1. The problem statement, all variables and given/known data Consider the simple circuit shown below, in which a light bulb with resistance R and a capacitor with capacitance C are connected in series with a battery and an open switch. The capacitor has a large capacitance and is initially uncharged. The battery provides enough power to light the bulb when it is connected directly to the battery. What happens when the switch is closed? [URL=http://imageshack.us][PLAIN]http://img169.imageshack.us/img169/229/question7lg1.gif[/URL][/PLAIN] A) Nothing B) The light bulb glows with a constant intensity. C) The light bulb gradually gets brighter as time passes. D) The light bulb starts out at its brightest and then gradually gets dimmer as time passes. [my second guess] E) The light bulb blinks on and off. [my 1st guess] 2. Relevant equations no equation necessary 3. The attempt at a solution My 1st thoughts: Well i thought that the current would flow from the battery to the capacitor, where it would be stored. When the capacitor discharged it would send the current through the lightbulb, causing it to flash on and off. Apparently this is faulty thinking, as E is not the correct answer. my second thoughts: electrons flow from regions of low potential, such as those found on a - battery terminal to regions of higher potential. Therefore, the electrons will "flow" from the negative battery terminal across the closed switch and through the filament of the lightbulb, thus illuminating it. Next, the electrons will flow into the capacitor where they will be stored. As the capacitor's stored charge builds, it "sucks" more and more available current from the lightbulb, causing it to dim. Feel free to poke any holes in my theory.