Please Explain when an RC Circuit Opens - Effect on a Light bulb

[ncm2]

Homework Statement

Initially the switch is closed in the circuit shown and the light bulb is shining brightly. What happens when the switch is opened? (diagram in attached link)
1. The bulb goes out immediately.

2. The bulb gets gradually brighter.

3. The bulb gets gradually dimmer.

4. It cannot be determined.

5. The bulb continues to shine without a change in brightness.

Homework Equations

Vc=Vo*exp(-t/RC)

The Attempt at a Solution

I believe that the capacitor should cause the light to gradually dim. However, I also think that the open switch should mean no current flow, which would mean the bulb would immediately go out.

Any explanation is much appreciated

 

[LowlyPion]

I can't see the picture.

I suppose that the capacitor is in parallel with the light, and the voltage source is DC. (If different then ignore my comments.)

The charge in the capacitor will continue to flow, despite no current from the power source, and the light will - depending on how much charge is available in the capacitor - dim slowly or quickly, as it discharges through the light (current) until extinct.

You can see this on some devices like computers when you pull the plug the power on light remains lit and then dims.

 

[ncm2]

I will try to explain the diagram for you, but it sounds like it probably will dim gradually.

The diagram is a box, with a line through the middle, creating a figure 8 shape. Voltage source is on the left, which is connected in series to the light bulb which is on the top line, to the left of the middle line. The capacitor runs on a wire down the middle. Then the Switch is on the far side of the box from the voltage source.

Hopefully that helps a bit. I don't know if it matters that the switch is not in series with the capacitor?

 

[LowlyPion]

It sounds like the capacitor is in || with the lightbulb.

If there is a path remaining for the cap to discharge - one side to the other through the lightbulb, then that's what it will do - dim.

 

[tmvphil]

i could be wrong (noob here) but it seems to me that (from the now provided diagram) that the capacitor will be fully charged on the opening of the switch, giving no time for the current to ramp down so it will shut off quickly?