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,

\[p = \frac{36 - 25 . 584}{36} \times 100\]

\[ \Rightarrow p = 28 . 93 \approx 29 %\]