I have seen this statement in a question paper. I was asked to prove it but I can not find the logic.
Electronic – Why starting current in star-delta starter becomes 1/√3 times that of DOL starter
electric-machineinduction motormotor
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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.
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: -
Here's the SE question where I borrowed the final picture from.