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1.) Depends on the charged voltage.
2.) Depends on the discharge current to the LED.
3.) Depends on the LED driver circuit
It depends on what the circuit looks like.
If the circuit is something overly simple, like just a resistor and a LED in series, then it will only be able to use a fraction of the capacitor's stored energy; i.e that available between the initial voltage, and the minimum forward voltage drop for the LED. Also for a simple circuit, the light output will not be constant. It will be bright at first, then dim as the capacitor voltage drops to the LEDs minimum forward voltage.
For a circuit containing some kind of power converter, you should be able to get more run time. To get an estimate for that, just divide the capacitors stored energy (U = 0.5*C*V^2) by the rate at which your circuit is using energy, i.e. dissipating power, in watts = joules/second. For a truthful estimate, include the power being wasted by the power converter circuit. For example, if your load draws 1 watt, 60% efficient converter would imply load and converter together are drawing 1/0.6 = 1.67 watt.
How much power (or how many mA's and at what voltage) do the LEDs draw? There are a lot of useful formulas that you can use to figure it out. To get accurate numbers though, we need to know what LEDs, and what driving circuitry.
I can tell you that it would probably be a VERY long time, if you are talking about a single super bright LED driven and like 0.5mA @ 2.5V with a high efficiency buck regulator
According to answer in this thread https://www.instructables.com/answers/How-many-mAh... one Farad = 0,2777mAh/V
So for example if your LED needs 20mA at 2V your capacitors should be charged at a higher voltage (4V for example) since the voltage drops when discharging and they will be able to run your LED for about 13 hours. But thats only an estimation and I don't even have much knowledge in this domain so I can be wrong. The best would be to try it and you'll see. Check out the other thread for further details.
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Posted:Feb 7, 2015
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