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Equivalent resistance between A and B

Short answer

Network has two 4 Ω in parallel (2 Ω) and 4 Ω ∥ 3 Ω (12/7 Ω). Add series resistors: 3 + 2 + 12/7 + 1 = 54/7 ≈ 7.71 Ω.

Find the equivalent resistance in between point A and B in figure below. The corresponding resistance values are indicated in figure. All the resistances are in Ohm (Ω). [Figure: terminals A and B; 3 Ω and 4 Ω in the upper path; two 4 Ω resistors in vertical positions; 1 Ω and 3 Ω in the lower path.]

Reading the figure

Let N₁ be the junction after the first 3 Ω, N₂ the top right junction, and B′ the junction between the 1 Ω and the lower 3 Ω.

  • A to N₁: 3 Ω.
  • N₁ to N₂: the top 4 Ω.
  • N₁ to N₂ again: the left vertical 4 Ω, whose lower end is wired along the bottom to N₂ (the lower wire goes to the right edge and up to N₂).
  • N₂ to B′: the right vertical 4 Ω (its lower end goes to B′) and the lower 3 Ω (its right end is joined to N₂ by the outer wire).
  • B′ to B: 1 Ω.
 A --[3]-- N1 ==[4]|[4]== N2 ==[4]|[3]== B' --[1]-- B

Step 1: N₁ to N₂ (two 4 Ω in parallel)

4 ∥ 4 = 2 Ω

Step 2: N₂ to B′ (4 Ω in parallel with 3 Ω)

4 × 3/(4 + 3) = 12/7 = 1.714 Ω

Step 3: Add the series resistors

R_AB = 3 + 2 + 1.714 + 1 = 7.714 Ω

Answer: R_AB = 54/7 ≈ 7.71 Ω

✓ Answer checked by Claude Opus 5.5 (high effort)

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