---
title: "Equivalent resistance between A and B"
subject: "Basic Electrical Engineering"
chapter: 2
key: "semester_1/basic_electrical_engineering:equivalent-resistance-ab-3-4-4-4-1-3"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/equivalent-resistance-ab-3-4-4-4-1-3
site: BCT Engineering
---

# Equivalent resistance between A and B

Basic Electrical Engineering · Chapter 2: DC circuits

- Asked in: 2067 Asoj, 6 marks
- Video: [Equivalent Resistance (Solved Problem 1)](https://www.youtube.com/watch?v=jTjEMBe4NpY) — Neso Academy
- 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-ab-3-4-4-4-1-3
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-2

## Question

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.]

## 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.

## Answer

#### 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.

```text
 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**

## Check yourself

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

1. The two 4 ohm resistors between N1 and N2 combine to what?
   A. 8 ohm
   B. 4 ohm
   C. 1 ohm
   D. 2 ohm, since they are in parallel
   Answer: D. Equal resistors in parallel give half: 4||4 = 2 ohm. (Key point: Parallel pair)

2. What is the 4 ohm in parallel with the 3 ohm?
   A. 12/7 = 1.714 ohm
   B. 7 ohm
   C. 1.0 ohm
   D. 3.5 ohm
   Answer: A. 4x3/(4+3) = 12/7. (Key point: Parallel 4 and 3)

3. How is R_AB obtained?
   A. 3 + 4 + 4 + 1 = 12 ohm
   B. 3 + 2 + 1.714 + 1 = 7.71 ohm
   C. 2 + 1.714 = 3.71 ohm
   D. 1/3 + 1/2 + 1/1.714 + 1
   Answer: B. The parallel blocks are in series with the 3 ohm and 1 ohm. (Key point: Series sum)

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