It seems to me that the calculations in the original post are not quite right.
> 1F = 1V * 1 Coulomb.
The Farad is a Coulomb per Volt or Q/V not Q * V
This starting point inserts errors in the calculations that follow.
For example:
> Thus 1F = 1V * 1 Amp Second.
The unit V * Amp * sec is equal to watts * sec = energy in joules. A Farad is not a joule of energy.
Also note that the voltage drops in a capacitor as you draw current from it, so you need some circuitry to keep the voltage up to keep your QRP rig happy. As a result there is probably some practical level where the capacitor still has energy left, but it can not be effectively used any more. (The rig would still work for a while as voltage dropped, but in the example given the voltage is considered constant at 12 Volts.)
A better approach to this calculation might be to start with the total energy stored in the 5000 F cap that was charged to 12 V. The energy stored in a cap is:
Energy = (1/2) * C * (voltage squared) where C is in Farads, V is in volts, and Energy is in Joules
In the example that was used, the QRP rig takes a constant 12 volts dc at a 0.5 amp. This is 6 watts or 6 Joules per second. If this constant power could be drawn from the capacitor until exhausted, then the time it would last would directly follow by dividing the initial energy in the cap by the rate of energy removal. However, the added circuitry mentioned above between the cap and the rig would have losses, minimum useable cap voltage, etc.
73, Richard, KG4VOX
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