Engineering How Do You Determine Brightness in Series and Parallel Circuits?

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To determine brightness in series and parallel circuits, it's essential to understand that parallel circuits have lower resistance than series circuits, leading to different current distributions. In the example provided, the brightness order for bulbs is influenced by how current splits at junctions; bulbs with less resistance receive more current and thus shine brighter. The initial guesses for brightness rankings were mostly incorrect, with specific corrections noted, such as bulb C being brighter than A and B due to its position in the circuit. The discussion emphasizes the importance of analyzing current flow and resistance to accurately rank brightness. Understanding these principles is crucial for solving circuit problems effectively.
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 resistence 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
 
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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
 
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