Skip to main content

Equivalent resistance between A and B

Short answer

Convert the delta A–n1–n2 (1 Ω, 5 Ω, 4 Ω) to a star: RA=0.4 Ω, Rn1=0.5 Ω, Rn2=2 Ω. Combine series Rn1+R4=2.5 Ω in parallel with Rn2+R5=5 Ω to get 1.667 Ω. Total RAB = 0.4 + 1.667 = 2.067 Ω.

Find the resistance between the terminals A and B in the circuit segment below. [Figure: R1 = 1.00 Ω in series from A to a node, R2 = 5.00 Ω continuing, R5 = 3.00 Ω to B; R3 = 4.00 Ω connected as a bypass above from A-side node to the node before R5; R4 = 2.00 Ω connected as a bypass below.]

Reading the figure

Nodes: A (left), n₁ (between R1 and R2), n₂ (right of R2, left of R5), B (right). Branches: A–n₁ = R1 = 1 Ω; n₁–n₂ = R2 = 5 Ω; A–n₂ = R3 = 4 Ω (top bypass); n₁–B = R4 = 2 Ω (bottom bypass); n₂–B = R5 = 3 Ω.

This is a bridge, so series-parallel rules alone do not work. Use a delta-star transformation.

Step 1: Delta A–n₁–n₂ (1 Ω, 5 Ω, 4 Ω) to star

Sum = 1 + 5 + 4 = 10 Ω

  • R_A = (R1 × R3)/10 = (1 × 4)/10 = 0.4 Ω
  • R_n1 = (R1 × R2)/10 = (1 × 5)/10 = 0.5 Ω
  • R_n2 = (R2 × R3)/10 = (5 × 4)/10 = 2.0 Ω

Step 2: Series-parallel

  • Path through n₁: 0.5 + R4 = 0.5 + 2 = 2.5 Ω
  • Path through n₂: 2.0 + R5 = 2 + 3 = 5 Ω
  • Parallel: 2.5 × 5/(2.5 + 5) = 1.667 Ω

Step 3: Add the leg at A

R_AB = 0.4 + 1.667 = 2.067 Ω

Answer: R_AB ≈ 2.07 Ω (31/15 Ω)

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Basic Electrical Engineering chapter 2 questions →