Physics, asked by pranavsjoshi4, 5 months ago

Pouler of the bulb.)
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on what factor dose the resut ance of
sonlutor depend? Explain
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Answers

Answered by Sakshithelearner
0

Answer:

Both of these are fairly obvious if you think about it. A cold filament gives off no light at all, and then as you heat it up it gradually starts to glow red and then gets increasingly brighter with temperature. In theory this only happens up to a maximum, which would be achieved at around 5000˚K. Above this temperature UV radiation would become predominant. However in practice metals melt long before this temperature is reached. We use Tungsten for most filaments as this metal has a very high melting point, but this is still only 3700˚K, and tungsten sublimation (the solid evaporating, going straight to gas) is a problem before this temperature is reached. In practice therefore the brightness of a bulb is a tradeoff with how long the filament will last. Typical filaments will operate at around 2800˚K.

As far as surface area is concerned, then obviously the more filament surface there is at a given temperature then the more light will be produced. This is complicated by the fact that filaments are tightly coiled, usually at least double coiled, which reduces the effective surface area compared to the same length of filament being stretched out uncoiled. So in practice the effective surface area is determined by a combination of the surface area of the wire and the degree to which this is coiled.

The temperature in turn is determined by how much electrical power is being dissipated in the filament, and the effective surface area once again. The temperature reached is an equilibrium, reached when the total rate of energy radiated from the filament (including infrared and UV, not just visible light) is equal to the total rate of energy dissipated. Radiation can be increased by higher temperature and/or increased effective surface area.

Hence an equally good alternative answer would be that the brightness is determined by a combination of the electrical power input and the filament effective surface area.

Answered by Msunkown
1

Answer:

In an L-C ocsillation maximum charge on capacitor can be Q

0

. If at any instant electrical energy is

4

3

th of the magnetic energy then the charge on capacitor at that instant is

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