Y / Δ converter — 3-phase power phase

Use the calculator

Pick whether your input is in Y or Δ form, enter line voltage, line current, and (optionally) per-phase impedance. The calculator returns line-to-line, line-to-neutral, both phase currents, and both equivalent phase impedances.

The four Y / Δ formulas

In wye, the line voltage is the vector sum of two line-to-neutral voltages 120° apart, giving the √3 factor. So 230 V phase-to-neutral becomes 400 V phase-to-phase; 277 V becomes 480 V; 120 V becomes 208 V.

At each junction in a delta winding, two phase currents merge into one line current. The vector sum gives line current = √3 × phase current. The internal winding sees less current than the line — useful for sizing the wire inside the motor.

For three equal impedances connected in delta, the equivalent wye has one-third the per-leg impedance. Used in motor control: same physical winding gives three times the impedance to start (in Y) and one-third the impedance to run (in Δ).

For unbalanced (unequal) impedances. Each Y impedance equals the product of the two adjacent Δ impedances divided by the sum of all three. Cyclic for the other terminals (b, c). Reverse: Z_Δ,ab = (Z_Y,a · Z_Y,b + Z_Y,b · Z_Y,c + Z_Y,c · Z_Y,a) / Z_Y,c.

How to convert Y ↔ Δ, step by step

1. Identify the connection. Y (star) has a common neutral point and three phase windings. Δ (delta) has windings connected end-to-end in a closed triangle, no neutral. Most North American distribution: delta primary, wye secondary. Most industrial motors: delta-wound stator with wye option for soft start.
2. Use the right voltage convention. For Y: V_LL = √3 · V_LN (line-to-line is 1.73× line-to-neutral). For Δ: V_LL = V_phase (the line voltage IS the phase voltage). Mixing conventions is the most common error in 3-φ math.
3. Use the right current convention. For Y: I_phase = I_line (the same current flows through the line and the winding). For Δ: I_phase = I_line / √3 (line current splits between two windings at each junction).
4. Convert impedance with the Δ-Y transformation. For balanced loads: Z_Y = Z_Δ / 3. A 30 Ω-per-phase delta load is equivalent to a 10 Ω-per-phase wye load. The line current is the same; only the per-winding voltage changes.
5. Pick the connection that matches your task. Y for 4-wire systems needing both 3-φ and 1-φ loads (residential service). Δ for 3-wire industrial systems with no neutral. Wye start / delta run for soft-starting motors (~⅓ starting current).
6. Compute power the same in both. Total 3-φ power S = √3 · V_LL · I_line is the same regardless of connection. The internal voltage and current per winding differ, but at the line terminals they look identical.

Comparison table — Y vs Δ at the line terminals

Worked example: 100 hp motor on 480 V Δ

A 100 hp 480 V 3-phase motor wound in delta has 124 A line current at full load (NEC Table 430.250). Compute the per-winding current, the wye-equivalent line voltage, and the wye-equivalent winding impedance assuming a per-phase impedance of 1.5 Ω in delta.

Variants and special cases

Open delta (V-V)

Three-phase service from only two transformers connected in V-shape. Delivers ~57.7 % of the equivalent 3-transformer bank rating. Used for small services where one of three transformers fails — emergency operation only, not a permanent design.

Zig-zag transformer

A specialised winding pattern that creates a neutral point in a 3-wire delta system without requiring a Y secondary. Used for ground-fault protection in industrial delta systems and for harmonic mitigation.

Corner-grounded delta

One of the delta corners is intentionally grounded — gives the equivalent of a single-phase system with a 480 V phase-to-ground while preserving 3-phase distribution. Used historically; rare in modern practice because phase-to-ground voltages on the ungrounded phases are full V_LL = 480 V.

The Kennelly transformation

A network of three impedances arranged in a triangle is electrically equivalent to a three-pointed star network if the impedance of each star arm equals the product of the two adjacent triangle sides divided by the sum of all three. This equivalence holds for any combination of impedances, balanced or otherwise.

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Frequently asked questions

What is the difference between Y and Δ in 3-phase?

Y (wye / star) has three phase windings joined at a common neutral point. The line voltage between two phases is √3 times the phase-to-neutral voltage. Δ (delta) connects three windings in a closed triangle; the line voltage equals the winding voltage but the line current is √3 times the winding current. Y has a neutral wire and is used where 1-phase loads share the system; Δ has no neutral and is more common in industrial motor circuits.

How do I convert delta impedance to wye?

Z_Y = Z_Δ / 3 for balanced loads. A 60 Ω delta load equals a 20 Ω wye load. The transformation lets you work an entire mixed-connection 3-φ network in either form, then convert back at the end. For unbalanced impedances, the full Kennelly formula applies (see Eq. 04 below).

Why is the line current √3 × phase current in delta?

At each junction in a delta winding, two phase currents merge into one line current. By Kirchhoff's current law: I_line = I_phase_a − I_phase_c (vector subtraction of two currents 120° apart). The result has a magnitude of √3 times either single phase current. For Y this doesn't happen — the line is the only path for that phase's current.

Can a motor run on either Y or Δ?

Yes — many industrial motors are dual-wound and shipped in delta with optional wye terminal jumpers. Wye-start, delta-run uses the higher impedance of wye on start (lower current, lower torque) then switches to delta for normal running (higher torque, full current). Reduces inrush 3× compared to direct-online start. Modern soft-starters and VFDs make Y-Δ switching less common.

What's the formula for line voltage in a wye system?

V_LL = √3 × V_LN. With V_LN = 230 V (European standard), V_LL = 400 V. With V_LN = 277 V (North American commercial), V_LL = 480 V. With V_LN = 120 V (NA light commercial), V_LL = 208 V. The √3 factor comes from the geometric phasor sum of two voltages 120° apart.

Why is the neutral point important in a Y system?

The neutral provides a return for unbalanced 1-φ loads sharing the 3-φ system. In a balanced load the neutral carries zero current — the three phase currents 120° apart sum to zero. Out-of-balance loads send the difference back through the neutral. NEC requires a neutral conductor in 3-φ systems with phase-to-neutral 1-φ loads (residential and most commercial); industrial 3-wire systems can run delta with no neutral.

What is the Δ-Y transformation in network theory?

The Kennelly Δ-Y (or Y-Δ) transform converts a triangle of impedances to an equivalent star, and back. Used in network analysis to simplify mixed-topology circuits before applying mesh or nodal analysis. The balanced-load case Z_Y = Z_Δ / 3 is a special case where all three impedances are equal.

Sources and methodology

1. Kennelly, A.E. The equivalence of triangles and three-pointed stars in conducting networks. Electrical World, 1899.
2. IEEE. IEEE Std 100 — The Authoritative Dictionary of IEEE Standards Terms, 7th Ed.
3. Hughes, A. Electric Motors and Drives — Fundamentals, Types and Applications, 5th Ed., Newnes, 2019. Y-Δ starting and motor windings.
4. NFPA. NEC NFPA 70, 2023 Ed. Article 250 — system grounding (Y vs Δ implications).