---
title: "Temperature coefficient of resistance"
subject: "Basic Electrical Engineering"
chapter: 1
key: "semester_1/basic_electrical_engineering:temperature-coefficient-resistance-wire-4-6"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/temperature-coefficient-resistance-wire-4-6
site: BCT Engineering
---

# Temperature coefficient of resistance

Basic Electrical Engineering · Chapter 1: General Electric System

- Asked in: 2072 Kartik, 8 marks
- Video: [Effect of Temperature on Resistance | Electrical and Electronics Engineering](https://www.youtube.com/watch?v=yYD5X0jttvw) — Magic 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/temperature-coefficient-resistance-wire-4-6
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-1

## Question

Define temperature coefficient of resistance. The resistance of a certain length of wire is 4.6Ω at 20°C and 5.88Ω at 80°C. Determine (a) The temperature coefficient of resistance of the wire at 0° (b) The resistance of the wire at 60°C.

## Short answer

Temperature coefficient of resistance is fractional change in resistance per °C. Using R = R0(1+α0 t) with R20=4.6Ω, R80=5.88Ω: subtract to get R0α0=0.021333, then R0=4.173Ω, α0=0.00511 per °C. R60 = 4.173 + 60×0.021333 = 5.45Ω. (a) α0 ≈ 0.00511/°C, (b) R60 ≈ 5.45Ω.

## Answer

#### Definition
The **temperature coefficient of resistance (α)** is the fractional change in resistance per degree rise in temperature, measured from a stated reference temperature. For a rise from t₀: R_t = R₀ [1 + α₀ (t − t₀)]. Unit: per °C.

#### (a) α at 0°C
Let R₀ be the resistance at 0°C and α₀ its coefficient.

R₂₀ = R₀ (1 + 20 α₀) = 4.6
R₈₀ = R₀ (1 + 80 α₀) = 5.88

Subtract: 60 R₀ α₀ = 1.28, so R₀ α₀ = 0.021333

From the first: R₀ = 4.6 − 20 × 0.021333 = 4.1733 Ω

α₀ = 0.021333/4.1733 = 0.005112 per °C

#### (b) Resistance at 60°C
R₆₀ = R₀ (1 + 60 α₀) = 4.1733 + 60 × 0.021333 = 5.453 Ω

**Answer: (a) α₀ ≈ 0.00511 per °C; (b) R₆₀ ≈ 5.45 Ω**

## Check yourself

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

1. The temperature coefficient of resistance is best defined as:
   A. The fractional change in resistance per degree rise from a stated reference temperature
   B. The total resistance of the wire at 0 degC
   C. The rise in temperature per ohm
   D. The power lost per degree
   Answer: A. alpha is the change in resistance per unit resistance per degC, with respect to a reference temperature. (Key point: Definition)

2. To find alpha at 0 degC from R20 = 4.6 and R80 = 5.88, what is done?
   A. Divide 5.88 by 4.6 and use it as alpha
   B. Write R20 = R0(1 + 20 a0) and R80 = R0(1 + 80 a0) and solve the two simultaneously
   C. Use only R80 = R0(1 + 80 a0) with R0 = 4.6
   D. Average the two resistances
   Answer: B. Two unknowns, R0 and alpha0, need two equations; subtracting gives 60 R0 a0 = 1.28. (Key point: Setting up equations)

3. What is alpha0 approximately?
   A. 0.0213 per degC
   B. 0.0021 per degC
   C. 0.00511 per degC
   D. 0.0128 per degC
   Answer: C. R0 a0 = 0.021333 and R0 = 4.1733 ohm, so a0 = 0.00511 per degC. (Key point: Result for alpha)

4. R60 = ?
   A. 4.6 ohm
   B. 5.88 ohm
   C. 5.0 ohm
   D. 5.45 ohm
   Answer: D. R60 = R0 + 60 R0 a0 = 4.1733 + 1.28 = 5.453 ohm. (Key point: Resistance at 60 degC)

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