IM and sync motor current adjust with load

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.