circuits and power question

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ihatebluescrubs

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Question:

Two light bulbs, one 60 watt bulb (higher resistance) and one 100 watt bulb (lower resistance) are placed in series. A current is run through them. Which bulb glows brighter?

Answer:
This is a paradox. You would think that the 100 watt bulb would light, but you have to reason out what is going on. A 60 watt bulb has greater resistance and when plugged in alone with the same current glows dimmer. A 100 watt bulb with less resistance draws more current and glows brighter by itself. But in series, the bulb with more resistance would draw more current, thus gets more voltage, so the 60 watt bulb glows brighter, has more power.

I have a hard time understanding these parts:
-the answer basically says increase power means decreased resistance. Doesn't Power=(I^2)R? So increased resistance means increased power?

-If a circuit is in series, shouldn't the current be the same at all points? So why would "the bulb with more resistance draw more current"?

Thanks!

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The advertised power of the bulbs is the power when a single bulb is connected to a source with a nominal voltage. It's easier to use P=V^2/R here - it's equivalent to P=RI^2 but with V constant it should be easier to tell that for bulb with higher power will have lower resistance. So when you're buying a 100 W light bulb, you're actually buying a 110^2/100=121 Ohm bulb. 100 W is just a convention, telling you what the power will be under given circumstances.

When you have two light bulbs in series, the numbers for the power are completely different. The resistance on the other hand will remain the same. And you are absolutely correct, the current through each of the bulbs will be the same. Now is a good time to use P=RI^2. Since I is fixed, you can tell that the bulb with higher resistance (and lower power rating) will be brighter.

Saying that the bulb draws more current when connected in series is wrong. There is a larger voltage drop over it and it consumes more power but the current is the same.
 
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