Electronic – Why starting current in star-delta starter becomes 1/√3 times that of DOL starter

electric-machineinduction motormotor

I have seen this statement in a question paper. I was asked to prove it but I can not find the logic.

Best Answer

The whole idea is that a smaller voltage is delivered to the motor windings during start up.

The end game is that the motor (or load) is configured as a delta winding formation. In other words, after the load is up and running, each winding receives line voltage.

However, during start-up the windings are put into star formation and each winding will receive 57.7% of the AC voltage.

enter image description here

When starting the motor (above) KM3 and KM1 activate whilst KM2 is deactivated. You should be able to see that KM1 terminates the motor windings into a star point thus the supply voltage to each winding is only 57.7%.

After a short while KM1 deactivates and KM2 activates. This puts all three motor windings into a delta configuration thus each winding receives 100% line voltage.

Here's a picture that shows (or implies) that math: -

enter image description here

Here's the SE question where I borrowed the final picture from.