Simple Curcuit Question (Theory)

[skwz]

Homework Statement

http://img153.imageshack.us/img153/8941/physicscircuits1ms3.th.png
a.) The circuit above shows four identical bulbs connected to an ideal battery. Rank the bulbs in order from brightest to dimmest

http://img509.imageshack.us/img509/4655/physics3ya0.th.png
b.) Suppose that switch has been added to the circuit as shown. The switch is initially closed. When the switch is opened, will the current through bulb A increase, decrease, or remain the same?

Homework Equations

Inet = 0 @ a junction (Kirchhoff's Junction rule)

The Attempt at a Solution

a.) A = D > B = C
For a, Bulbs A and D have the same current going through both of them since they are in series. Therefore they are equally bright. For the parallel network between bulbs A and D, the current going through the current equals that of the current going through bulbs A and D. The current is split between both bulbs B and C then (identical, so same resistance) and they are half as bright as bulbs A and D

b.) Here is where I'm having trouble. I'm confused about the placement of the switch and the fact does current flow through bulb C still? If the switch were placed before bulb C , then the resistance in the circuit would increase since there is one less bulb in parallel right, meaning less current and a decrease in current through bulb A. But with the placement of the switch after bulb C, I thought, by Kirchhoff's junction rule that the current entering a junction must equal the current leaving or flowing out. Wouldn't bulb B now have more current flowing through it while Bulb A remains unchanged? Any help is appreciated.

 

[kamerling]

a) is OK
b) do you mean it matters if the switch is above or below bulb C? In both cases, there can be no current going through bulb C if the switch is open.
Kirchhof's junction rule tells you that I_A_closed = I_B_closed + I_C_closed
and I_A_open = I_B_open
I don't think you can draw any conclusions from that.
(I_A_open is the current through lamp a if the switch is open)

 

[skwz]

If no current is going through bulb C then does that mean that overall resistance in the circuit is increased since their resistances now add (bulbs A,B, and D are in series now)?

 

[kamerling]

If no current is going through bulb C then does that mean that overall resistance in the circuit is decreased since their resistances now add (bulbs A,B, and D are in series now)?

I think you mean increased

 

[skwz]

Woops... yeah meant increased! So if the equivalent resistance increases, then the total current in the circuit decreases and thus current through each bulb decreases?

 

[kamerling]

Woops... yeah meant increased! So if the equivalent resistance increases, then the total current in the circuit decreases and thus current through each bulb decreases?

The current through B doesn't decrease because it now gets all the current that flows through the circuit.

 

[skwz]

Ok gotcha. So the current through Bulb B increases since it now gets the current that flows through the circuit but this current is decreased for A and D as the equivalent resistance is now increased because all 3 bulbs are in series. So the total current in the circuit decreases, yet this is more than the current that bulb B had going through it before the switch was opened?