---
title: "Equivalent resistance between A and B"
subject: "Basic Electrical Engineering"
chapter: 2
key: "semester_1/basic_electrical_engineering:resistance-ab-r1-r5-bridge"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/resistance-ab-r1-r5-bridge
site: BCT Engineering
---

# Equivalent resistance between A and B

Basic Electrical Engineering · Chapter 2: DC circuits

- Asked in: 2072 Kartik, 4 marks
- Video: [Find the equivalent resistance between points a and b](https://www.youtube.com/watch?v=q1oer_qoBnk) — WNY Tutor
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/resistance-ab-r1-r5-bridge
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-2

## Question

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.]

## 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.

## Answer

#### 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)**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Why is delta-star conversion used here?
   A. It has a current source
   B. It has too many resistors
   C. The network is a bridge and cannot be reduced by series-parallel rules alone
   D. It has a capacitor
   Answer: C. No pair of resistors is purely in series or parallel, so a transformation is needed. (Key point: Why delta-star)

2. In delta to star, each star resistor equals
   A. sum of the two adjacent delta resistors
   B. product of the two adjacent delta resistors divided by their sum (10 Ω)
   C. product of all three divided by the sum
   D. delta resistor divided by 3 only
   Answer: B. R_A = R1 x R3/10 = 4/10 = 0.4 Ω. (Key point: Star leg)

3. After conversion, the parallel part is 1.667 Ω. R_AB is
   A. 0.4 + 1.667 = 2.07 Ω
   B. 1.667 Ω
   C. 0.4 x 1.667 Ω
   D. 5 Ω
   Answer: A. The star leg at A is in series with the parallel combination. (Key point: Final addition)

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