Confusion on Farad calculation

AI Thread Summary
The discussion centers on the discharge capacity of capacitors, specifically how voltage relates to capacitance and current. A 1 Farad capacitor can generate 1 volt at 1 amp for one second, while a 0.00005 Farad capacitor can theoretically generate 20,000 volts under the same conditions. The key equation, Q = C·V, indicates that current and time do not directly factor into the voltage calculation. When considering pre-charged capacitors, drawing 1 amp for one second from a 1F capacitor results in a minimal voltage drop, while the same action on a 0.00005F capacitor leads to a significant voltage drop. This highlights the importance of capacitance in determining voltage changes during discharge.
jerbeast
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Hi,

Please apologize the ignorant question, but I am confused on applying the discharge capacity of a capacitor.

If Farads = Coulombs/Volt, then a hypothetical 1 Farad capacitor could generate 1 volt at 1 amp for one second. But by the same formula, a .00005 Farad capacitor could generate 20,000 volts at 1 amp for one second.

What am I missing?

Thanks for your patience!
Jerry
 
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jerbeast,
You have it correct, but your wording is a little unusual. See if this version makes more sense to you...

Starting with uncharged capacitor:

If I pump 1A of current for 1 sec into a 1F capacitor, its voltage will be pumped up to 1V.

If I pump 1A of current for 1 sec into a .00005F capacitor, its voltage will be pumped up to 20,000v.
 
jerbeast said:
If Farads = Coulombs/Volt, then a hypothetical 1 Farad capacitor could generate 1 volt at 1 amp for one second. But by the same formula, a .00005 Farad capacitor could generate 20,000 volts at 1 amp for one second.

What am I missing?

Hi jerbeast! http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif

Q = C·V[/size]

There is no mention of current or time in that equation.
 
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the_emi_guy said:
jerbeast,
You have it correct, but your wording is a little unusual. See if this version makes more sense to you...

Starting with uncharged capacitor:

If I pump 1A of current for 1 sec into a 1F capacitor, its voltage will be pumped up to 1V.

If I pump 1A of current for 1 sec into a .00005F capacitor, its voltage will be pumped up to 20,000v.
That's a good way of looking at it. Perhaps OP could instead consider two capacitors that start off already charged. To nominate a figure, let's imagine each is pre-charged to 40kV.

Now, if you draw 1A for 1 sec from the 1F capacitor, its plate voltage will fall by just 1V, and the capacitor will still be almost fully charged, at 39,999V.
If you draw 1A for 1 sec from the .00005F capacitor, its voltage will fall by 20kV. So that capacitor will as a result become half discharged.
 
Thanks for the quick responses. That helped considerably.
 
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