--- title: "Filament lamp current at switch-on" subject: "Basic Electrical Engineering" chapter: 1 key: "semester_1/basic_electrical_engineering:lamp-inrush-current-temperature" source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/lamp-inrush-current-temperature site: BCT Engineering --- # Filament lamp current at switch-on Basic Electrical Engineering · Chapter 1: General Electric System - Asked in: 2074 Asoj, 6 marks; 2072 Chaitra, 6 marks (asked 2 times) - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/lamp-inrush-current-temperature - Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-1 ## Question A 60 watt, 240 V incandescent filament lamp is switched on at 20°C. The operating temperature of the filament is 2000°C. Determine the current taken by the lamp at the instant of switching ON. The temperature coefficient of resistance of the filament material is 0.0045°/k. ## Short answer Hot resistance at 2000°C is 960 Ω. Using α=0.0045, cold resistance at 20°C is 96.87 Ω. At switch-on, current = 240 V / 96.87 Ω ≈ 2.48 A. ## Answer A metal filament has a positive temperature coefficient, so its resistance is much lower when cold. At the instant of switching on, the filament is at 20°C, so the current is far larger than the normal running current. #### Hot resistance (at 2000°C) Rated power and voltage give the running resistance: R_hot = V²/P = 240²/60 = 960 Ω #### Cold resistance (at 20°C) The question does not state the reference temperature for α, so take α = 0.0045 per °C at 20°C: R_t = R₂₀ [1 + α(t − 20)] 960 = R₂₀ [1 + 0.0045 × (2000 − 20)] 960 = R₂₀ × 9.91 R₂₀ = 960/9.91 = 96.87 Ω #### Switch-on current I = V/R₂₀ = 240/96.87 = 2.48 A The running current is 60/240 = 0.25 A, so the switch-on surge is about 10 times larger. **Answer: Current at the instant of switching on ≈ 2.48 A** (If α = 0.0045 is taken at 0°C instead, R₂₀ = 960 × 1.09/10 = 104.6 Ω and I ≈ 2.29 A.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is the current at the instant of switching on a filament lamp much larger than its running current? A. The supply voltage is higher at the instant of switching on B. The cold filament has a much lower resistance because metals have a positive temperature coefficient C. The cold filament has a higher resistance, so more power is drawn D. The filament inductance releases stored energy Answer: B. Metal resistance rises with temperature, so at 20 degC the resistance is far lower and I = V/R is far larger. (Key point: Cold vs hot resistance) 2. How is the hot (running) resistance of the 60 W, 240 V lamp found, and what is it? A. V^2/P = 960 ohm B. V/P = 4 ohm C. P/V^2 = 0.00104 ohm D. V*P/2 = 7200 ohm Answer: A. R = V^2/P = 240^2/60 = 960 ohm. (Key point: Hot resistance from rating) 3. With R_hot = 960 ohm at 2000 degC and alpha = 0.0045 per degC at 20 degC, which relation gives the cold resistance R20? A. 960 = R20 [1 + 0.0045 x 2000] B. 960 = R20 [1 - 0.0045 x 1980] C. 960 x [1 + 0.0045 x 1980] = R20 D. 960 = R20 [1 + 0.0045 x 1980] Answer: D. R_t = R20 [1 + alpha (t - 20)] with t - 20 = 1980, giving R20 = 960/9.91 = 96.87 ohm. (Key point: Using the temperature formula) 4. What is the approximate current at switch-on, and how does it compare with the running current of 0.25 A? A. 2.48 A, about 10 times larger B. 0.25 A, the same C. 24.8 A, about 100 times larger D. 2.48 A, about 10 times smaller Answer: A. I = 240/96.87 = 2.48 A, which is about 10 times the 0.25 A running current. 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