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Electrical Engineering DISCUSSION
Our supply is described as 230/400 V. I understand that 230 V is measured from any phase to neutral and 400 V between two phases, but I expected two 230 V phases to give 460 V between them, not 400 V.
Where does the factor of 1.732 come from, and how do voltage and current relate for star and delta connected loads? I keep mixing up which quantity gets multiplied by √3.
You would get 460 V if the two voltages were 180° apart, as in a split-phase system. In a three-phase system they are 120° apart, and the voltage between two lines is the phasor difference of the two phase voltages. Drawing the two phasors and the line between their tips gives an isosceles triangle, and the geometry works out to V_line = 2 × V_phase × cos 30° = √3 × V_phase.
So 230 V × 1.732 = 398 V, rounded to 400 V. The same construction shows that each line voltage leads the corresponding phase voltage by 30°.
For loads, remember which quantity is shared. In star, each element is between a line and the star point, so it sees the phase voltage, and the line current is the element current: V_line = √3 × V_ph, I_line = I_ph. In delta, each element is directly between two lines and sees the line voltage, while each line feeds two elements: V_line = V_ph, I_line = √3 × I_ph.
Either way total power is P = √3 × V_line × I_line × cos φ. A balanced 10 kW heater on 400 V draws 10,000 / (1.732 × 400) = 14.4 A per line whether it is wired in star or delta internally.
A numeric example makes the difference concrete. Take three 40 Ω resistors. In star each has 230 V across it: 5.75 A and 1.32 kW per resistor, 3.97 kW in total, with 5.75 A in each line. In delta each has 400 V across it: 10 A and 4 kW per resistor, 12 kW in total, with 17.3 A in each line.
The same elements take three times the power and three times the line current in delta. That factor of three is the basis of star-delta motor starting, and it is also why connecting a star-rated heater in delta destroys it.
The neutral follows from the same phasor arithmetic. With a balanced star load the three currents are equal and 120° apart, so they sum to zero and the neutral carries nothing; a balanced three-phase motor needs no neutral at all. With unbalanced single-phase loads the neutral carries the phasor sum. For example 10 A on two phases and nothing on the third gives 10 A in the neutral, not 20 A and not zero.
One exception to watch: third-harmonic currents from electronic loads are in phase on all three lines, so they add in the neutral instead of cancelling, and the neutral of a lighting or IT distribution board can carry more current than a phase.
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