Originally posted by Pirate88179
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Bill who told you that a SuperCap would continue to accept charge if the voltage across it remained the same?
Here are some basic facts that apply to all caps (SuperCaps included);
Q = CV
C = Q/V
V = Q/C
V is the voltage on the cap. C if the capacity in Farads and Q is the charge or Coulombs.
The energy in the cap can be found with a basic formula J = 1/2(CV^2) where J is Joules or WattSeconds. If you want WattHours then WHr = J/3600
So if yo have a cap of 650F @ 2.7V you get J=1/2(650*(2.7^2)) = 2.369E3 or 2,369 WattSeconds. Divide this by 3600 and 2369/3600= 6.581E-1 Watt Hours or 0.6581 = 658milliwatt hours.
Yes if you direct short this cap through the meter it should burn something up, but the fuse will not necessarily respond as the initial pulse is very short.
This is not much energy as the power is determined over time so you must use V*I*T.
The stored energy in a cap can also be found by W=CV^2/2 which I showed above.
Note that one Joule or WattSecond is only 2.78E-7 KwHr.
Hope this helps.
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