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Phase and line voltage in star

Short answer

In a star system, line voltage equals √3 times phase voltage, and it leads the corresponding phase voltage by 30°. Derivation: V_RY = V_RN - V_YN gives magnitude √3 V_ph and a 30° leading phase shift.

Develop relation between phase voltage and line voltage in 3-φ star connected system.

In a star connection the three phase windings (or load branches) share a common neutral point N, and R, Y, B are the line terminals.

  • Phase voltages (line to neutral): V_RN, V_YN, V_BN, each of magnitude V_ph, 120° apart.
  • Line voltages (between lines): V_RY, V_YB, V_BR, each of magnitude V_L.

Derivation

Apply KVL through the neutral from R to Y:

V_RY = V_RN − V_YN = V_RN + (−V_YN)

Take V_RN = V_ph∠0° and V_YN = V_ph∠−120° (RYB sequence).

V_RY = V_ph(1 + j0) − V_ph(−0.5 − j0.866) = V_ph (1.5 + j0.866)

|V_RY| = V_ph √(1.5² + 0.866²) = V_ph √3

Angle = tan⁻¹(0.866/1.5) = 30°

Phasor diagram

   V_BN (+120°)                    ↗ V_RY (+30°)
       ↖              −V_YN ↗   ↗
          ↖           (+60°)  ↗
             ↖           ↗ ↗
                ↖     ↗ ↗  30°
                   ●─────────────────► V_RN (0°)
                ↙
             ↙
   V_YN (−120°)

−V_YN is V_YN reversed (at +60°). Adding it to V_RN by the parallelogram rule gives V_RY, which bisects the 60° angle between V_RN and −V_YN, so it leads V_RN by 30°.

(Using geometry: the triangle formed by V_RN, V_YN and V_RY is isosceles with an apex angle of 120° at N, so V_RY = 2 V_ph cos 30° = √3 V_ph.)

Result: V_L = √3 V_ph, and each line voltage leads its corresponding phase voltage by 30°. The line current equals the phase current in a star connection.

✓ Answer checked by Claude Opus 5.5 (high effort)

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