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Star-delta equivalent resistance A to B

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.

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.]

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)

✓ Answer checked by Claude Opus 5.5 (high effort)

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