Resistance and the flow in a circuit

[MitsuShai]

Diagram: http://i324.photobucket.com/albums/k327/ProtoGirlEXE/bulbs.jpg

Assume each bulb is identical and have the same resistance to the flow.


I don't know how to determine the amount of flow into and out of bulbs B, E, A, C, and D. (amount as in large flow or small flow)

Suppose that the flow out of bulbs B and E is of size 1 glow (don't mind the unit, this is suppose to be conceptual), how large is the flow through bulbs B and E and why.

So B and E are in parallel, but I don't entirely know how to figure this question out. Is the flow through the bulbs B and E bigger than the flow out because there's resistance in the bulb?
My TA said this, "The flow is the same for two bulbs in parallel. If the flow was different going into the bulb from the flow coming out, then that would imply electricity gets used up and we learned that that isn't true from the last lab."
So I understand what she is saying but, what about bulbs A,C, and D? I initially thought that all the bulbs have the same flow at this point but she also said that the resistance effect the flow, but it effects the flow throughout that branch (I don't know what she means by this).

Now I'm just confused. So the flow IN AND OUT of bulbs B and E are the same because they're parallel. Ok, so the flow must effect bulbs A,C, and D. Bulbs C and D are in series, so the flow must have changed there, but what do I say about bulb A? A is parallel to C and D, so it's the same, but C and D are different?

 

[MitsuShai]

Also, position-wise, I think that bulbs A,C, and D must have greater flow than bulbs B and E, but I don't know how to explain this in "physics terms."

 

[WatermelonPig]

Brightness is basically the energy dissipation in a bulb. P = I^2*R. Find the current through each wire segment and compute power for each. For instance, let the emf = V and each bulb has a resistance R. I would begin by finding the equivalent resistance of the circuit.

 

[MitsuShai]

Brightness is basically the energy dissipation in a bulb. P = I^2*R. Find the current through each wire segment and compute power for each. For instance, let the emf = V and each bulb has a resistance R. I would begin by finding the equivalent resistance of the circuit.

No, it's suppose to be conceptual. There's no numbers involved.

 

[MitsuShai]

I understand the B and E network now, but I still don't understand the A,D,C network...

 

[WatermelonPig]

Well, as B and E are parralell and of the same resistance, the flow through is equal, so they equal brightness. Also, the flow into B and E is equal to the flow from above. Also, the flow through A is twice that of the flow through C and D.