Just a basic question, how much energy is one Farad? Is there a way to calculate how many Joules there is in one Farad?
Does this depends on something or is it always a fixed amount?
If ive worked things out correctly its:
Joule = Farad * Voltage^2
(see: http://www.evergreen.edu/biophysics/technotes/electron/basics.htmhttp://www.evergreen.edu/biophysics/...ron/basics.htm)
The big question is, why do you ask?
Many thanks Stefan, much appreciated.
Its just something Ive been thinking of as a weapon for our robot. Very exprimental at the moment, hence I dont want to give out to much information about it.
If you have MSN or something Id be happy to share some of my ideas, but not on a public forum.
Tell you now, a Farad is a big thing to generate.
I have ICQ, you could always mail me ofcourse
Remember that EMP wheapons are not allowed.
Oh yeah, havent thought of that.
Its not an EMP weapon, though it does utilize a couple of big magnets (and Im not talking about solenoids).
I just happend to come by a similar project to what Im doing, and apparantly it should be more like this:
1/2CV^2
Where C is the capacitans in Farads, and V is the voltage in Volts.
If id have to venture a guess, Im thinking youre trying to accellerate something. Or does this have anything to do with you trackdrive system?
The factor 1/2 probably has to do with the inefficiency you experiance when dischargeing capacitors.
BTW: youre 7Ah NiMh batteries are incapable of delivering high currents.
Youre probably right about the inefficiency of discharging capacitors, or Im totally out of ideas.
Youre also correct on the fact that Im trying to accelerate something, but Im not sure about what results to expect. It may work great, which would mean that weve finally got an original weapon for our robot.
Or it may be rubbish, in which case Ill keep experimenting with it at home. In either case, Ive got something to do for the next few weeks.
As for the Ni-Mh batteries, were using NCC60s as speed controllers, it wont require a high current!
The efficiency in charge and discharge of a capacitor should be very high as long as it doesnt have to store the charge for too long (when internal resistances will cause it to slowly discharge itself).
The factor of 1/2 is down to the pure physics of it - its nothing to do with efficiency. (in fact, the equation is the result of an integration of C x V with respect to voltage, in the same manner as integrating momentum (m x v) with respect to velocity results in kinetic energy - 1/2mv^2).
I assume you are using supercaps since you are talking capacitance in Farads. Just remember that these generally have a maximum voltage of 5.5V. They WILL blow up if you exceed this! You could try putting them in series to get around this, but this would reduce the overall capacitance (2 1F caps in series = 0.5Farad).
What are you trying to do exactly?
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