---
title: "IM and sync motor current adjust with load"
subject: "Electrical Machine"
chapter: 5
key: "semester_4/electrical_machine:im-current-adjust-with-load"
source: https://www.bctengineer.com/q/semester_4/electrical_machine/im-current-adjust-with-load
site: BCT Engineering
---

# IM and sync motor current adjust with load

Electrical Machine · Chapter 5: Three Phase Induction Machines

- Asked in: 2073 Magh, 4+4 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machine/im-current-adjust-with-load
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machine/chapter-5

## Question

How does induction motor and synchronous motor adjust the current according to the change in load? Explain briefly using mathematical expression.

## Short answer

In an induction motor, increased load raises slip s, increasing rotor current I₂ = sE₂/√(R₂²+s²X₂²) and stator current I₁ = I₀ + I₂'. In a synchronous motor, load increases the torque angle δ, raising armature current I_a = (V−E_b)/Z_s, with constant speed.

## Answer

Both machines draw more stator current when the mechanical load increases. The mechanism differs because the induction motor changes its speed (slip) and the synchronous motor changes its load angle.

#### Induction motor
1. The load torque rises, so the rotor slows down and the slip s = (N_s − N)/N_s rises.
2. The relative speed between the field and the rotor increases, so the rotor emf and frequency rise: E₂ᵣ = sE₂, f₂ = sf.
3. The rotor current rises:

**I₂ = sE₂ / √(R₂² + s²X₂²)** (while the rotor power factor cos φ₂ = R₂/√(R₂² + s²X₂²) falls slightly, because sX₂ grows)

4. By transformer action, the stator draws an extra load component that balances the rotor mmf:

**I₁ = I₀ + I₂'**, where I₂' = I₂ · (rotor turns/stator turns) and I₀ is the no-load (magnetising) current, which is nearly constant.

5. The developed torque T = (3/ω_s) · sE₂²R₂/(R₂² + s²X₂²) rises with s until it equals the load torque. The motor settles at a new slip with a larger current.

#### Synchronous motor
The speed is fixed at N_s, so slip is not available. With constant supply voltage V and constant field excitation, the back emf E_b is constant in magnitude.

1. When the load increases, the rotor momentarily slows and falls behind the stator field. The load (torque) angle δ between V and E_b increases.
2. The net voltage across the armature impedance becomes larger:

**E_r = V − E_b** (phasor difference), **I_a = E_r / Z_s** with Z_s = R_a + jX_s.

3. The armature current I_a grows in magnitude and its phase changes, so the motor takes more electrical power:

**P = (3 V E_b / X_s) · sin δ** (neglecting R_a)

4. The motor again runs at synchronous speed, but with a larger δ and a larger I_a. The maximum power is at δ = 90° (pull-out). Beyond that the motor loses synchronism.

| Point | Induction motor | Synchronous motor |
|---|---|---|
| Variable that changes | slip s | load angle δ |
| Speed | falls slightly | stays at N_s |
| Current equation | I₂ = sE₂/√(R₂²+s²X₂²), I₁ = I₀ + I₂' | I_a = (V − E_b)/Z_s |

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. When the load torque on an induction motor rises, what variable changes to let it draw more current?
   A. Supply frequency rises
   B. Slip rises as the rotor slows
   C. Field excitation is increased
   D. Load angle increases at constant speed
   Answer: B. The rotor slows, slip rises, rotor emf sE2 and rotor current rise, and the stator draws more current. (Key point: Mechanism in induction motor)

2. How does the stator of an induction motor respond to the increased rotor current?
   A. I1 = I0 + I2', balancing the rotor mmf by transformer action
   B. I1 falls so flux stays constant
   C. I1 = I2 with no magnetising component
   D. Stator current stays fixed and only the rotor current rises
   Answer: A. The stator draws a load component I2' to balance rotor mmf, added to the nearly constant magnetising current I0. (Key point: Stator current relation)

3. In a synchronous motor, which quantity changes when the load increases?
   A. Slip
   B. Speed, which falls permanently
   C. Load angle delta between V and Eb
   D. Supply voltage V
   Answer: C. Speed is fixed at Ns, so the rotor falls back and the load angle delta increases, raising the voltage across Zs. (Key point: Synchronous motor mechanism)

4. Which expression gives the armature current of a synchronous motor?
   A. Ia = (V + Eb)/Zs
   B. Ia = (V - Eb)/Zs as a phasor difference
   C. Ia = sE2/sqrt(R2^2+s^2X2^2)
   D. Ia = V/Eb
   Answer: B. The net voltage Er is the phasor difference V - Eb, divided by Zs = Ra + jXs. (Key point: Armature current of synchronous motor)

## Related in Electrical Machine

- [Induction motor current with load; stator connection](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Aim-current-adjust-load-stator-connection) — Electrical Machines, chapter 5
- [6-pole slip-ring IM no-load current](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machine%3Aim-6pole-slipring-standstill-no-load-current) — Electrical Machine, chapter 5
- [Induction motor equivalent circuit from no-load and blocked rotor tests](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Aim-noload-blocked-400v-equivalent) — Electrical Machines I, chapter 5
- [3-phase IM principle; speed less than sync](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machine%3Athree-phase-im-principle-slip) — Electrical Machine, chapter 5
- [No-load and blocked rotor tests of induction motor](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Aim-noload-blocked-rotor-test-explain) — Electrical Machines I, chapter 5
- [Full-load torque, rotor current and copper loss of 440 V motor](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Aim-440v-1425rpm-rotor-impedance) — Electrical Machines I, chapter 5

## Related in other subjects

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- [Induction motor plugging: braking current and torque](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Ainduction-motor-plugging-braking-current) — Utilization of Electrical Energy, chapter 3
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- [Synchronous motor drawing lagging and leading current](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machines_ii%3Amotor-lagging-leading-current) — Electric Machines II, chapter 2, asked 4×