1. The problem statement, all variables and given/known data The circuit has identical bulbs and an ideal battery. When bulb D is unscrewed from its socket (the socket remains in its place), what would happen to the brightness of bulb F? 2. Relevant equations V = IR 3. The attempt at a solution Since bulb D is unscrewed from it socket and the socket remains in its place, we can replace bulb D with a switch that is open. Assigning resistance of 1 for each bulb, the right side of circuit has 2R and the current for bulb F would be I = V/2R. When the bulb D is back in its place, the resistance of the right side of the circuit would be 5R/3 and the current for bulb F would be I = (3V/5R)/(2/3) = 9V/10R. This shows that when bulb D is unscrewed from its socket, bulb F would have less brightness. When I entered the above explanation it turned out to be wrong. Can anyone explain why the brightness of the bulb increases?