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Question
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.
Solution
Case-I : When the supply voltage is 220 V.
Power consumed by the bulb = 100 W
Excess power = 100 − 40 = 60 W
Power converted to light = 60% of 60 W = 36 W
Case-II : When the supply voltage is 200 V.
Power consumed = \[\frac{200}{220} \times 100= 82.64 W\]
Excess power = 82.64 − 40 = 42.64 W
Power converted to light = 60% of 42.64 W = 25.584 W
Percentage drop in light intensity,
\[p = \frac{36 - 25 . 584}{36} \times 100\]
\[ \Rightarrow p = 28 . 93 \approx 29 %\]
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