Analyzing Brightness in Series Lightbulb Circuits

[SFphysics101]

Hi
Two lightbulbs in series, one with 50W one with 100W which is brighter. I have two different solutions and can't see my error. Using PR=V^2 and I^2=P/V 50/√50R1=100/√100R2 with the same current and R2=2R1. Using PR=V^2 and the same voltage across both bulbs yields 50R1=100R2 or R1=2R2. Which is the appropriate solution when the bulbs are in series? Why are they different?

 

[sophiecentaur]

Hi and welcome. This looks like homework and should probably be posted elsewhere.
But it could be a more practical question and so:
With any circuit question you cannot assume anything that you haven't already been given or that you haven't yet calculated. The order of doing things has to be right.
Two resistors in series will 'share' the volts in the ratio R1/(R1+R2) : R2/(R1+R2). Potential Divider Equation - look it up.You can see that the 50W bulb has higher volts across it so it will be brighter. By how much, is a harder question. They will definitely not have the same volts across them (that daft result is arrived at by making invalid assumptions!).
You need to remember that tungsten filaments change their resistance by a factor of around 1:10 as they heat up. That makes it very hard to predict reliably what will happen.
If you assume that their resistances don't change (i.e that they don't actually glow), you can work out their resistances from the operating voltage and specified powers.
For real filaments, the 100W bulb will start off with 1/3 of the volts across it but, as they both heat up, the 50W bulb, with 2/3 volts will get hotter and its resistance will go proportionally higher than the 100W (lower resistance) bulb so it will hog more than its fair share of the volts. The 50V bulb will be a bit dimmer than normal but brighter than it would with 2/3 of the supply volts.

 

[SFphysics101]

When I said same voltage I meant a nominal voltage applied to each bulb individually to find the resistance. Other solutions use 120V as this nominal voltage and R=V^2/P. This finds the resistance at the voltage and power ratings of the bulb. Using this method gives a higher resistance for the 50W bulb. I cannot actually see through a voltage divider that the 50W bulb has higher resistance, but the resistance should not change with the voltage. I do know the voltage across each bulb will differ and the current will be the same through both bulbs, hence my second solution above which does not confirm the first. BTW I am out of school, or I'd ask a professor.

 

[berkeman]

Thread closed for Moderation...

 

[berkeman]

Thread will remain closed. OP has been reminded of the PF schoolwork rules, and will repost in the Homework Help forums.