2
$\begingroup$

I increased the potential difference across and incandescent bulb and noted the change in voltage and current, which, of course, changed proportionally. Why does increased voltage increase the temperature (and thus, the colour temperature) of an incandescent bulb? Does it cause more electron collisions and thus more energy is transferred to the lattice of the conduction filament, heating it up? And a further question on this, when photons are released, are they all within the visible light spectrum or do some contribute to thermal radiation as well?

$\endgroup$
4
$\begingroup$

Why does increasing the voltage across and incandescent bulb increase its temperature?

Because incandescent bulbs produce light by virtue of being hot, and not through some other process like those used by other light bulbs. For a black body we know that the total amount of power radiated is:$$ P \propto T^4. $$ While for a real object, like the wire in a light bulb, the relationship won't be exactly this, it will still follow that pattern that more power output (and thus, input) means higher temperature.

Does it cause more electron collisions and thus more energy is transferred to the lattice of the conduction filament, heating it up?

Yes, this is a description of how the energy is transferred into the heat of the filament.

And a further question on this, when photons are released, are they all within the visible light spectrum or do some contribute to thermal radiation as well?

No. In fact, most of the photons are thermal and a substantial fraction of the energy is infrared, too. There will also be a small amount released as ultraviolet photons, too. Real wires aren't exactly black-bodies, but the spectrum (a description of the fraction of energy in each frequency) they produce is reasonably close to Planck's Law.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.