A switch is closed to charge a 1 farad (yes, one FARAD) capacitor with a 3 volt battery (actually two 1.5 volt batteries in series). The capacitor has 1.5 volt light bulbs on each side of it in the circuit, as shown in the circuit drawing above.
Consider the following series of questions:
Q: What will happen when the battery is connected (switch turned to left position):
(a) both bulbs will light and stay lit,
(b) both bulbs will go on momentarily,
(c) only one bulb will light and stay lit (if so, which one?),
(d) only one bulb will go on momentarily (if so, which one?),
(e) neither bulb will go on at all,
(f) something else will happen (if so, what?).
A: (b), Both bulbs will go on momentarily as the capacitor charges, then they will fade out. Click below for video
Q: What will happen when the switch is opened (center switch position)?
A: Nothing: the capacitor remains charged and no current flows.
Q: What will happen when the switch is closed to the right, completing the circuit including the capacitor and the two light bulbs?
A: The capacitor will discharge through the bulbs, turning them on momentarily while current is flowing, with the intensity decreasing as the current falls to zero. Click below for video.