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Star-Delta Transformation Calculator

Some three-terminal networks cannot be reduced by series and parallel rules alone. Converting a delta into an equivalent star — or the reverse — unlocks them, and the same transformation appears throughout three-phase work.

Star ↔ Delta Transform

Convert three delta (Δ) resistances between terminals A, B, C into the equivalent star (Y) network.

R_A (star)
R_B (star)
R_C (star)
  1. 1.Given delta: R_AB = 10 Ω, R_BC = 20 Ω, R_CA = 30 Ω
  2. 2.FormulaΣ = R_AB + R_BC + R_CASubstituteΣ = 10 + 20 + 30ResultΣ = 60 Ω
  3. 3.FormulaR_A = (R_AB × R_CA) ÷ ΣSubstituteR_A = (10 × 30) ÷ 60ResultR_A = 5 Ω
  4. 4.FormulaR_B = (R_AB × R_BC) ÷ ΣSubstituteR_B = (10 × 20) ÷ 60ResultR_B = 3.3333 Ω
  5. 5.FormulaR_C = (R_BC × R_CA) ÷ ΣSubstituteR_C = (20 × 30) ÷ 60ResultR_C = 10 Ω
Common trap: Star values are always smaller than the delta they replace (for equal arms, exactly ⅓). If your converted values came out bigger going Δ → Y, you applied the formulas backwards.

The formula

Δ→Y: R_a = (R_ab × R_ca) / (R_ab + R_bc + R_ca) | Y→Δ: R_ab = R_a + R_b + (R_a R_b / R_c)

R_ab, R_bc, R_ca
the three sides of the delta (ohms)
R_a, R_b, R_c
the three arms of the star, meeting at a common node (ohms)

Worked example

A delta with R_ab = 10 Ω, R_bc = 20 Ω, R_ca = 30 Ω.

  1. Sum of the sides: 10 + 20 + 30 = 60 Ω
  2. R_a = (10 × 30) / 60 = 5 Ω
  3. R_b = (10 × 20) / 60 = 3.33 Ω
  4. R_c = (20 × 30) / 60 = 10 Ω

Star arms of 5 Ω, 3.33 Ω and 10 Ω replace the delta exactly.

Where you'll use it

Unbalanced bridge problems, three-phase load conversion, and any network where series-parallel reduction stalls. For a balanced delta the star arms are simply one third of each side.

The laws behind it

Parts this applies to

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