How Do You Determine Brightness in Series and Parallel Circuits?

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SUMMARY

The discussion focuses on determining the brightness of bulbs in series and parallel circuits, emphasizing that parallel circuits exhibit lower resistance than series circuits. The user, Xhanger, initially provided incorrect brightness rankings for several configurations of bulbs. A key correction highlighted that bulb C is brighter than bulbs A and B due to its position in the circuit, which allows it to receive a greater share of the total current. The explanation clarifies that current distribution in parallel circuits leads to varying brightness levels among bulbs based on their resistance.

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Xhanger
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Wats up fellow forumers. I just signed up for physics forum, looks like an awsome forum that I will be as active as I can in. In class I got these circuit questions for review and I just don't know if I know how to figure them out. They are series/parallel brightness questions. It says to order them from brightest to least bright, ex. (A=B>C>D=E). I know that parallel are less resistance than series.

Homework Statement


Here is a link to the circuits re-drawn in Paint
http://i129.photobucket.com/albums/p201/Codester93/CIRCUITS.jpg"



Homework Equations


Parallel Resistance < Series Resistance


The Attempt at a Solution


Here is what I think they are ( but I guess I am wrong at pretty much all of them)
1.A=B>C>D=E
2.A>D>B=C>E=F
3.G=F>A=B=C>D=E
4.C>A=B
5.E=B>C=D>A (I didn't really know how to do this one at all, pure guess)
6.B>F>D=E>C=A

If you could just tell me which ones are wrong and right. And with the ones that are wrong, could you please give me an explanation of why mine is wrong and yours is right. The explanation doesn't have to be long, just get the point over.

Any questions, comments, suggestions just post. I will be looking over this very often as I need this info with-in a few days ^^.

Thanks,
Xhanger
 
Last edited by a moderator:
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Welcome to physicsforums! Let's start with the first circuit, shall we?
Assuming all bulbs have the same resistance, then...

Xhanger said:
1.A=B>C>D=E

Should be C>A=B=>D=E

You got the fact that C > D=E correct, but consider this. The overall circuit has one total current. When it meets the first branch AB, it splits into 2 equal currents, and meets up again at the end of the branch. When the current hits the following branch, it splits into 2 currents again, but more so across the C bulb. Current is lazy and prefers the path of least resistance, and bulb C has less resistance than D+E. So more current goes through the bulb C, more than half of the total current, which makes the bulb brighter than A and B.

Are you comfortable with that explanation? I want to make sure you get that before I move on to the other parts.
 
Yeh I get that, could someone please check the rest, it's due tomarrow. >.<
And yes they are all the same resistance
 
Last edited:

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