Three-phase loads and sources are connected in one of two ways: star (Y) or delta (Δ). The two connections behave very differently — in star the line voltage is √3 times the phase voltage, in delta the line current is √3 times the phase current — and mixing them up is one of the most common mistakes in three-phase work. This post explains both, with a worked example on the same load.
Figure 1: Star (Y) connection (a) and delta (Δ) connection (b) of a balanced three-phase load. (Schematic, not to scale.)
Star (Y) connection
In star, one end of each phase load is joined at a common neutral point (N), and the other ends are the three line terminals (A, B, C). Each load sees the voltage between its line terminal and the neutral — the phase voltage, Vph.
- Line voltage: VL = √3 × Vph (the line voltage is the phasor difference of two phase voltages, 120° apart)
- Line current: IL = Iph (each line feeds exactly one phase load)
- Neutral: available — so single-phase loads can be connected between any line and neutral (230 V on a 400 V system)
- On a 400 V line: Vph = 400 / √3 = 231 V
Delta (Δ) connection
In delta, the three phase loads are joined end to end in a triangle, and the line terminals are the three corners. Each load is connected directly across two lines, so it sees the full line voltage.
- Phase voltage: Vph = VL (each load is across two lines)
- Line current: IL = √3 × Iph (each line current is the phasor difference of two phase currents)
- Neutral: none — only three wires
- On a 400 V line: Vph = 400 V
Power is the same formula in both cases: P = 3 × Vph × Iph × cos φ = √3 × VL × IL × cos φ.
Worked example — the same 20 Ω load in star and in delta
A balanced resistive load of 20 Ω per phase is connected to a 400 V, three-phase supply. Compare the two connections.
Star connection: Vph = VL / √3 = 400 / 1.732 = 231 V. Phase current Iph = Vph / R = 231 / 20 = 11.55 A, and line current IL = Iph = 11.55 A. Power P = 3 × 231 × 11.55 = 8000 W ≈ 8.0 kW. (Check: √3 × 400 × 11.55 = 8002 W ✓)
Delta connection: Vph = VL = 400 V. Phase current Iph = 400 / 20 = 20 A, and line current IL = √3 × 20 = 34.64 A. Power P = 3 × 400 × 20 = 24,000 W = 24 kW.
The delta connection draws three times the power of the star connection for the same load — 24 kW against 8 kW — because each load sees the full line voltage instead of the phase voltage. That is exactly why induction motors are started in star (one-third the starting current and torque) and switched to delta for running: the star-delta starter.
Key points to remember
- Star: VL = √3 × Vph, IL = Iph, neutral available; on 400 V, Vph = 231 V.
- Delta: Vph = VL, IL = √3 × Iph, no neutral; on 400 V, Vph = 400 V.
- Worked example (20 Ω per phase, 400 V): star → 11.55 A line, 8.0 kW; delta → 34.64 A line, 24 kW.
- Delta draws three times the power of star for the same load — the basis of the star-delta motor starter.
- Power in both cases: P = √3 × VL × IL × cos φ.