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## Main Question or Discussion Point

Hey.

If i Have a 3x400VAC system, and I want to calculate the voltage drop over the cable "vectorially" - or what you call it :) - how do I do this?

U.nominal=400V

I have tried this method:

[itex]\Delta[/itex]U.conductor=I.1*Z.cable1+I.2*Z.cable2+etc. and then I will have to find the new voltage ->

[itex]\Delta[/itex]U.phase=U.conductor*[itex]\sqrt{3}[/itex]

U.new=U.nominal-[itex]\Delta[/itex]U.phase

When I'm calculating this with angles, I actually get a higher voltage, after the voltage drop :D

The voltage drop is cause by a electrical motor starting.

In school we learn it this way:

[itex]\varphi[/itex].motor=80 degrees

U.new=U.nominal-[itex]\sqrt{3}[/itex]*I1*Length.cable1*(R.cable1*cos[itex]\varphi[/itex]+X.cable1*sin[itex]\varphi[/itex])+[itex]\sqrt{3}[/itex]*I2*Length.cable2(R.cable2*cos[itex]\varphi[/itex]+X.cable2*sin[itex]\varphi[/itex])+etc.

But that is not with vectors.

Hope you understand?

Thanks in advance.

If i Have a 3x400VAC system, and I want to calculate the voltage drop over the cable "vectorially" - or what you call it :) - how do I do this?

U.nominal=400V

I have tried this method:

[itex]\Delta[/itex]U.conductor=I.1*Z.cable1+I.2*Z.cable2+etc. and then I will have to find the new voltage ->

[itex]\Delta[/itex]U.phase=U.conductor*[itex]\sqrt{3}[/itex]

U.new=U.nominal-[itex]\Delta[/itex]U.phase

When I'm calculating this with angles, I actually get a higher voltage, after the voltage drop :D

The voltage drop is cause by a electrical motor starting.

In school we learn it this way:

[itex]\varphi[/itex].motor=80 degrees

U.new=U.nominal-[itex]\sqrt{3}[/itex]*I1*Length.cable1*(R.cable1*cos[itex]\varphi[/itex]+X.cable1*sin[itex]\varphi[/itex])+[itex]\sqrt{3}[/itex]*I2*Length.cable2(R.cable2*cos[itex]\varphi[/itex]+X.cable2*sin[itex]\varphi[/itex])+etc.

But that is not with vectors.

Hope you understand?

Thanks in advance.

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