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Equivalent resistance RAB, ladder network

Short answer

Left branch reduces to 10 Ω, middle branch is 5 Ω, right branch reduces to 10 Ω. The three branches in parallel give R_AB = 1/(1/10 + 1/5 + 1/10) = 2.5 Ω.

Find the equivalent resistance across the terminals A and B, RAB. [Figure: network with terminals A and B; resistors of 6 Ω, 3 Ω, 2 Ω, 3 Ω, two 4 Ω, 5 Ω, 10 Ω, 5 Ω on the right, and 6 Ω, 6 Ω, 2 Ω, 6 Ω in the lower branches, node D at the bottom.]

Reading the figure

Terminals A (top) and the node D (bottom, drawn as the second terminal). The network has several branches between A and D.

Left branch: 6 Ω ∥ 3 Ω (at the top left), then in series with two 4 Ω in parallel, then in series with 6 Ω (bottom) to D.

  • 6 ∥ 3 = 2 Ω
  • 4 ∥ 4 = 2 Ω
  • Left branch = 2 + 2 + 6 = 10 Ω

Middle branch: 5 Ω from A straight down to D.

Right branch: from A, 2 Ω to the top right node T. From T:

  • the 10 Ω going down to the bottom node, and
  • the path 3 Ω + 5 Ω + 2 Ω = 10 Ω (right side) to the same bottom node.

10 ∥ 10 = 5 Ω

Then the bottom node returns to D through two 6 Ω in parallel: 6 ∥ 6 = 3 Ω.

Right branch = 2 + 5 + 3 = 10 Ω

Combine the three branches between A and D

1/R_AB = 1/10 + 1/5 + 1/10 = 0.1 + 0.2 + 0.1 = 0.4

R_AB = 1/0.4 = 2.5 Ω

Answer: R_AB = 2.5 Ω

✓ Answer checked by Claude Opus 5.5 (high effort)

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