The resistor is to limit current through the LED which is a "constant
voltage" device like any diode.

The LED will start dimming the instant power is removed.  I would expect it
to dim rapidly at first and then more slowly.  It might retain a minimal
glow for some time -- LEDs will visibly emit with microamps of current
running through them.

LEDs dim perfectly well.  If you put 2-watt LEDs in your instrument panel
they would dim similarly to the 2-watt incandescent lamps they replace --
at high brightness.  At low brightness they would still dim much more
slowly than the incandescents because the latter have to be red hot before
they begin to emit, which takes considerable current.  But if you replace
your two watt panel lamps with 0.2 watt LEDs, the dimming rheostat meant to
work with a total of ~15 watts will have little effect on the 1.5 watt load.

The formula for energy in watt-seconds (aka joules) stored in a capacitor
is Estored = Cfarads x Evolts^2 / 2.  So your 4,000 uF cap charged to
twelve volts would in theory contain enough energy to run a two watt
festoon bulb for about three tenths of a second.  In practice it would run
longer than that because it would only be using the full two watts at the
instant power was disconnected.

The way the commercial circuits work is by using a timing circuit (no doubt
containing a timing capacitor) to operate a transistor that switches
battery current on and off.

Yrs,
d

On Mon, Jul 31, 2017 at 3:45 PM, Neil N <musomuso@gmail.com> wrote:

> Yes. I too think the resistor is for voltage drop. Parts list
> specifies "an LED" so.....
>
> My sense is that voltage stored in the cap would naturally taper off
> but kind of moot if I used an LED(s).
>
>