Confusion on Farad calculation

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    Calculation Confusion
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Discussion Overview

The discussion centers around the application and understanding of the discharge capacity of capacitors, specifically the relationship between capacitance (in Farads), voltage, and current. Participants explore the implications of the formula Farads = Coulombs/Volt and how it relates to different capacitor sizes and their behavior under current flow.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • Jerry expresses confusion about the application of the formula relating Farads, Coulombs, and Volts, particularly in the context of different capacitor sizes and their ability to generate voltage under current.
  • Some participants clarify that if 1A of current is applied for 1 second to a 1F capacitor, it will reach 1V, while the same current applied to a 0.00005F capacitor would result in a voltage of 20,000V.
  • Another participant emphasizes that the equation Q = C·V does not include current or time, suggesting that this may be a source of confusion for Jerry.
  • A later reply introduces a scenario with pre-charged capacitors, illustrating how drawing current affects their voltage differently based on their capacitance, but does not resolve the initial confusion about the formula.

Areas of Agreement / Disagreement

Participants generally agree on the calculations related to the discharge of capacitors, but there is some disagreement regarding the interpretation of the formula and its implications, indicating that the discussion remains unresolved in terms of clarity for Jerry.

Contextual Notes

There are limitations in the discussion regarding the assumptions made about the initial charge of capacitors and the role of time and current in the discharge process, which are not fully explored.

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.
 
Last edited by a moderator:
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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