The coil of an electric bulb requires 40 W to begin glowing. When more than 40 W is supplied, 60 % of the surplus power is converted into light, while the remainder turns into heat. The bulb consumes 100 W when operated at 220 V . Determine the percentage drop in light intensity at a point if the supply voltage is reduced from 220 V to 200 V .

Step-by-step solution

The power consumed by the bulb is given by P = V^2 R . At V₁ = 220 V , the power consumed is P₁ = 100 W . The light power output at 220 V is: L₁ = 0.6 (P₁ - 40) = 0.6 (100 - 40) = 36 W At V₂ = 200 V , assuming the resistance R remains constant, the power consumed is: P₂ = P₁ ( V₂ V₁ )^2 = 100 ( 200 220 )^2 = 100 100 121 = 10000 121 W The light power output at 200 V is: L₂ = 0.6 (P₂ - 40 ) = 0.6 ( 10000 121 - 40 ) = 0.6 ( 10000 - 4840 121 ) = 0.6 5160 121 = 3096 121 W The light intensity is directly proportional to