Delta Line and Phase Current

IL=3 IφI_{L} = \sqrt{3} \, I_{\varphi}

Worked example: 10 A per winding → 17.32 A line — press Try an example to run it live, then adjust anything.

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Delta Line and Phase Current explained

IφIL

Delta is wye's mirror image: the windings sit directly across the lines, so phase voltage equals line voltage, but each line conductor collects current from two windings that are 120° apart — giving the same √3, this time on the current. A delta motor winding carrying 10 A per coil draws 17.32 A in each line lead.

This is exactly why a wye-delta starter works. Started in wye, each winding sees only VL/√3, so the winding current drops by √3 and the line current by three; the motor starts on about a third of its delta inrush and a third of the torque, then switches to delta for full running duty. The trap in the field is the six-lead terminal box: reconnect a 400 V delta motor as wye and it runs at 58% voltage and stalls under load.

Delta Line and Phase Current formula

IL=3 IφI_{L} = \sqrt{3} \, I_{\varphi}
Where
  • ILI_{L}= Line current (A)
  • IφI_{\varphi}= Phase (winding) current (A)

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