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I know that a 230/400V 3phase motor requires 230V on each coil. No more and no less. So on a 3*230V grid it would be wired in Delta, on a 3*400V it would wired in Y.

I went to evening school to learn about this and my course on the subject has several diagrams on this, most resembling these 2: enter image description here enter image description here

We also put this in practice and wired the motors and tested errors etc.

But thinking about it now I noticed that we never grounded the center point of these connections. How can a Y-config at 3*400V produce 230V on the coil?

Is there a mathematical way of calculating this? It's confusing since there's 3 voltages that are all averaged at 400V but measured at any given moment they're 3 very different voltages connected to eachother.

I have however also seen images online of a delta/Y connections that do ground the centerpoint: enter image description here And here it totally makes sense that the coil get 230V because we know that L-N voltage is sqrt(3) of L-L voltage.

I feel like I'm missing something here.

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Your understanding so far is correct. If you draw your phasor diagrams to scale you can solve this from basic trigonometry.

enter image description here

Figure 1. Phasor diagram for delta and star connections

You should, when you have finished, see that the phase to phase voltages are $ \sqrt 3 $ times the phase to star voltages.

But thinking about it now I noticed that we never grounded the center point of these connections. How can a Y-config at 3*400V produce 230V on the coil?

Now realise that without a neutral connection the star / wye point will only be in the centre if all three phases are balanced. If one phase is more highly loaded then the star point will be pulled towards that phase.

The reason that we have a neutral star point at all is that the currents into this point all sum to zero at every instant.

Is there a mathematical way of calculating this? It's confusing since there's 3 voltages that are all averaged at 400V but measured at any given moment they're 3 very different voltages connected to each other.

I tend to think graphically and my illustration below may help.

enter image description here

Figure 2. (1) The red phase current is the geometric sum of the black and blue. (2) When black and blue are equal and opposite the red must be zero. (3) When black falls to zero the other two must sum to zero.

I hope that helps.

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  • $\begingroup$ I understand everything you said(I had to brush up on my trigonometry though) and I think I understand now. It's the convergance point that is always neutral because the sum of the forces is always 0. I guess that this is also an application of kirchoff's law? $\endgroup$ – Rakward Aug 22 '17 at 9:24
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Simply put a balanced load does not need, but may have, a neutral connection.

Then the effective resistance of any supplied phase is its resistance plus the parallel resistance of the other two as the current is shared.

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