---
title: "Resistance of wire after drawing"
subject: "Basic Electrical Engineering"
chapter: 1
key: "semester_1/basic_electrical_engineering:wire-redrawn-resistance-change"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/wire-redrawn-resistance-change
site: BCT Engineering
---

# Resistance of wire after drawing

Basic Electrical Engineering · Chapter 1: General Electric System

- Asked in: 2069 Asar, 4 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/wire-redrawn-resistance-change
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-1

## Question

A 1km length of wire having a diameter of 11.7mm and a resistance 0.031Ω is drawn so that its diameter becomes 5mm. What does its resistance become?

## Short answer

Volume constant: l2/l1 = (d1/d2)². R ∝ l/A ∝ (d1/d2)⁴. R2 = 0.031 × (11.7/5)⁴ ≈ 0.93 Ω.

## Answer

When a wire is drawn out, its volume stays constant, so length rises as area falls. Resistance R = ρl/A.

#### Step 1: Area and length change
Volume constant: A₁l₁ = A₂l₂

A ∝ d², so A₂/A₁ = (5/11.7)² = 0.1826

l₂/l₁ = A₁/A₂ = (11.7/5)² = 5.476

l₂ = 1000 × 5.476 = 5476 m (new length)

#### Step 2: New resistance
ρ is unchanged, so

R₂/R₁ = (l₂/l₁) × (A₁/A₂) = (11.7/5)² × (11.7/5)² = (11.7/5)⁴ = 29.98

R₂ = 0.031 × 29.98 = 0.929 Ω

**Answer: New resistance ≈ 0.93 Ω (about 30 times larger)**

## Check yourself

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

1. What stays constant when a wire is drawn out to a smaller diameter?
   A. Its length
   B. Its volume
   C. Its cross-sectional area
   D. Its resistance
   Answer: B. Material is not lost, so A1 l1 = A2 l2; the length increases as area falls. (Key point: Constant volume)

2. If diameter goes from 11.7 mm to 5 mm, the length changes by a factor of:
   A. (11.7/5) = 2.34
   B. (5/11.7)^2 = 0.183
   C. (11.7/5)^2 = 5.48
   D. (11.7/5)^4 = 30
   Answer: C. Area is proportional to d^2, and length is inversely proportional to area, so l2/l1 = (11.7/5)^2. (Key point: Length change)

3. R2/R1 is:
   A. (11.7/5)^2 = 5.48
   B. (11.7/5)^4 = about 30, giving R2 = 0.93 ohm
   C. (5/11.7)^4 = 0.033
   D. 1, since rho is unchanged
   Answer: B. Length rises by 5.476 and area falls by the same factor, so R rises by 5.476^2 = (11.7/5)^4 = 29.98. (Key point: Resistance change)

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