PW-A3 · Three-phase AC

Balanced three-phase power: √3 V_L I_L cosφ

Three-phase power is just three times one phase. Rewritten in line values it collects into √3 V_L I_L cosφ regardless of connection. Build the intuition for where that √3 comes from and why Y and Δ give the same formula.

Sweep the power factor and watch the delivered power

Drag the gauge to change the power factor cosφ. The apparent power S = √3 V_L I_L stays fixed, but the active power actually delivered, P = √3 V_L I_L cosφ, shrinks and grows with the power factor.

Drag the gauge to set the power factor.

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

One phase of power, times three

When balanced, all three phases do identical work. So adding one phase of power V_p I_p cosφ three times gives the total: P = 3 V_p I_p cosφ. Here cosφ is the phase angle between each phase voltage and current — the load power factor.

Y substitution · phase into line

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

Δ substitution · same destination

In delta it was reversed: V_p = V_L, I_p = I_L/√3. Substituting: P = 3 · V_L · (I_L/√3) · cosφ = √3 V_L I_L cosφ. The same formula as Y, not one symbol different. Whether √3 comes from the voltage or the current, the result is one.

ObserveP = 3 V_p I_p cosφ

The total is three times one phase.

ChooseY: V_p = ?

In Y the phase voltage is the line voltage over √3.

Fill inP = 3 · (V_L/√3) · I_L cosφ = ?

Shaving 3 to √3 gives √3 V_L I_L cosφ.

On your ownΔ: P = ?

Delta arrives at the same √3 V_L I_L cosφ.

A formula that needs no connection

So with only the line voltage, line current and power factor, you find the power without knowing whether it is Y or delta. 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 share the same √3 shape. This is why nameplate ratings in the field are always written in line values.

Back to the first screen

On the gauge the apparent S held fixed while only the active P moved with the power factor, because in P = √3 V_L I_L cosφ only cosφ changed. That √3 is the one you got by subtracting two phasors in A2, and it sits in the same seat for Y and for delta. No need to memorize the connection: line values and the power factor are enough.

The balanced three-phase power is three times one phase. In line values, regardless of connection, P = √3 V_L I_L cosφ. Apparent S = √3 V_L I_L and reactive Q = √3 V_L I_L sinφ share the same √3 shape. In Y the √3 comes from the voltage, in delta from the current, but combined they are the same.

The next step

With the formula in hand, the next unit (PW-A4) watches the three phasors rotate together. In a balanced set the sum of the three instantaneous powers does not pulse with time but stays constant. Why the rotating phasors make smooth, constant power is exactly why three-phase beats single-phase.