How many amps are there

[Estiban]

How many amps are there for 1 killowatt three phase supply?

 

[telectrix]

about 1.5A assuming a PF of 1.

 

[Lucien Nunes]

And assuming the supply is 400V line-line.,

 

[Estiban]

Thanks but I thought three phase was divide single phase by 1.72.
4.3amps is 1kw single phase so how did you guys get 1.5 and not 2.5 a three phase ?

 

[Lucien Nunes]

I (line) = P / V (line) / sqrt(3)
=1000 / 400 / 1.73
=1.45A

By convention we use the line-line voltage in 3-phase calculations (typically 400V in Europe) instead of the line-neutral voltage (typically 230V). But if you want to think of it in relation to an equal size of single-phase load operating on the line-neutral voltage, consider dividing it into three loads of 1/3 the size, taking 1/3 of the current, star-connected between each phase and neutral.

 

[Estiban]

I (line) = P / V (line) / sqrt(3)
=1000 / 400 / 1.73
=1.45A

By convention we use the line-line voltage in 3-phase calculations (typically 400V in Europe) instead of the line-neutral voltage (typically 230V). But if you want to think of it in relation to an equal size of single-phase load operating on the line-neutral voltage, consider dividing it into three loads of 1/3 the size, taking 1/3 of the current, star-connected between each phase and neutral.

So kw rating is always the same amount of amps regardless of single or three phase just its split via three phase

 

[DefyG]

So kw rating is always the same amount of amps regardless of single or three phase just its split via three phase

No the constants are Load 1kW and 1ph voltage 230v OR
Load 1kW and 3ph voltage 400v
The variable is
The current or 'amps' varies as suggested previously.

 

[pc1966]

So kw rating is always the same amount of amps regardless of single or three phase just its split via three phase

The easiest way to look at it is the star-connected case, then you have:

• single phase = 1000 / 230 = 4.35A
• three phase = 1000 / (230 * 3) = 1.45A (i.e. single phase / 3)

However, if you have a delta-connected load and no neutral then it is less obvious how a star-connected system might be used for calculations, and results are as @Lucien Nunes just showed. Which is the same.