Using complex numbers to model 3 phase AC

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The discussion focuses on calculating line-to-line (L-L) voltage in a three-phase AC system using complex numbers and phasor diagrams. The initial voltage is given as 230V line-to-neutral (L-N), and the relationship between L-L and L-N voltages involves the square root of three (√3), which is exact in an ideal balanced system. The calculations show that the L-L voltage is approximately 400V, derived from the relationship between the angles of the phasors. The conversation highlights the importance of vector addition in phasors, clarifying that L-L voltage is determined by the voltage difference rather than a simple sum. Overall, the discussion emphasizes the theoretical aspects of three-phase voltage calculations and the implications of phase balance.
Guineafowl
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TL;DR
How to derive the voltages seen in a three phase, Y connected transformer.
Assume a transformer as above, with 230V L-N, and I want to work out the L-L voltage. A phasor diagram will show me that the voltages are 120° out of phase.

(230∠0°) + (230∠120°) = (230cos0 + j230sin0) + (230cos120 + j230sin120) = 230 + (-115 + j199.2)

115 + j199.2 = 230∠60

What I’m looking for is, of course, 400V. I know √3 is involved, but how, and is that an exact calculation or a useful rule of thumb?

This is not homework.
 
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The specified 230 Vac is RMS, so the signal amplitude is 230 * √2 * Sin( ).
 
Guineafowl said:
Summary:: How to derive the voltages seen in a three phase, Y connected transformer.

is that an exact calculation or a useful rule of thumb?
How exact do you need to be? In reality, the three phases are never exactly balanced.

Edit: But in the ideal balanced case, ##\sqrt{3}## is exact.
 
Last edited:
anorlunda said:
How exact do you need to be? In reality, the three phases are never exactly balanced.
It’s not important, just a theoretical question really.I think I have it worked out. Because, on the phasor diagram, the L-N voltages oscillate either side of the origin (0V), the angle between each phasor and the nearest is actually 60°.

So:
VL-L = (230∠0°) + (230∠60°) = (230) + (230cos60 + j230sin60)

= (230) + (115 + j199.2) = 345 + j199.2

|VL-L| = 398.4 V (∠30°)

EDIT: And because the magnitude of L-L voltage depends on cos 30°, and that of L-N depends on cos 60°, the ratio of voltages (400:230) is similar to the ratio of cos 30:cos 60, which is 1.73, or ≈√3.
 
Last edited:
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When L1 is at zero volts, L2 will be at +60°, and L3 will be at –60°.
The L2 to L3 voltage will then be 2 * Sin( 60° ) = 2 * √(3/4) = √3.
 
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According to vectorial diagram:
 

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Babadag said:
According to vectorial diagram:
Thanks - I see you’ve ignored the (-) sign when adding the vectors. I assumed that if the voltage magnitudes are of opposite sign they subtract. Is that not correct?
 
Guineafowl said:
I see you’ve ignored the (-) sign when adding the vectors.
Phasors point from the origin, neutral or ground.
The Line to Line voltage is therefore not the sum, it is the voltage difference.
 
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