I have a machine (a machine with seven three-phase motors) with this information on its nameplate:

3 Phases

400 V, 50 Hz

100 A, 40 kW

I am wondering:

  1. Is 100 A for all three phases or only for one phase?
  2. What does it mean by 400V? Line to line voltage? Or the phase one?
  3. Is 40 kW the maximum power consumption? Or I should consider the reactive power and provide a larger power source such as 50 kVA? Also is it for all phases?
  4. What size of wire should I choose?

5. How was the power, 40 kW, calculated? 400V x 100A = 40kW? So what about P = sqrt(3) * VL * IL?

Environment: My machine has 7 three-phase motors. Local electricity line: 220V, 50Hz, each phase.

  • 3
    \$\begingroup\$ Given that you are asking these questions, the only responsible answer is: hire an industrial electrician who is familiar with the electrical codes in your jurisdiction and installation of industrial equipment. You mention 220V, 50Hz, but it's unclear exactly what you are describing. It's possible that your service is actually 400V 3-phase, with each phase 230V to neutral, but there's not enough information here for us to know. There are a lot if issues here and information needed that really need to be addressed/determined prior to moving forward. \$\endgroup\$
    – Makyen
    Aug 7, 2018 at 19:34
  • \$\begingroup\$ I am curious to know what it means. So I thought here is the right place. I meant three-phase lines with each phase almost 220V to neutral. So each line to line voltage is 400V approximately. \$\endgroup\$
    – Pana
    Aug 8, 2018 at 4:21

4 Answers 4


Is 100 A for all 3 phases or only for one phases?

That will be each phase.

What does it mean by 400V? Line to line voltage? or the phase one?

It is line to line voltage.

Is 40 kW the maximum power consuming? Or I should consider the reactive power and provide a larger source power such as 50 kW?

An easy way to calculate is to use the phase to neutral voltage and calculate for one phase.

  • One phase will have \$ P = \frac {P_{TOT}}{3} = \frac {40}{3} = 13.3 \ \text{kW} \$.
  • The phase to neutral voltage will be \$ {V_{L-L}}{\sqrt 3} = {400}{\sqrt 3} = 234\ \text V\$.
  • If the load is purely resistive then \$ I = \frac {P}{V} = \frac {13300}{234} = 57 \ \text A \$.

The 100 A rating, therefore, must be to cover start-up current and reactive power.

Also is it for all phases?

Yes. The power rating is for the whole machine.

What size of wire shall I choose?

Check your local regulations which should have tables to guide you. The distance from the supply will be a factor and whether any power-factor correction capacitors are placed at the feed or load end of the cable.

  • \$\begingroup\$ Can you please give an example in this conditions: Germany, 50 m distance and PF = 0.85. \$\endgroup\$
    – Pana
    Aug 7, 2018 at 17:19
  • 3
    \$\begingroup\$ Sorry, but I think I've done enough for you. It's time for you to do some work. \$\endgroup\$
    – Transistor
    Aug 7, 2018 at 17:22
  • \$\begingroup\$ It seems to me that 40 kW is not the rated mechanical output and is input real power which multiplied with efficency percentage to yield mechanical output power. Is that correct? \$\endgroup\$
    – Pana
    Aug 7, 2018 at 17:23
  • \$\begingroup\$ See electrical-engineering-portal.com/…. \$\endgroup\$
    – Transistor
    Aug 7, 2018 at 17:25
  • \$\begingroup\$ Please pay attention that I asked about a machine with seven three-phases motors not a single motor. Are both the same in reading nameplate? \$\endgroup\$
    – Pana
    Aug 7, 2018 at 17:31

The "100A" is the breaker rating for each. (phase)
The "400V" is line-line.
The "40kW" is max. steady load rating.
The size of the wire is AWG 6 max. ( 101A rated @ some temp )

The choice of wire gauge depends on local industrial codes and feed length. The voltage drop depends on the resistance of the wire and the load in amps.

Power consumption in Watts and VARs depends on pf, efficiency, mechanical load and acceleration which briefly exceeds by 300% steady-state max on start up or more.

  • \$\begingroup\$ Thanks. What did you mean by breaker rating? \$\endgroup\$
    – Pana
    Aug 7, 2018 at 15:29
  • 1
    \$\begingroup\$ The current rating for a safety breaker is the alotted current designed to withstand continuously. They use this number for Breaker rating current sharing. So add up all equip running at the same time should not exceed the breaker rating. \$\endgroup\$ Aug 7, 2018 at 15:45
  • \$\begingroup\$ Useful. Thanks. So my machine draws the maximum current, I = 40,000 / 400 / sqrt(3) = 58A from each phase? \$\endgroup\$
    – Pana
    Aug 7, 2018 at 16:06
  • \$\begingroup\$ No. That does not consider efficiency and power factor. See my answer. \$\endgroup\$
    – user80875
    Aug 7, 2018 at 16:11

100A is the current for each of the three phases. V X A X sqrt(3) = 69.3 kVA. V X A X sqrt(3) X power factor = input kW. 40 kW is the rated mechanical output power assuming that the marking on the motor conforms to international standards. Input power X efficiency = output power.

If the 40 kW rating is marked on the machine, it is the input power for the entire machine including the sum of the power inputs to all of the motors plus whatever else uses electric power in the machine. The power factor is then 40/69.3 = 0.577, a rather low power factor, but not unreasonable.

  • \$\begingroup\$ My machine has 7 three-phases motors. So 40kW is the rated mechanical power for all seven motors? \$\endgroup\$
    – Pana
    Aug 7, 2018 at 16:19
  • \$\begingroup\$ I was assuming a single motor with the ratings stater. See revised answer. \$\endgroup\$
    – user80875
    Aug 7, 2018 at 18:49

kW is definitely a power rating of some sort, because otherwise it would say kVA.

Hard to say WHAT power, though. Maybe it is output power? The output power of a three phase motor is:

P = sqrt(3) * I * V * PF * eff

Where P is the output power of the motor, I is the nameplate current, V is the nameplate voltage, PF is the power factor, and eff is the efficiency. Let's re-arrange it so solve for PF * eff.

PF * eff = P / (sqrt(3) * I * V) PF * eff = 40,000 / (1.73 * 100 * 400) = 1/1.73 = 0.577

Is that plausible? It is a bit on the low side. With 7 motors, the motor size must be around 7.5 HP. I would expect efficiency for a motor that size to be 90% and power factor maybe 0.8, so that the product would be 0.72, not 0.577. So, it is a mystery.

In order to make the math work, the power factor and efficiency would be something like 75 or 76 percent each.

The other explanation is to say that the input power really is 40,000W, and the kVA really is sqrt(3) * 400 * 100, in which case the power factor would be a miserable 0.577. This is possible. It is just kind of a low power factor.


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