seegongsik
PW-A3 · Three-phase AC

Balanced three-phase power: √3 V_L I_L cosφ

Three-phase power is just 3 times the power of one phase, and in line quantities it comes together as √3 V_L I_L cosφ regardless of the connection. Get a feel for where that √3 comes from and why Y and Δ arrive at the same formula.

Change the power factor and watch the power delivered

Dragging the gauge changes the power factor cosφ. The apparent power S = √3 V_L I_L stays the same, but the active power actually delivered, P = √3 V_L I_L cosφ, shrinks and grows with the power factor.

Drag left and right on the gauge to change the power factor continuously; the length of the filled bar is the active power as a fraction of the apparent power.
Active power P (against apparent S)
P = √3 V_L I_L · 0.85
Power factor cosφ = 0.85 · φ = 32° · P/S = 0.85

Power of one phase × 3

When balanced, the three phases do the same work, so adding the power of one phase, V_p I_p cosφ, three times gives the total power P = 3 V_p I_p cosφ. Here cosφ is the cosine of the phase difference between each phase's voltage and current, that is, the power factor of the load.

Y substitution · phase into line

In a Y we had V_p = V_L/√3 and I_p = I_L (A2). Substituting gives P = 3 · (V_L/√3) · I_L · cosφ = √3 V_L I_L cosφ, and shaving the leading 3 down to √3 is exactly the √3 from A2.

Δ substitution · same destination

In a delta it was the other way round, V_p = V_L and I_p = I_L/√3. Substituting gives P = 3 · V_L · (I_L/√3) · cosφ = √3 V_L I_L cosφ, the same formula as for Y, not one symbol different. Only whether √3 comes from the voltage or the current differs; the result is one.

ObserveP = 3 Vp Ip cosφ
The total is 3 times one phase.
ChooseY: Vp = ?
In a Y the phase voltage is 1/√3 of the line voltage.
Fill inP = 3 · (VL/√3) · IL cosφ = ?
Shaving 3 to √3 gives √3 VL IL cosφ.
On your ownΔ: P = ?
Δ also arrives at the same √3 VL IL cosφ.

A formula that needs no connection

So with the line voltage and current and the power factor, the power can be found without knowing whether the connection is Y or Δ. Apparent S = √3 V_L I_L, active P = √3 V_L I_L cosφ and reactive Q = √3 V_L I_L sinφ all have the same √3 form, which is also why the rated voltage and current on a three-phase nameplate are always given as line values. One caution: the φ in these formulas is not the angle between line voltage and line current but the angle between one phase's voltage and current, the angle of the load impedance. The line voltage and line current are a further 30 degrees apart.

Back to the first screen

On the gauge the apparent S stayed fixed while only the active P followed the power factor, because in P = √3 V_L I_L cosφ only cosφ changed. That √3 is the √3 obtained in A2 by subtracting two phasors, and it takes the same seat whether the connection is Y or Δ. There is no need to memorize connections; the line quantities and the power factor are enough.

Balanced three-phase power is 3 times the power of one phase, and written in line quantities it is P = √3 VL IL cosφ regardless of the connection. Apparent S = √3 VL IL and reactive Q = √3 VL IL sinφ have the same √3 form; in a Y the √3 comes from the voltage and in a Δ from the current, but put together they are identical. φ is the angle between one phase's voltage and current.
The next step

With the formulas in place, the next unit (PW-A4) looks at the three phasors rotating together. In balanced three-phase, the sum of the instantaneous power of the three phases does not wobble with time but stays constant, and why rotating phasors make smooth, constant power is one of the reasons three-phase is used more widely than single-phase.