oscarrod5 said:
I understand phase voltage (phase to neutral) well, but I'm still confused by what exactly the potential difference is between any 2 phases in 3 phase power. If you were to try to find the potential difference where 2 sine wave phases cross, then at that instantaneous point, the potential difference looks to be 0.
https://en.wikipedia.org/wiki/Mathe...ric_power#/media/File:3_phase_AC_waveform.svg
You can use the standard Mathematics expression to find the Potential Difference. notably, the difference between the two Sine functions. Sin A - Sin B = 2 x Cos [(A+B)/2] x Sin [(A-B)/2]
We used to recite that as " sine minus sine equals two cos a half sum, sine a half difference." (With similar expressions for sin plus sin, cos plus cos and cos minus cos)*
The phases in -phase electricity are 120 degrees difference, so you are looking at two expressions like Sin(A) and Sin(A+120). (if I could type pi I could do that in radians rather than degrees)
"a half sum" equals (A +60), but cos(A+60) is just a phase shifted Sin(A)
"a half difference" equals 120, but sin(120) = √3 / 2
so 2 x {cos half sum) x (sin half difference) effectively means √3 sin(A), so the 3-phase PD is √3 times the single phase PD.
*
sin plus sin equals two sine a half sum, cos a half difference
sin(A) + sin(B) = 2 . sin[(A+B)/2] . cos[(A-B)]/2
cos plus cos equals two cos a half sum cos a half difference
cos(A) + cos(B) = 2 . cos[(A+B)/2] , cos[(A-B)/2]
cos minus cos equals two sine a half sum sine a half difference reversed.
cos(A) - cos(B) = 2 . sin[(A+B)/2] . sin[(B-A)/2]