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How do you change a lightbulb?

How does the efficiency of a lightbulb change with input power?

  • Hi, Just wondering about the above question for a physics assignment, thought you guys might be able to help. I'm doing a hypothetical experiment where I change the input voltage into a lightbulb (and therefore input power), and work out its efficiency at each voltage (using specific heat capacity, temperature etc.) But I was wondering what trend I should see, will the bulb become more efficient as it gets brighter? Or the opposite? and why? Thanks in advance for any advice you can give :) Dan

  • Answer:

    Efficient in what? You need to define "efficiency". In terms of using all the supplied electrical energy, it is 100% "efficient". EDIT: You need to look at Planck's Blackbody Radiation Law. Loosely speaking, as you increase the voltage, the temperature of the filament increases (gets brighter) and the light output, which was reddish at low voltage, becomes whiter. The efficiency increases If the filament didn't melt (ideal filament!) and you kept ranking up the voltage, eventually the peak of the light would shift to the UV etc. The efficiency decreases. Specific heat plays no role, just the temperature. To actually calculate the efficiency, you need to integrate Planck's Law over the visible spectrum ( it will become a function of temperature only) and divide by the input total power.

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After about a second or so, the bulb is operating in a 'steady-state' - meaning that the temperature is steady. So the specific heat capacity doesn't make any difference as the temperature isn't changing. What happens in the steady-state is that the bulb converts electrical energy to electromagnetic radiation - infra-red and visible light, which are radiated away. This radiation is called 'black-body' radiation - you can look ot up. At small currents, the steady-state temperature is low and nearly all the energy is radiated as infra-red; there is virtually no visible light produced and the (optical) efficiency is effectively zero. If the current increases, the steady-state temperature increases;the proportions of infra-red decrease and the proportion of visible light increases - you see the filament glow. The efficiency has increased. As current increases, the filament gets hotter and efficiency continues to increase. However, the melting point of the filament (usually tungsten) prevents you using even higher currents and temperature to improve the efficiency - the filament would melt/vaporuise. Take a look at the link to see how the proportion of visible light to infra red changes with temperature- but you have to understand the graph in the middle.. For low temperatures, most of the energy is in the infra-red region.

Steve4Physics

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