---
title: "Delta-star and star-delta conversion"
subject: "Basic Electrical Engineering"
chapter: 2
key: "semester_1/basic_electrical_engineering:delta-star-star-delta-conversion"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/delta-star-star-delta-conversion
site: BCT Engineering
---

# Delta-star and star-delta conversion

Basic Electrical Engineering · Chapter 2: DC circuits

- Asked in: 2070 Asar, 4 marks
- Video: [Delta to Wye and Wye to Delta Transformations | Engineering Circuit Analysis | (Solved Examples)](https://www.youtube.com/watch?v=XtciQU4KMSg) — Question Solutions
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/delta-star-star-delta-conversion
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-2

## Question

Explain with neat diagram and write the equations for Delta- Star Conversion and for Star-Delta Conversion.

## Short answer

Delta-to-star: R_A = R_AB·R_CA/(R_AB + R_BC + R_CA); similarly for R_B and R_C. Star-to-delta: R_AB = (R_A R_B + R_B R_C + R_C R_A)/R_C; similarly for R_BC and R_CA. These equivalences allow bridge circuit reduction.

## Answer

A delta (Δ) network of three resistors can be replaced by an equivalent star (Y) network, and the reverse, so that the resistance between each pair of terminals stays the same. This lets bridge circuits be reduced by series-parallel rules.

```text
        Delta                      Star
          A                          A
         / \                         |
    Rab /   \ Rca                    Ra
       /     \                       |
      B-------C                      O
         Rbc                       /   \
                                 Rb     Rc
                                 /       \
                                B         C
```

#### Delta to star
Let the delta resistances be R_AB, R_BC, R_CA. Each star resistor equals the product of the two delta resistors touching that terminal divided by the sum of all three:

- R_A = R_AB × R_CA/(R_AB + R_BC + R_CA)
- R_B = R_AB × R_BC/(R_AB + R_BC + R_CA)
- R_C = R_BC × R_CA/(R_AB + R_BC + R_CA)

For equal delta resistors R_Δ, R_Y = R_Δ/3.

#### Star to delta
Let the star resistances be R_A, R_B, R_C. Let S = R_A R_B + R_B R_C + R_C R_A.

- R_AB = S/R_C
- R_BC = S/R_A
- R_CA = S/R_B

(Each delta resistor is the sum of products divided by the star resistor opposite to it.) For equal star resistors R_Y, R_Δ = 3 R_Y.

#### Derivation idea
Equate the resistance between each pair of terminals with the third terminal open. For example, between A and B: R_A + R_B = R_AB ∥ (R_BC + R_CA). Solving the three equations gives the formulae above.

#### Example
A delta of 30 Ω, 30 Ω, 30 Ω gives a star of 10 Ω, 10 Ω, 10 Ω.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For delta to star, R_A =
   A. R_AB + R_CA
   B. (R_AB x R_BC)/(R_BC + R_CA)
   C. S/R_C
   D. R_AB x R_CA/(R_AB + R_BC + R_CA)
   Answer: D. The star resistor at a terminal is the product of the two adjacent delta resistors over the sum of all three. (Key point: Delta to star)

2. For star to delta, R_AB =
   A. R_A + R_B
   B. S/R_A
   C. S/R_C with S = R_A R_B + R_B R_C + R_C R_A
   D. R_A R_B/R_C
   Answer: C. Sum of products divided by the star resistor opposite to that delta resistor. (Key point: Star to delta)

3. For equal resistors, the star is
   A. R_Y = 3 R_delta
   B. R_Y = R_delta/3 (and R_delta = 3 R_Y)
   C. R_Y = R_delta
   D. R_Y = R_delta/2
   Answer: B. A 30 Ω delta gives a 10 Ω star. (Key point: Equal resistors)

4. The formulae are derived by
   A. equating the resistance between each pair of terminals in both networks with the third terminal open
   B. equating voltages across the sources
   C. equating power only
   D. assuming a current source
   Answer: A. Equivalence requires equal terminal resistance for each pair. (Key point: Derivation)

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