1. The problem statement, all variables and given/known data 2. Relevant equations I know that power of a lightbulb/resister is I^2 * R 3. The attempt at a solution OK. Here's my attempt: Let's consider the situation where we do not close the lightbulb and assume that the wires and switch are ideal because they have no resistance. I have a Total Itot current flowing through the circuit loop, and everyone is happy. Now comes along some random guy and he closes the switch. The total current in this case will be flowing from the positive side of the battery to the first lightbulb, go through it, and then reach the junction. My impression was that the current will then split itself, with some going into the branch of bulb2 and some going through the switch before the two sub-currents meet up and then head back to the battery's terminal. The equivalent resistance of the initial configuration without the switch will be 2*R (R in this case is the resistance of each of the batteries), but connecting the switch with the ideal resistance of zero will have the Req also equal 2*R. Thus, the total current in the circuit would remain the same as ElectromotiveForce = I * Requivalent, and Requivalent is going to remain the same. However, due to the loss of some current going through the switch, the current going through bulb2 would decrease, and thus by Power = I^2 * R, it's power (brightness) would decrease while the current through bulb1 which equals Itotal will remain the same. The explanation that I read says that the switch will short out the current, with all of the current flowing through the new pathway created by the closed switch, and none will flow through bulb2, making it go out. With that, the circuits total resistance decreases, so more current flows through bulb1 making it brighter, so we would have B as the answer. Why is it the case that all of the current goes through the pathway caused by the switch with even a tiny portion of the current not going through bulb2 at all? Thanks in advance.