--- title: "Equivalent resistance RAB, ladder network" subject: "Basic Electrical Engineering" chapter: 2 key: "semester_1/basic_electrical_engineering:equivalent-resistance-rab-3-6-4-5-ohm" source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/equivalent-resistance-rab-3-6-4-5-ohm site: BCT Engineering --- # Equivalent resistance RAB, ladder network Basic Electrical Engineering · Chapter 2: DC circuits - Asked in: 2070 Asar, 6 marks - Video: [Electrical Engineering: Basic Laws (18 of 31) Finding the Equivalent Resistor Ex. 3](https://www.youtube.com/watch?v=3HSIKumW9SE) — Michel van Biezen - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/equivalent-resistance-rab-3-6-4-5-ohm - Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-2 ## Question 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.] ## 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. ## Answer #### 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** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The left branch is A. 6 + 3 + 4 + 4 + 6 Ω B. 9 + 8 + 6 Ω C. 1.5 Ω D. (6||3) + (4||4) + 6 = 2 + 2 + 6 = 10 Ω Answer: D. Each parallel pair is reduced first, then the series elements are added. (Key point: Left branch) 2. From T, the 10 Ω is in parallel with the path 3 + 5 + 2 = 10 Ω, giving A. 20 Ω B. 10 Ω C. 5 Ω D. 0 Ω Answer: C. 10 || 10 = 5 Ω. (Key point: Right branch) 3. The right branch total is A. 2 + 5 + 6 = 13 Ω B. 2 + 5 + 3 = 10 Ω C. 5 Ω D. 2 + 3 Ω Answer: B. 2 Ω at the top, 5 Ω parallel part, and 6||6 = 3 Ω at the bottom. (Key point: Right branch total) 4. The three branches 10 Ω, 5 Ω, 10 Ω between A and D are A. in parallel, giving 2.5 Ω B. in series, giving 25 Ω C. in parallel, giving 5 Ω D. in parallel, giving 0.4 Ω Answer: A. 1/R = 0.1 + 0.2 + 0.1 = 0.4, so R = 2.5 Ω. 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