Please wait for the animation to completely load.

In the animation you can close and open switches to see what happens to the lightbulb. Restart.

When the animation begins, the capacitor is initially charged. Push the "play" button and then open/close the switches. Watch what happens to the lightbulb. When the lightbulb goes out (is dark), throw the switches again. What happens? You should observe that the lightbulb is bright initially and always eventually goes out (even when there is a battery in the circuit). Notice, too, that the lightbulb is the brightest right after you throw the switches. This means that the current is the biggest. When the bulb is out, however, the current is zero. From these observations, what then is the voltage across the bulb as a function of time after the switch is thrown?

Show the graph of voltage vs. time. The voltage across the lightbulb (green), the voltage across the capacitor (red) and the total voltage across both (blue) are shown. What does the graph look like for the situation when the capacitor is charging (capacitor and resistor in series with the battery)? Notice that the lightbulb voltage plus the capacitor voltage equal the total voltage and that current flows (the lightbulb lights) until the capacitor voltage equals the battery voltage. What does the graph look like when the capacitor is discharging (battery not in the circuit with the capacitor and the resistor)? Notice that the capacitor and resistor voltages are equal and opposite so that the sum of their voltages is zero. A negative voltage across the bulb simply means that current is flowing in the other direction as the capacitor discharges down to 0 V. So as the capacitor charges, the current flows from the battery through the resistor and charge builds up on the capacitor, but as the capacitor discharges, the current flows from the capacitor through the resistor until the capacitor has no charge on its plates.

Illustration authored by Anne J. Cox.
Script authored by Wolfgang Christian.