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Capasitor-inductors dc

  1. Apr 26, 2014 #1
    Ok i started working with capasitors and inductors and i have a question.
    The capasitor is an open circuit to dc,right?

    Now the questions:
    a)If the current of capasitor is 0 how can it store energy?I know that we use this:1/2 C V^2 but it seems weird that with just stable voltage we store energy.

    also
    b)if in a branch we have a resistance and then a capasitor(and lets say the current goes from the resistance to the capasitor),the current that pass through the resistance is 0 as well because of the capasitor?
     
  2. jcsd
  3. Apr 26, 2014 #2

    gneill

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    Staff: Mentor

    Energy is stored in electric or magnetic fields. In a capacitor it's the electric field between the plates that stores the energy. In an inductor it's the magnetic field that surrounds the conductor that stores the energy.

    In a circuit with a DC supply, the steady state condition (a long time after power is applied) is achieved when the capacitor has reached its final charge and potential difference. Current will have ceased by then. When power is first applied it will take some amount of time for the capacitor to charge through the resistor. During that period current will flow through the resistor. This period is also called the "transient period" (when things are changing) in contrast with the "steady state" when all changes have died away.
     
  4. Apr 26, 2014 #3
    Any pair of conductors separated in space forms a capacitor, and to put some equal and opposite charge on each conductor takes work. That work is converted into electric potential energy in the capacitor.

    For the second question, yes, a resistor connected in series with a capacitor wool admit no current under a static voltage load.

    The AC/time-varying case is more complex.
     
  5. Apr 26, 2014 #4
    So the power(if not called this way i mean work/time) of a capasitor in this steady state will be 0?

    Also how the polarity of a capasitor is determined?
     
  6. Apr 26, 2014 #5

    gneill

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    Staff: Mentor

    At steady state nothing is changing so the work being done must be zero, hence the power is zero also.

    The ideal capacitor has no inherent polarity. It will happily accept potential differences of either polarity across its terminals.

    Certain real-life capacitors rely on chemical films as their dielectrics (so called electrolytic capacitors for example) which can be sensitive to the applied polarity due to chemical reactions being driven by the potential. You only want to charge them with the specified polarity or risk the dielectric degrading and failing (with possible explosive force!).
     
  7. Apr 26, 2014 #6
    Because the voltage across the inductor is 0,i suppose that the power is zero as well?

    To get the amount of energy stored in a capacitor-inductor,we get E=1/2CV^2-E=1/2LI^2 respectively.So i suppose that ohm's law cannot be applied here cause if it were,the energy stored would be 0.What is the reason for this?
     
  8. Apr 26, 2014 #7

    gneill

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    Staff: Mentor

    Sure.

    Ohm's law applies only to resistance, not capacitance or inductance (until, that is, you get into the realm of AC circuits where reactive components have "impedance", but that's a tale for another day).

    Why would you want to take the difference between the energies? They are both potential energy.

    The energy stored in a reactive component isn't a signed quantity per-se. It's how that potential energy is made use of when its released back into the circuit that determines how you interpret its sign in your calculations. Usually one considers the total potential energy to be the sum of the stored energies, which may be a constant over time if there is no resistance in the circuit to dissipate energy.
     
  9. Apr 26, 2014 #8
    Ok everything is clear to me now.Expect me if i face a problem about capacitors-inductor for AC circuits.
     
  10. Apr 29, 2014 #9
    Hey.i have a question.

    Lets say i am having a branch with a resistance and a capacitor.As we know the capacitor is an open circuit in "steady" state,right?

    The question is what will happen to the resistance.I mean because of the capacitor,i erase the branch and everything it has with him?
     
  11. Apr 29, 2014 #10

    gneill

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    Staff: Mentor

    Supposing that your branch consists of a resistor and capacitor in series, then at steady state there will be no current flowing through that resistor and no changes of voltage across the capacitor, so yes, then you can ignore the whole branch.
     
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