Delta-Wye Transformation Calculator
Delta-Wye (Δ-Y) Transformation Calculator
Convert a three-terminal resistor network between its delta (Δ, mesh) and wye (Y, star) forms in either direction, for unequal arms or balanced ones — and check the answer against the resistance you would actually measure between the terminals.
Delta-wye transformation
a delta of Rab = 100 Ω, Rbc = 200 Ω, Rca = 300 Ω
The two transformations
Ra = RabRca ÷ ΣΔ, Rb = RabRbc ÷ ΣΔ, Rc = RbcRca ÷ ΣΔ
Y → Δ, with ΣP = RaRb + RbRc + RcRa:
Rab = ΣP ÷ Rc, Rbc = ΣP ÷ Ra, Rca = ΣP ÷ Rb
balanced: RY = RΔ ÷ 3, RΔ = 3 RY
- R ab
- the delta arm between terminals a and b — it touches neither the star point nor terminal c
- R a
- the wye arm from terminal a to the star point
- star point
- the common node of the wye. It exists in the Y and not in the Δ, so nothing you measure there transfers
Worked example
a delta of Rab = 100 Ω, Rbc = 200 Ω, Rca = 300 Ω
ΣΔ = 100 + 200 + 300 = 600 Ω
Ra = RabRca ÷ ΣΔ = 100 × 300 ÷ 600 = 50 Ω
Rb = 100 × 200 ÷ 600 = 33.33 Ω
Rc = 200 × 300 ÷ 600 = 100 Ω
Check it: from a to b with c disconnected, the wye gives Ra + Rb = 83.33 Ω
and the delta gives Rab ∥ (Rbc + Rca) = 100 ∥ 500 = 83.33 Ω — the same, which is what makes them equivalent
The balanced case, both directions
| Each delta arm RΔ | Each wye arm RΔ ÷ 3 | Δ from a wye of the same value | Measured across two terminals, Δ | Measured across two terminals, Y |
|---|---|---|---|---|
| 1 Ω | 0.3333 Ω | 3 Ω | 0.6667 Ω | 0.6667 Ω |
| 10 Ω | 3.3333 Ω | 30 Ω | 6.6667 Ω | 6.6667 Ω |
| 47 Ω | 15.6667 Ω | 141 Ω | 31.3333 Ω | 31.3333 Ω |
| 100 Ω | 33.3333 Ω | 300 Ω | 66.6667 Ω | 66.6667 Ω |
| 1,000 Ω | 333.3333 Ω | 3,000 Ω | 666.6667 Ω | 666.6667 Ω |
What it is for, and where the factor of three comes from
Some resistor networks will not come apart. Series and parallel reduction works by finding two elements that share a node with nothing else attached, or that share both nodes — and in a bridge, or a three-terminal network with something across the middle, no such pair exists. The delta-wye transformation is the tool that breaks the deadlock: it swaps a triangle of three resistors for a star of three different ones that behaves identically at the three terminals, and the star almost always leaves series and parallel pairs that were not there before.
The classic example is a loaded bridge. Four arms with a fifth resistor across the middle: nothing is in series with anything, nothing is in parallel with anything. Replace the upper triangle — the two top arms and the middle resistor — with its wye, and the network becomes one resistor in series with two parallel branches, which is a two-line calculation. That is how you get the total resistance a supply sees when a detector is connected across a Wheatstone bridge, and it is checked against a full nodal solve in this page’s test suite.
The relations, and why they are the shape they are. Going Δ → Y, the arm at a terminal is the product of the two delta arms that meet at that terminal, over the sum of all three. Going Y → Δ, the arm between two terminals is RaRb + RbRc + RcRa over the wye arm at the terminal it does not touch. Both fall out of writing the three terminal-pair resistances for each network and demanding they match. From a to b with c open, the wye gives Ra + Rb and the delta gives Rab ∥ (Rbc + Rca); do that for all three pairs and solve the three simultaneous equations, and the formulas above are what comes out. The headline on this page is that first measurement, computed from both sides, so you can see the two agree.
Where R ÷ 3 comes from. Put Rab = Rbc = Rca = R into the Δ → Y relation and it gives R × R ÷ (3R) = R ÷ 3 for every arm. That is a derivation, not a rule to memorise, and it is worth doing because the factor of three is the source of a great deal of three-phase confusion. A balanced delta load of R per phase is equivalent to a star load of R ÷ 3 per phase; at the same line voltage it therefore draws three times the current and three times the power. That single fact is why a motor can be started in star and run in delta. The three-phase power calculator works in the same two connections from the supply side — line and phase voltages and currents — where this page is about the resistances themselves.
What the transformation does not preserve. Only the three terminals. The star point is a node that exists in the wye and simply is not there in the delta, so a voltage measured at it, or a neutral connected to it, has no counterpart. The power dissipated in the individual arms is different too — the total is the same, but which resistor is getting hot is not, so never use a transformed network to size components. The transformation is exact for linear resistances; for impedances at a single frequency the same algebra holds with complex numbers, which is how it is used in matching networks as the π-to-T transformation. For the loop-by-loop alternative when a network has two meshes, see the two-loop mesh solver.
Frequently asked questions
What are the delta to wye conversion formulas?
With ΣΔ = Rab + Rbc + Rca: Ra = Rab·Rca ÷ ΣΔ, Rb = Rab·Rbc ÷ ΣΔ, Rc = Rbc·Rca ÷ ΣΔ. Each wye arm is the product of the two delta arms meeting at its terminal, over the sum of all three.
What are the wye to delta conversion formulas?
With ΣP = Ra·Rb + Rb·Rc + Rc·Ra: Rab = ΣP ÷ Rc, Rbc = ΣP ÷ Ra, Rca = ΣP ÷ Rb. Each delta arm divides by the wye arm at the terminal it does not touch.
Why is a wye resistor one third of the delta resistor?
Only when all three are equal. Put Rab = Rbc = Rca = R into Ra = Rab·Rca ÷ ΣΔ and you get R × R ÷ 3R = R ÷ 3. For unequal arms there is no factor of three and the general formulas are needed.
What is the delta-wye transformation used for?
Reducing a network that no series or parallel step will touch — a bridge with a load across it is the standard case — and converting between the two ways of connecting a three-phase load. It is also known as the pi-to-tee transformation in filter and matching work.
Does it work for capacitors and inductors?
At a single frequency, yes: the same algebra with impedances instead of resistances. The arithmetic is then complex, and the result only holds at that frequency, so a transformed network is not a design you can build and expect to behave the same across a band.
Is the star point the same node before and after?
There is no star point in a delta at all. The equivalence is between the three terminals only; anything measured at the star point, or connected to it such as a neutral, has no counterpart in the delta form.
Related calculators
References
- Nilsson JW, Riedel SA. Electric Circuits, 11th ed. Pearson, 2019. §3.7 “Delta-to-Wye (Pi-to-Tee) Equivalent Circuits”. Section title confirmed against Pearson’s own published front matter for the eleventh edition.
- Kennelly AE. The equivalence of triangles and three-pointed stars in conducting networks. Electrical World and Engineer, vol. 34, no. 12, pp. 413–414, 1899. The original statement, which is why the transformation is also called the Kennelly star-mesh transform.
- Horowitz P, Hill W. The Art of Electronics, 3rd ed. Cambridge University Press, 2015. Chapter 1: resistor networks, Thévenin equivalents, and the limits of series-parallel reduction.