---
title: "Length of heater wire from conductivity"
subject: "Basic Electrical Engineering"
chapter: 1
key: "semester_1/basic_electrical_engineering:heater-wire-length-conductivity"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/heater-wire-length-conductivity
site: BCT Engineering
---

# Length of heater wire from conductivity

Basic Electrical Engineering · Chapter 1: General Electric System

- Asked in: 2067 Asoj, 6 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/heater-wire-length-conductivity
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-1

## Question

You are using a heater 220V/1KW made of a material of conductivity 16666.67 mho/cm. The wire has thickness of 0.5mm. What could have been the length of the given wire used in building heating element?

## Short answer

R = V²/P = 48.4 Ω. Resistivity ρ = 1/σ = 6×10⁻⁷ Ω·m. Area A = π(0.5×10⁻³)²/4 = 1.9635×10⁻⁷ m². Length l = RA/ρ = 15.84 m ≈ 15.8 m.

## Answer

The heater rating gives its resistance; the wire dimensions and conductivity then give the length.

#### Step 1: Resistance of the element
R = V²/P = 220²/1000 = 48.4 Ω

#### Step 2: Resistivity
σ = 16666.67 mho/cm = 16666.67 × 100 S/m = 1.6667 × 10⁶ S/m

ρ = 1/σ = 6 × 10⁻⁷ Ω·m (that is 60 μΩ·cm)

#### Step 3: Cross-sectional area
d = 0.5 mm, so A = π d²/4 = π (0.5 × 10⁻³)²/4 = 1.9635 × 10⁻⁷ m²

#### Step 4: Length
R = ρ l/A, so

l = R A/ρ = 48.4 × 1.9635 × 10⁻⁷/(6 × 10⁻⁷) = 15.84 m

**Answer: Length of wire ≈ 15.8 m** (assuming the wire is round with 0.5 mm diameter)

## Check yourself

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

1. How is the heater resistance found from 220 V / 1 kW?
   A. R = V^2/P = 48.4 ohm
   B. R = P/V = 4.5 ohm
   C. R = V x P
   D. R = V/P = 0.22 ohm
   Answer: A. R = 220^2/1000 = 48.4 ohm. (Key point: Resistance from rating)

2. Converting sigma = 16666.67 mho/cm to SI and finding resistivity gives:
   A. rho = 6e-5 ohm-m
   B. rho = 6e-7 ohm-m
   C. rho = 6e-9 ohm-m
   D. rho = 1.67e4 ohm-m
   Answer: B. sigma = 1.6667e6 S/m, so rho = 1/sigma = 6e-7 ohm-m. (Key point: Unit conversion)

3. The cross-sectional area of the 0.5 mm diameter wire is:
   A. 0.25 mm^2
   B. d^2 = 0.25e-6 m^2
   C. pi d^2/4 = 1.96e-7 m^2
   D. pi d = 1.57e-3 m^2
   Answer: C. A = pi (0.5e-3)^2/4 = 1.9635e-7 m^2. (Key point: Area)

4. The length of wire is:
   A. 1.58 m
   B. 158 m
   C. 0.158 m
   D. 15.8 m
   Answer: D. l = R A / rho = 48.4 x 1.9635e-7 / 6e-7 = 15.84 m. (Key point: Length)

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