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Star and delta connection relations
Short answer
In star connection, V_L = √3 V_ph and I_L = I_ph; line voltage leads phase voltage by 30°. In delta connection, V_L = V_ph and I_L = √3 I_ph; line current lags phase current by 30°. Neutral is available only in star.
What are the two ways of connecting a 3-phase system? Draw their phasor diagrams and write down the relationship between phase and line voltages and phase and line current for these system.
The two ways of connecting a three-phase system are star (Y) and delta (Δ).
1. Star connection
One end of each of the three phase windings is joined at a common neutral point N. The other ends are the line terminals R, Y and B.
R ●
\
● N ── (neutral, 4th wire)
/ \
Y ● ● B
Phasor diagram (RYB sequence, V_RN as reference):
V_BN (+120°) V_RY = V_RN − V_YN
↖ ↗ (leads V_RN by 30°)
↖ ↗
↖ ↗
↖ ↗ 30°
●─────────────────► V_RN (0°)
↙
↙
V_YN (−120°)
Relations:
- Line voltage V_L = √3 V_ph (each line voltage leads its phase voltage by 30°)
- Line current I_L = I_ph
2. Delta connection
The windings are joined end to start to form a closed loop. The lines are connected to the three junctions.
R
/ \
/ \
Y─────B
Phasor diagram of currents (I_RY as reference). The line current is I_R = I_RY − I_BR:
I_BR (+120°)
↖
↖
●──────────────────► I_RY (0°)
↙ ↘
↙ ↘
I_YB (−120°) ↘ I_R = I_RY − I_BR (lags I_RY by 30°)
For a balanced load the magnitude of I_R is √3 I_ph, and it lags the phase current I_RY by 30°.
Relations:
- Line voltage V_L = V_ph
- Line current I_L = √3 I_ph
Summary
| Quantity | Star | Delta |
|---|---|---|
| Line voltage | √3 V_ph | V_ph |
| Line current | I_ph | √3 I_ph |
| Phase angle between line and phase quantity | V_L leads V_ph by 30° | I_L lags I_ph by 30° |
| Power (balanced) | √3 V_L I_L cos φ | √3 V_L I_L cos φ |
| Neutral | available | not available |
For both, P = 3 V_ph I_ph cos φ, where φ is the angle between phase voltage and phase current.
✓ Answer checked by Claude Opus 5.5 (high effort)
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