1. The problem statement, all variables and given/known data For an ideal battery (r = 0 Ω) closing the switch in the figure below does not affect the brightness of bulb A. In practice, bulb A dims just a little when the switch closes. To see why, assume that the 1.5 V battery has an internal resistance r = 0.44Ω and that the resistance of a glowing bulb is R = 4Ω. 2. Relevant equations 3. The attempt at a solution I'm trying to wrap my head around this physically before I do the math. Brightness depends on power. We know that P=IV or v^2/R or R I^2 The battery keeps a constant voltage. When the switch closes the current should be going 50/50 to A and B because they have the same resistance. Now that means that the current should be SMALLER through A when the switch is closed. If the current is smaller and V is constant, how can it have the same brightness?!