star-delta transformation

star-delta transformation

1. What Star-Delta Are

Star (Y) = three resistors $R_Y1,R_{Y2},R_{Y3}$ meeting at a common central node (Y shape). Delta (Δ) = three resistors $R_Δ12,R_Δ23,R_Δ31$ forming a triangle (Δ shape) between three outer nodes (1,2,3). They are equivalent if resistance between any two outer nodes is same with third node isolated.

Transformation allows simplify circuits where $R$ network is neither series nor parallel (e.g., bridge).

2. Formulas — Delta to Star (Δ→Y)

Given Δ resistors $R_{ab}, R_{bc}, R_{ca}$ between nodes a-b, b-c, c-a, equivalent Y resistors from each outer node to centre $n$:

i.e., Y = (product of adjacent Δ)/sum of Δ.

3. Star to Delta (Y→Δ)

Given Y $R_a,R_b,R_c$ from nodes a,b,c to centre:

where

i.e., Δ = sum of all pairwise Y products / opposite Y.

4. Balanced Case (all equal) — Simplifies

If all Δ are equal $R_Δ$, then each Y is:

If all Y equal $R_Y$, then each Δ is:

This 1:3 ratio is exam shortcut.

5. Worked Examples

(a) Balanced Δ with 30Ω each → Y each =10Ω ✓

(b) Δ: R_ab=6Ω, R_bc=6Ω, R_ca=3Ω? Actually Δ sum=6+6+3=15: R_Ya=6·3/15=1.2Ω, R_Yb=36/15=2.4Ω, R_Yc=18/15=1.2Ω ✓

(c) Y: Ra=1Ω,Rb=2Ω,Rc=3Ω: Σ=1·2+2·3+3·1=2+6+3=11: R_ab=11/3≈3.67Ω, R_bc=11/1=11Ω, R_ca=11/2=5.5Ω ✓

(d) Bridge imbalance: Delta between nodes 1-2-3 can be converted to Y to make series/parallel obvious then compute R_eq.

(e) Power? Equivalent preserves resistance, not power? But same external behaviour.

6. Exam Points

• State Δ→Y $R_Y = (product of adjacent Δ)/ΣΔ$ and Y→Δ $R_Δ = Σ(R_Y products)/opposite$ — 5-7 marks, very frequent (almost certain to ask with 3 values).
• Balanced case $R_Y=R_Δ/3$ and $R_Δ=3R_Y$ — 5 marks, very frequent.
• Solve $R_{eq}$ for bridge by delta-star conversion — 7-10 marks, very frequent (common numerical: find equivalent resistance).