1. The problem statement, all variables and given/known data a.) The circuit above shows four identical bulbs connected to an ideal battery. Rank the bulbs in order from brightest to dimmest 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.