---
title: "Field winding final working temperature"
subject: "Basic Electrical Engineering"
chapter: 1
key: "semester_1/basic_electrical_engineering:dc-motor-field-winding-temperature"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/dc-motor-field-winding-temperature
site: BCT Engineering
---

# Field winding final working temperature

Basic Electrical Engineering · Chapter 1: General Electric System

- Asked in: 2071 Chaitra, 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/dc-motor-field-winding-temperature
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-1

## Question

The field winding of d.c. motor connected across 230 V supply takes 1.15 A at room temperature of 20°C. After working for some hours the current falls to 0.26 A, the supply voltage remaining constant. Calculate the final working temperature of field winding. Resistance temperature coefficient of copper at 20°C is 1/254.5.

## Short answer

Cold resistance 200Ω (230V/1.15A), hot 884.6Ω (230V/0.26A). With α20=1/254.5 per °C, R_t = R20[1+α20(t−20)] gives 4.423 = 1 + (t−20)/254.5, so t−20 = 871.2°C, t ≈ 891°C. Final working temperature ≈ 891°C.

## Answer

The field winding is copper, so its resistance rises with temperature. Compare the resistance at 20°C with the resistance when hot.

#### Step 1: Resistances
R₂₀ = V/I = 230/1.15 = 200 Ω
R_t = 230/0.26 = 884.6 Ω

#### Step 2: Temperature
α₂₀ = 1/254.5 per °C

R_t = R₂₀ [1 + α₂₀ (t − 20)]
884.6/200 = 1 + (t − 20)/254.5
4.423 = 1 + (t − 20)/254.5
t − 20 = 3.423 × 254.5 = 871.2

t = 891.2°C

**Answer: Final temperature ≈ 891°C**

The given currents imply a very large rise in resistance, so the answer is far above the normal working temperature of a field winding. The numbers are used exactly as stated in the question.

## Check yourself

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

1. How are the cold and hot resistances found?
   A. R = V x I for each
   B. R = V/I at each current: 200 ohm and 884.6 ohm
   C. R = I/V for each
   D. R = V/I using 1.15 A for both
   Answer: B. Supply is constant at 230 V, so R20 = 230/1.15 = 200 ohm and R_t = 230/0.26 = 884.6 ohm. (Key point: Resistances from V and I)

2. Why does the current fall after the motor has run for some hours?
   A. The supply voltage drops
   B. The winding heats up, so copper resistance rises
   C. The winding cools down
   D. The field winding inductance rises
   Answer: B. Copper has a positive temperature coefficient; as it warms, resistance rises and current falls at constant voltage. (Key point: Why current falls)

3. With alpha20 = 1/254.5, which equation gives the temperature?
   A. 884.6 = 200 [1 + (t - 20)/254.5]
   B. 884.6 = 200 [1 + t/254.5]
   C. 200 = 884.6 [1 + (t - 20)/254.5]
   D. 884.6 - 200 = (t - 20)/254.5
   Answer: A. R_t = R20 [1 + alpha20 (t - 20)] gives t - 20 = 871.2 so t = 891.2 degC. (Key point: Temperature formula)

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