1. The problem statement, all variables and given/known data http://img153.imageshack.us/img153/8941/physicscircuits1ms3.th.png [Broken] 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 [Broken] 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? 2. Relevant equations Inet = 0 @ a junction (Kirchhoff's Junction rule) 3. 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.