PW-A2 · Three-phase AC

Line and phase values: where √3 comes from

That the line voltage is √3 times the phase voltage is not a number to memorize but the geometry of a difference of two phasors. Build the intuition for why √3 appears and on which side — voltage or current — it lands.

Subtract two phase voltages into a line voltage

Two phase voltages V_a and V_b spin together. Their difference (the gold arrow) is the line voltage. Change the separation until that difference is exactly √3 times as long.

Tap to switch the separation angle.

Magnitude of the difference, relative to phase

|V_ab| ≈ 1.73 · V_p

Exactly 120 degrees. The difference is √3 times as long and points 30 degrees ahead. This is the line voltage.

√3 · Y line voltage

√3 · balanced ·

Y connection · √3 on the voltage

In Y each line runs straight to one end of a single phase coil, so the line current is just the phase current. But the voltage between two lines (line voltage) is the difference of two phase voltages, so it is √3 times — the difference of two phasors 120 degrees apart. The 380V line voltage of Korean low-voltage distribution is exactly √3 times the 220V phase voltage.

Δ connection · √3 on the current

In delta it is reversed. Each line sits across both ends of a single coil, so the line voltage is just the phase voltage. Instead the current leaving on one line is the difference of the two phase currents meeting at that node, so the same √3 lands on the current. Whichever side forms the difference decides where √3 attaches.

One geometry, two faces

In the end √3 is a single fact: subtract two equal phasors 120 degrees apart and the difference is √3 times as long. Y shares a common point (the neutral) so the voltages form the difference; delta shares a common node so the currents form the difference. That is why √3 shows up on the voltage in Y and on the current in delta.

ObserveV_ab = V_a − V_b

The line voltage is the difference of two phase voltages.

Choose|V_ab| = 2 V_p sin(?)

Difference magnitude = 2 V_p sin(separation/2).

Fill in|V_ab| = 2 V_p · ? = √3 V_p

sin 60 degrees = √3/2, so √3 V_p.

On your ownΔ: |I_line| = ?

In delta the same √3 lands on the current.

Back to the first screen

The gold difference arrow grew to √3 times the phase voltage exactly when the separation was 120 degrees. That difference is the line voltage of Y. In delta the very same subtraction happens to the currents, so the line current is √3 times the phase current. What to remember is not the number √3 but the single picture: subtracting two phasors 120 degrees apart gives √3.

The line voltage is the difference of two phase voltages. The difference of two equal phasors 120 degrees apart is √3 times as long and leads by 30 degrees. Y connection: line voltage = √3 · phase voltage, line current = phase current. Δ connection: line current = √3 · phase current, line voltage = phase voltage. √3 always attaches to the side that forms the difference.

The next step

With line and phase values in hand, the next unit (PW-A3) looks at balanced three-phase power. Add one phase of power three times to get P = 3 V_p I_p cosφ, and rewriting it in line values gives the famous P = √3 V_L I_L cosφ. The √3 you built here moves to the center of the power formula.