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 ? >> > |
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