Induction motor current with load; stator connection

How the motor adjusts its current with load

An induction motor behaves like a transformer with a rotating secondary.

1. At no load, the rotor runs near $N_s$ and the slip is tiny, so the rotor emf $sE_2$ and the rotor current are very small. The stator takes only the no-load current $I_0$ (mostly magnetising, power factor 0.1 to 0.2).
2. When the shaft load increases, the load torque exceeds the electromagnetic torque and the rotor slows down.
3. The slip $s$ increases, so the relative speed between the field and the rotor increases. The rotor emf $sE_2$ and its frequency $sf$ increase, so the rotor current $I_2 = \frac{sE_2}{\sqrt{R_2^2+(sX_2)^2}}$ increases. The rotor power factor also improves.
4. The torque therefore rises until it matches the load, and the motor settles at a new speed with slightly higher slip.
5. The rotor mmf opposes the stator flux (Lenz's law) and tends to reduce it. To keep the flux constant ($\phi \propto V_1$), the stator draws an extra current $I_2'$:

$$\vec I_1 = \vec I_0 + \vec I_2'$$

So the stator current rises automatically with load, and the input power factor improves with load.

Effect of stator connection (slip-ring motor)

The stator can be connected in star or delta (the rotor is normally star). For the same line voltage $V_L$:

• A delta-connected winding is used for normal running; it takes more current but delivers full torque.
• Star connection at starting reduces the starting current to one third and also gives one third of the torque (star-delta starter for delta-rated motors).
• The rotor connection does not change the stator current directly. With a star-connected rotor, the phase emf is $E_{L}/\sqrt3$, where $E_L$ is the emf between slip rings; this sets the rotor current and the values of the external resistance.