total charge lifted by the "charge escalator" 1. The problem statement, all variables and given/known data A 1.5 V flashlight battery is connected to a wire with a resistance of 5.00 ohms. The figure shows the battery's potential difference as a function of time. The slope is linear going from (0, 1.5 V) to (2 hrs, 0). 2. Relevant equations I = [tex]\Delta[/tex]V / R I = dQ / dt 3. The attempt at a solution I = 1.5 V / 5 ohms = 0.3 A 0.3 A = dQ / (2 hrs x 60 x 60) --> dQ = 2160 C This is not the correct answer, and I don't understand why. Thanks for your help.