---
title: "Star-delta equivalent resistance A to B"
subject: "Basic Electrical Engineering"
chapter: 2
key: "semester_1/basic_electrical_engineering:star-delta-resistance-ab-6-4-3-5"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/star-delta-resistance-ab-6-4-3-5
site: BCT Engineering
---

# Star-delta equivalent resistance A to B

Basic Electrical Engineering · Chapter 2: DC circuits

- Asked in: 2074 Asoj, 6 marks
- Video: [Determine the equivalent resistance between A and B, using star delta transformation](https://www.youtube.com/watch?v=1y6njIvQP0g) — Engg-Course-Made-Easy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/star-delta-resistance-ab-6-4-3-5
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-2

## Question

For the circuit shown in below figure, determine the resistance between points A and B using star / delta transformation theorem. [Figure: terminal A at top; a bridge-like network with 6 Ω and 4 Ω in the upper branches, 3 Ω and 5 Ω in the lower branches of the bridge, connected down through three legs 5 Ω, 8 Ω and 4 Ω to terminal B at the bottom.]

## Short answer

Transform the star (3,5,8 Ω) to delta, combine parallel resistors, then transform the upper delta (6,4,9.875 Ω) to star, and calculate series-parallel. R_AB ≈ 4.23 Ω.

## Answer

#### Reading the figure
Nodes: A (top, between 6 Ω and 4 Ω), L (left), R (right), M (middle, between 3 Ω and 5 Ω), B (bottom). The three legs 5 Ω, 8 Ω and 4 Ω all end at B.

Branches: A–L 6 Ω, A–R 4 Ω, L–M 3 Ω, M–R 5 Ω, L–B 5 Ω, M–B 8 Ω, R–B 4 Ω.

#### Step 1: Star at M (3 Ω to L, 5 Ω to R, 8 Ω to B) to delta
Sum of products = 3×5 + 5×8 + 8×3 = 15 + 40 + 24 = 79

- R_LR = 79/8 = 9.875 Ω (opposite leg 8 Ω)
- R_RB = 79/3 = 26.33 Ω (opposite leg 3 Ω)
- R_LB = 79/5 = 15.8 Ω (opposite leg 5 Ω)

#### Step 2: Combine in parallel with existing branches
- L–B: 5 ∥ 15.8 = 3.798 Ω
- R–B: 4 ∥ 26.33 = 3.473 Ω

#### Step 3: Delta A-L-R (6, 4, 9.875 Ω) to star
Sum = 6 + 4 + 9.875 = 19.875 Ω

- R_A = 6×4/19.875 = 1.208 Ω
- R_L = 6×9.875/19.875 = 2.981 Ω
- R_R = 4×9.875/19.875 = 1.987 Ω

#### Step 4: Series-parallel
Path via L: 2.981 + 3.798 = 6.779 Ω
Path via R: 1.987 + 3.473 = 5.460 Ω
Parallel = 6.779 × 5.460/(6.779 + 5.460) = 3.024 Ω

R_AB = 1.208 + 3.024 = 4.23 Ω

**Answer: R_AB ≈ 4.23 Ω** (checked by node analysis)

## Check yourself

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

1. In converting a star to a delta, each delta resistor equals:
   A. Sum of products of star resistors divided by the star resistor opposite to it
   B. Sum of star resistors divided by two
   C. Product of two adjacent star resistors divided by the sum
   D. Sum of the two adjacent star resistors
   Answer: A. R_delta = (sum of pairwise products)/(opposite star leg); here sum of products is 79. (Key point: Star to delta formula)

2. In converting delta (6, 4, 9.875 ohm) to star, R_A = 6 x 4/19.875 means:
   A. Product of the two adjacent delta resistors over the sum of all three
   B. Sum of the adjacent resistors over the product
   C. Product of all three over the sum
   D. The average of the delta resistors
   Answer: A. Each star leg is the product of the two delta resistors meeting at its node divided by the delta sum. (Key point: Delta to star formula)

3. For the star at M (3, 5 and 8 ohm), the delta resistor between L and R (opposite the 8 ohm leg) is:
   A. 79/3 = 26.33 ohm
   B. 79/8 = 9.875 ohm
   C. 79/5 = 15.8 ohm
   D. 16 ohm
   Answer: B. Opposite leg of L-R is the 8 ohm leg, so 79/8 = 9.875 ohm. (Key point: Star at M)

4. The resistance between A and B is approximately:
   A. 3.02 ohm
   B. 6.78 ohm
   C. 1.21 ohm
   D. 4.23 ohm
   Answer: D. R_A = 1.208 plus parallel of 6.779 and 5.460 (3.024) gives 4.23 ohm. (Key point: Final result)

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