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Field winding final working temperature

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.

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.

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.

✓ Answer checked by Claude Opus 5.5 (high effort)

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