With the given values and a two-watt festoon bulb, the cap has enough
energy to supply the lamp for a third of a second at full brightness.  In
practice it would fade down over a longer time, maybe a second?

Yrs,
d

On Mon, Jul 31, 2017 at 7:05 PM, David Beierl <dbeierl@attglobal.net> wrote:

> It's not a matter of "think".  It's basic physics.  The voltage from the
> capacitor will begin to drop the instant a load is applied, since it's only
> the pressure of so many electrons packed together that creates the
> voltage.  Start letting electrons out and the pressure drops in proportion
> to how many you let out.
>
> Yrs,
> d
>
> On Mon, Jul 31, 2017 at 6:24 PM, Alistair Bell <albell@shaw.ca> wrote:
>
>> Back to Neil's delayed off circuit. As it is what do you think will
>> happen? Will the cap output voltage drop ?
>>
>