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    gumaynor's Avatar
    gumaynor Posts: 1, Reputation: 1
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    #1

    Sep 4, 2013, 05:59 PM
    Capacitor bank mishaps
    1.If I have a capacitor with 500 pico ferrite how much volts or watts can I fit in it
    2.if I had enough capacitors and they all had 500 pico ferrite how long would it take to charge it to any where between 100 and 500 volts or watts using 1 or 3, 9 volt batteries
    3.how long would that take if I put the power source through a 1 volt to 10 volt transformer if that makes a difference
    smoothy's Avatar
    smoothy Posts: 25,495, Reputation: 2853
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    #2

    Sep 4, 2013, 07:31 PM
    I guess this is homework... because there would be no other reason for such questions.


    Since we don't do homework as clearly stated in the rules... what do you think the answer is and how did you arrive at it?
    ebaines's Avatar
    ebaines Posts: 12,130, Reputation: 1307
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    #3

    Sep 6, 2013, 06:41 AM
    I'm guessing this is not homework, since the questions as written show a lack of understanding of how capacitors work.

    First of all, the unit of measurement for a capacitor is "farad," not "ferrite."

    A capacitor stores a charge, and the amount of voltage of that charge depends on the voltage source connected to it. If you connect a 1.5 volt battery it will charge to 1.5 volts. Thus your questions about charging to 100 volts using a 1 volt battery makes no sense.

    The total energy that a capacitior can store is:

    . Thus for a 500pF capacitor being charged with a 1.5 volt battery the total energy stored is:



    which is a very small number. The peak current flow out of the capacitor depends on the resistance of the circuit it discharges through - if using 100 ohm resistor the initial current flow is V_o/R = 15 mA. A capacitior discharges to near zero in about 5RC seconds, or in this case 5(100)(500E-9) = 250 microseconds.

    As for using a bank of capacitors - when you put capacitiors in parallel in a circuit the capacitance values add, so ten 500pF caps in parallel would give . The peak voltage stays the same but the total energy stored goes up by a factor of ten, and the discharge time is lengthened by a factor of ten as well.

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