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I have read answers pertaining to this same question, but I could not understanf why the phase voltage is equal to line voltage. This answer says:

In a delta system, the two points required to measure a line voltage also happen to be the same two points connected across the phase.

What I don't understand is why aren't the rest two phases affecting this voltage since they both are connected to one point each and are out of phase. What am I failing to see?

I think I have not been able to express what my confusion is. I do understand since the points across which phase voltage is measured is also the point the lines are connected and I do agree with that, but if I compare the voltage for a delta configuration and other for an isolated AC supply I get the same values for the voltage. I don't understand why.

I took this snip at a certain moment and the volatge measured by the voltmeter is equal

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  • \$\begingroup\$ Does this answer your question? Phase voltage vs line voltage in a delta configuration \$\endgroup\$
    – RoyC
    Apr 29, 2022 at 20:54
  • \$\begingroup\$ @RoyC No, it doesn't, like other answers this one too simply states that, the phase current is equal to line current or maybe I am not able understand. \$\endgroup\$
    – Darlington
    Apr 29, 2022 at 21:26
  • \$\begingroup\$ If line voltage is defined as being measured from A to B and phase voltage is defined as being measured form A to B they have to be the same. Which part of this don't you understand. \$\endgroup\$
    – RoyC
    Apr 29, 2022 at 21:42
  • \$\begingroup\$ @RoyC I have edited my question, I think now I express my question more clearly. \$\endgroup\$
    – Darlington
    Apr 30, 2022 at 9:56

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If I understand that what you are asking is "why are the sums of the other two phases equal to the phase I am looking at".

The other two phases are 120 degrees and 240 degrees shifted in phase from your "reference" phase if you add up sin(angle+120) and sin(angle+240) you get -sin(angle) this is consistent with your diagram and the way around your reference phase is shown connected. Hence the other two phases create exactly the same voltage as the phase you are looking at.

Following is a little table created in excel which illustrates this.

enter image description here

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  • \$\begingroup\$ I did get this result, but won't that imply they are out of phase by 180 degrees? \$\endgroup\$
    – Darlington
    May 1, 2022 at 18:29
  • \$\begingroup\$ No it just indicates which way round it is connected is you look at your diagram if the bottom and left phase are connected so they add up in series and the right hand phase is connected backwards. \$\endgroup\$
    – RoyC
    May 1, 2022 at 18:50
  • \$\begingroup\$ And since they are connected backwards, the potential difference, would, be nullified? \$\endgroup\$
    – Darlington
    May 2, 2022 at 20:11
  • \$\begingroup\$ There is no potential difference between the other two phases and the phase you are looking at. The two reinforce each other in terms of providing current. \$\endgroup\$
    – RoyC
    May 2, 2022 at 20:58

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