capacitor output

Discussion in 'General Electronics Chat' started by studystudystudy, Jul 20, 2010.

  1. studystudystudy

    Thread Starter Member

    May 21, 2010
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    i have 1F capacitor, what would the output voltage,

    same question, i have 1F capacitor, what is maximum energy can store, the store energy is in term of voltage or current?
     
  2. BMorse

    Senior Member

    Sep 26, 2009
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    What is the voltage rating of the capacitor?? What type of capacitor is it? What voltage are you using to "charge" the cap?

    B. Morse
     
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  3. beenthere

    Retired Moderator

    Apr 20, 2004
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  4. studystudystudy

    Thread Starter Member

    May 21, 2010
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    my capacitor is 1F and 5.5V. what does this two figure tell me?
     
  5. sage.radachowsky

    Member

    May 11, 2010
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    Energy is in terms of Joules, which is a Watt-Second. Watt is a measure of power. So 1 Watt for 1 second is 1 Joule.

    If you charge your 1F cap up to 5.5V, you will have a certain amount of energy stored in it. There is an easy formula to find out how much energy. But I won't tell you.
     
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  6. BMorse

    Senior Member

    Sep 26, 2009
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  7. studystudystudy

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    May 21, 2010
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  8. sage.radachowsky

    Member

    May 11, 2010
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    When current flows into a capacitor, it increases the charge.

    If you start with an empty capacitor (0V), it has no energy stored.
    If you flow current into it, the voltage increases.
    The rate depends on 2 things:
    1. the rate of current flow (i.e. the number of electrons per second)
    2. the capacity of the capacitor (i.e. how big is the tank, how many F?)

    Think of it like a tank of water.

    Suppose you have a 10 liter tank, and you put in water a 1 liter per second.
    After 10 seconds, it will be filled to capacity.

    The first liter of water is at the bottom. It contains the least stored energy. If you let it back out, it will flow out slowly.

    The last liter of water is higher, so it contains more energy. If you let out 1 liter from the full tank, then it will exit faster. It has more stored energy.

    That is why the formula gives more energy to each additional volt of charge on the capacitor.

    If you're just looking for the easy formula, then look here:
    http://en.wikipedia.org/wiki/Capacitor#Energy_storage
     
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  9. Blackbull

    Well-Known Member

    Jul 26, 2008
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    Speaking as an ignoramus: A capacitor must not be subjected to more than its stated voltage. If a capacitor of 1 farad has a p.d. of 1V across its terminals it will have a charge of 1 coulomb.
    Coulombs = Capacitance X Volts.
    It will charge 2/3 of 1V, then 2/3 of the remainder and so on; in theory it will never fully charge.
     
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