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QPages 218-219: Q11: HC Verma: Concepts of Physics - Volume 2 Thermal and Chemical Effects of Electric Current Pages 218-219: Q11 Solution: The coil of an electric bulb takes 40 watts to start glowing.…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often treat the 60% conversion factor as applying to the total supplied power, calculating light output as $0.6P$, instead of applying it only to the power exceeding the 40 W glow‑threshold ($P-40$). Always subtract the 40 W baseline first, then take 60% of the remainder for the light component.

The coil of an electric bulb takes 40 watts to start glowing. If more than 40 W are supplied, 60% of the extra power is converted into light and the remaining into heat. The bulb consumes 100 W at 220 V. Find the percentage drop in the light intensity at a point if the supply voltage changes from 220 V to 200 V.

Step-by-Step Verified Solution

**Answer:** Drop in light intensity = 29%. **Step-by-Step Explanation:** At 220 V: Excess power = 60 W. Light power = 0.6 _ 60 = 36 W. At 200 V: P=2002/484=82.64P = 200^2/484 = 82.64 W. Excess = 42.64 W. Light power = 0.6 _ 42.64 = 25.58 W. Drop = (36−25.58)/36=28.9%≈29%(36 - 25.58)/36 = 28.9\% \approx 29\% [citation:9].

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