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Jul 20, 2012 at 23:10 comment added W5VO @TonyStewart, try rereading the question. I think you're missing some details about the circuit from the original question. I will go ahead and draw a schematic since this is causing confusion. I agree that lightbulb thermal dynamics would probably cause issues if this was a real circuit, but it seems much more like a theoretical/homework question.
Jul 20, 2012 at 19:09 comment added D.A.S. Also the cold resistance is ~ 10% of hot resistance. so with contact bounce, you can expect the 60W bulb to get hot faster while the 100W bulb is slower in heating up since less than rated current and in equilibrium gets only 60/160 or 3/8 of the voltage (37.5%) and even less power. Proof is elementary.
Jul 20, 2012 at 19:01 comment added D.A.S. My question is did you realize you were burning the 60W bulb hotter than rated? and that running in parallel on 220V line would really make them fail quickly. YOu cannot run 2 different Watt rated bulbs in series at twice the rated voltage. One the lower power will have a larger voltage drop than rated and burn out faster. Since they share current, they will tend to run closer to similar power dissipation except the 60W bulb has a thinner filament and will run hotter. so"Power at the top half is 60W. Power at the lower half is 100W." is a false assumption.
Jul 20, 2012 at 5:12 comment added stevenvh @zebonaut - and fortunately the same result :-)
Jul 19, 2012 at 9:32 vote accept vvavepacket
Jul 19, 2012 at 9:20 history edited zebonaut CC BY-SA 3.0
added units into last eqtn
Jul 19, 2012 at 9:19 comment added zebonaut +1. A nice and elegant variation that shows how multiple approaches lead to equally good results.
Jul 19, 2012 at 9:02 history edited jippie CC BY-SA 3.0
added 218 characters in body
Jul 19, 2012 at 8:54 history answered jippie CC BY-SA 3.0