Series/Parallel Circuit Help

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  • Thread starter Xhanger
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In summary: A>D>B=C>E=FShould be A>D>B=C>E=F.This circuit also has 2 currents, but this time they are both going through bulb E. The current through A and D is the same, but the current through B is greater, so it is brighter. Again, the current is lazy and prefers the path of least resistance, so the bulb with the lesser resistance is brighter.
  • #1
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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  • #2
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.
 
  • #3
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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1. What is the difference between a series circuit and a parallel circuit?

A series circuit is a circuit in which all components are connected in a single loop, while a parallel circuit is one in which the components are connected in multiple branches. In a series circuit, the same current flows through each component, while in a parallel circuit, the current is divided among the different branches.

2. How do I calculate the total resistance of a series circuit?

In a series circuit, the total resistance is equal to the sum of the individual resistances. This can be calculated by adding all the resistances together using the formula Rtotal = R1 + R2 + R3 + ...

3. What is the formula for calculating total voltage in a parallel circuit?

In a parallel circuit, the total voltage is the same across each branch. The formula for calculating total voltage is Vtotal = V1 + V2 + V3 + ..., where V1, V2, V3, etc. are the voltages of each individual branch.

4. What happens to the total current when resistors are added in a parallel circuit?

When resistors are added in a parallel circuit, the total current increases. This is because each resistor provides another path for the current to flow, resulting in a larger overall current.

5. How do I determine the total power dissipated in a parallel circuit?

In a parallel circuit, the total power dissipated is equal to the sum of the power dissipated in each individual branch. This can be calculated using the formula Ptotal = P1 + P2 + P3 + ..., where P1, P2, P3, etc. are the powers dissipated in each branch.

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