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Due to change in main voltage, the temperature of an electric bulb rises from 3000K to 4000K. What is the percentage rise in electric power consumed?

Answer» When temperature , `T_(1) = 3000K` then
`E_(1) = sigma T_(1)^(4) = xx sigma (3000)^(4)`..(i)
When temperature, `T_(2) = 4000K`, then
`E_(2) = sigma T_(2)^(4) = xx sigma (4000)^(4)`..(ii)
% rise in electric power is `=(E_(2)-E_(1))/(E_1) xx100 = ((E_2)/(E_1)-1)xx100`
`=(((4000)^(4))/((3000)^(4))-1)xx100 = ((256)/(81)-1)xx100 = 216%`.


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