1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
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.

Open in App
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 = $\frac{200}{220}×100$ = 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, $p=\frac{36-25.584}{36}×100\phantom{\rule{0ex}{0ex}}⇒p=28.93\approx 29%$

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Building a Thermometer
PHYSICS
Watch in App
Explore more
Join BYJU'S Learning Program