---
title: "Nichrome strip resistor length and width"
subject: "Electric Machine Design"
chapter: 2
key: "semester_5/electric_machine_design:nichrome-strip-resistor"
source: https://www.bctengineer.com/q/semester_5/electric_machine_design/nichrome-strip-resistor
site: BCT Engineering
---

# Nichrome strip resistor length and width

Electric Machine Design · Chapter 2: Heating and cooling of electric machine

- Asked in: 2074 Asoj, 8 marks; 2072 Kartik, 6 marks (asked 2 times)
- Video: [ELECTRICAL HEATING  - PART - 03 - THREE PROBLEMS IN DESIGN OF HEATING ELEMENT](https://www.youtube.com/watch?v=4tFLfZXrAeM) — zenmurali Sachithanandam
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/electric_machine_design/nichrome-strip-resistor
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-2

## Question

A 250 volt, 1 kilowatt single element resistor is made from 0.2 mm thick nickel chromium strip. The temperature rise of the strip does not exceed 300°C. Calculate the length and width of the strip. Assume e = 0.9, Radiating efficiency = 0.75. Resistivity of nichrome = 1×10⁻⁶ Ω-m.

## Short answer

Resistance R = 250^2/1000 = 62.5 ohm. Radiating area needed: from Stefan-Boltzmann, q ≈ 4617 W/m^2, so l*w = 1000/(2*4617) ≈ 0.1083 m^2. From R = rho l/(w t), l/w = 12500. Solving yields length ≈ 36.8 m, width ≈ 2.94 mm.

## Answer

The strip must have the resistance that gives 1 kW at 250 V, and enough surface area to radiate 1 kW at the allowed temperature.

**Assumption:** ambient temperature 25 °C, so the strip temperature is 25 + 300 = 325 °C. Heat is lost by radiation only (Stefan's law), from both faces of the strip; the edges are neglected.

#### Step 1: resistance required
$$
R = \frac{V^2}{P} = \frac{250^2}{1000} = 62.5\ \Omega
$$

#### Step 2: heat dissipated per m² of surface
Stefan's law with emissivity $e$ and radiating efficiency $\eta$:

$$
\begin{aligned}
q &= 5.7\times10^{-8}\, e\,\eta\,(T_1^4 - T_2^4) \\
T_1 &= 325 + 273 = 598\ \text{K},\quad T_2 = 25 + 273 = 298\ \text{K} \\
q &= 5.7\times10^{-8}\times0.9\times0.75\times(598^4 - 298^4) \\
&= 3.8475\times10^{-8}\times(1.2788\times10^{11} - 0.0789\times10^{11}) \\
&= 4617\ \text{W/m}^2
\end{aligned}
$$

#### Step 3: surface area needed
For length $l$ and width $w$, the radiating surface is $2lw$:

$$
\begin{aligned}
2\,l\,w\,q &= 1000 \\
l\,w &= \frac{1000}{2\times4617} = 0.10830\ \text{m}^2 \quad (1)
\end{aligned}
$$

#### Step 4: length–width ratio from resistance
$$
\begin{aligned}
R &= \frac{\rho\, l}{w\,t} \\
\frac{l}{w} &= \frac{R\,t}{\rho} = \frac{62.5\times0.2\times10^{-3}}{1\times10^{-6}} = 12500 \quad (2)
\end{aligned}
$$

#### Step 5: solve (1) and (2)
$$
\begin{aligned}
l^2 &= 12500\times0.10830 = 1353.75 \\
l &= 36.79\ \text{m} \\
w &= \frac{l}{12500} = 2.943\times10^{-3}\ \text{m} = 2.94\ \text{mm}
\end{aligned}
$$

**Check:** $R = \dfrac{10^{-6}\times36.79}{2.943\times10^{-3}\times0.2\times10^{-3}} = 62.5\ \Omega$.

**Answer: length ≈ 36.8 m, width ≈ 2.94 mm** (for 25 °C ambient). With a 20 °C ambient the same method gives about 37.4 m and 2.99 mm.

## Check yourself

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

1. What resistance is needed for 1 kW at 250 V?
   A. 6.25 Ω
   B. 62.5 Ω
   C. 250 Ω
   D. 4 Ω
   Answer: B. R = V^2/P = 62500/1000 = 62.5 Ω. (Key point: Resistance)

2. Which law gives the heat dissipated per m2?
   A. Stefan's law with emissivity and radiating efficiency, using T1^4 - T2^4 in kelvin
   B. Ohm's law
   C. Newton's cooling law
   D. Fourier's law
   Answer: A. q = 5.7e-8 e eta (T1^4 - T2^4) with temperatures in kelvin. (Key point: Radiation law)

3. What two equations are solved for l and w?
   A. l w q = 62.5 and l/w = rho/(R t)
   B. l w = 1000 and l = w
   C. 2 l w = R and l/w = 1
   D. 2 l w q = 1000 and l/w = R t / rho
   Answer: D. Both faces radiate, so 2lwq = P, and R = rho l/(w t) gives l/w = Rt/rho = 12500. (Key point: Area and ratio)

4. What are the final dimensions?
   A. Length 62.5 m, width 5 mm
   B. Length 36.8 mm, width 2.94 m
   C. Length 36.8 m, width 2.94 mm
   D. Length 3.68 m, width 29.4 mm
   Answer: C. l^2 = 12500 x 0.1083, so l = 36.8 m and w = l/12500 = 2.94 mm. (Key point: Result)

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