with N_s in rps and per-phase rotor values.

At what slip is the torque maximum?

At s_m = R₂/X₂, where rotor resistance equals rotor reactance per phase under running condition.

Does the maximum torque depend on rotor resistance?

No. Rotor resistance only changes the slip at which the maximum torque occurs.

Why is torque proportional to the square of the voltage?

Rotor emf is proportional to flux, and flux is proportional to supply voltage. Torque depends on both the emf and the current it drives, so it varies as V².