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Can someone explain to me this physics question?

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NateRiver
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10 Apr 2013, 6:17 am

http://gyazo.com/670683fac1aeacf4b9341727b07f95b3

By the way, it's not homework. I'm practising for a physics challenge.



Apple_in_my_Eye
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10 Apr 2013, 7:00 am

I'm very sleepy, so this might be messed up, but...

I think the key is remember that in series the current through X & Y is the same,
and in parallel the voltage across them is the same.
And then to think about the power that each is emitting in each configuration.
I'll leave the rest of the details to you so as not to give away the answer.



ruveyn
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10 Apr 2013, 9:19 am

The answer is A



n2ogb
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10 Apr 2013, 10:06 am

The answer would be A if the question said both bulbs had the same brightness when connected directly to the battery. The question says both bulbs light up normally. Not all bulbs of the same voltage have the same brightness.



Chrisicus
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10 Apr 2013, 11:43 am

A, putting light bulbs in parallel splits the voltage/current equally so the bulbs get the same amount of brightness. But if the light bulbs are in series, the first bulb takes the bulk of the voltage/current so its brighter then the second bulb.



Stargazer43
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10 Apr 2013, 12:56 pm

Chrisicus wrote:
A, putting light bulbs in parallel splits the voltage/current equally so the bulbs get the same amount of brightness. But if the light bulbs are in series, the first bulb takes the bulk of the voltage/current so its brighter then the second bulb.


Only if you assume that the circuits to each bulb have equal resistance, which would probably be a valid assumption to make in this case. What you're looking to determine is power to each bulb, which is represented by any of the following equivalent equations: P=IV=(I^2)R=(V^2)/R

The voltage across each bulb will be the same, but the current/resistance may not be (but I'd say it's a reasonable assumption to assume they're equal)



Vectorspace
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10 Apr 2013, 5:03 pm

OK...

When the bulbs are in series, the current though them (denoted "I") is obviously the same. But the power that X gets is higher, as it's brighter.
With P=I²*R, this means that X's resistance is higher.

Now let's put them in parallel. Then we have P=U²/R with the voltage "U" applied to both bulbs. This means that lower resistance means more power, so the answer is C.

Why does it mention that they operate normally at the battery voltage? I can think of two explanations:
1. So the answer isn't "they'll blow up".
2. It suggests that that have the same efficiency (at least at battery voltage), so "brigher" equals "more power". Otherwise, they could just have the same resistance, and X could be brighter in both cases.