One can't be certain without knowing the voltage/ temperature relationship for the bulb/lamp (over the range of voltages of interest), but I'd be pretty surprised if the voltage reduction didn't 'win' (over resistance reduction) - so I think the answer will usually be (b) - but I may be wrong. Don't forget that, if resistance remained unchanged, power is proportional to E² (or V²), not E (or V).
Yes, in practice, I'm sure that must be the case, but ...
Indeed - and conversely, halving the voltage will decrease the current, but not to half the current, because of the reduction in resistance. Theoretically, if the temperature coefficient of resistance of the material were high enough, the current might not fall at all, or might even rise - but, as winston said, that would then result in an 'unstable' situation (essentially positive feedback). Whether any real-world material would/could do that, I don't know.
One might think so - but it's of note that the first response (albeit not a direct answer to the question) was actually wrong - and that there are a few "probably" qualifications to the correct answers! One suspects that there was probably "a reason" why RF asked the question!I can't believe the question is even being asked!
Quite - although that theoretically does not preclude the possibility that, over some range of voltages, something 'surprising' might happen. At the other extreme, if one progressively increased voltage, one would expect power to also progressively increase, but it would eventually fall to zero
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