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You really need to know the line voltage (VL) involved. Provided the load is balanced, to find the line current, you need to divide the load (in watts) by (1.732 x VL). Incidentally, the correct symbol for kilowatts is kW, not Kw!

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10y ago
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11y ago

1KW=1000*1.736*440*1(Ampere)*p.f (for 3 phase). Here power factor is required for solution.

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14y ago

The formula you are looking for is Amperage = kw x 1000/1.73 x Voltage x pf. pf is power factor. Use .9 as a reference for power factor.

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11y ago

That can be split into three 254 v single-phase circuits each carrying 40 amps and the total VA is 30.5 kVA. Therefore the power is up to 30.5 kW depending on the power factor of the load.

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11y ago

Here is the Formula for you to find out you yourself. Power=3IV*Power Factor of that motor. ( The value for I=Current and V=Voltage) must be phase values.

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12y ago

The formula you are looking for is I = kW x 1000/1.73 x E x pf. The power factor (pf) value of .9 will be used. Amps = 60 x 1000/343 = 60000/343 = 175 amps.

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Q: 20KW machine 480 volts 3 phase what is the amps?
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