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Capacitor working...

  1. May 16, 2015 #1
    when we apply DC to capacitor then its capacitive reactance becomes zero and it acts like a short circuit....if it acts like short circuit then a huge amount of current will flow through the capacitor..isn't it??? if so happens then what will happen to the dielectric present in between the two plates??and how can the capacitor store charge??
     
  2. jcsd
  3. May 16, 2015 #2
    It acts as an open, not a short.

    X = wL + 1/(wC); as w-->0, X--> ∞.
     
  4. May 16, 2015 #3
    taking transients in to account...when we suddenly close the switch connected to a capacitor,it behaves like open circuit or short circuit??
     
  5. May 16, 2015 #4

    NascentOxygen

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    Rapid changes in applied voltage are akin to a high frequency signal, and at high frequency the capacitor is like a low impedance or short-circuit.
     
  6. May 16, 2015 #5

    phinds

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    I think Jeff must have misunderstood the question. For an ideal capacitor fed by an ideal power source, the transient current at the time the ideal switch closes is infinite because it is, as you state, a short circuit. In the real world the current is just large, with the actual value depending on the resistance of the power supply and its ability to provide current.

    The dielectric doesn't care how fast your charge the cap, just how MUCH you charge it. To get dielectric breakdown, you have to put on too much voltage.
     
  7. May 16, 2015 #6
    Lets not over think or analyze this.. first capacitive reactance is a time based value... at T= 0 the capacitor has zero impedance ( zero "resistance") , at t = infinity it has infinite impedance. Next from a circuit based perspective current flows "though" the capacitor - but I have always found it better to think charge flows in (to one side) and opposite and equivalent charge flows out. Basically you are adding ( or reducing ) the charge in the cap.

    So any time we change the voltage applied to a Cap we are changing the strength of the electric field in the capacitor. It stores energy - an ideal cap does not dissipate any energy - for most cases the "ideal" model is all we need to consider.
     
  8. May 16, 2015 #7

    sophiecentaur

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    But that isn't the "DC" you consider in your first post. Its behaviour is not 'like' an open or short circuit. Its Capacitance, in conjunction with any (unavoidable) resistance in a circuit, gives a time constant that governs the rate of response to a step function.
     
  9. May 17, 2015 #8
    if the current is so large it means capacitor perfectly conducting current...if perfectly conducts then is it so that the dielectric inside the capacitor breaksdown completly..if not how can a large current flows through it...
     
  10. May 17, 2015 #9

    meBigGuy

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    As stated previously, the current flows to build up a field between the plates. When you disconnect the source, the field between the plates remain (the capacitor holds its charge).

    The actual effect is called displacement current, and is not a simple concept to fully understand
    http://en.wikipedia.org/wiki/Displacement_current#Current_in_capacitors

    Suffice it to say that current flow into a capacitor will build up a field between the plates. Once the potential of the field reaches the applied voltage, the current flow will stop.
    You can then draw current from the capacitor, which reduces the field.

    charge = capacitance X voltage

    For a simple overview:
    http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capchg.html
     
  11. May 17, 2015 #10

    phinds

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    Did you not read post #6 ? Did you not understand it?
     
  12. May 17, 2015 #11
    It is common to think electricity flows through wires. It is common but wrong.

    The energy of electricity is in a combination of electric fields and magnetic fields. Capacitors store energy as electric fields. Inductors store energy as magnetic fields. The wires guide the energy, but (mostly) do not contain it.

    So an empty capacitor has an instant voltage change applied (for some value of instant). The electrons from one plate start to flow away from it, building a positive charge on that plate. This charge attracts electrons to the other plate so that plate builds a negative charge.

    Initially each plate is uncharged, so there are lots of uncommitted electrons just sitting there. They are easy to move. But as the field develops between the plates, there are less and less uncommitted electrons. This limits the current. So the current starts out large, but gets smaller as the plates charge. But notice that at no point do any electrons flow through the dielectric. They flow from the plate to the wire on one side and from the wire to the plate on the other.

    Of course there is no such thing as an instant voltage change, so the current will never be infinite, though it can be quite large. But once the plates get charged it stops.
     
  13. May 17, 2015 #12

    sophiecentaur

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    I have a real problem with these posts about "ideal" components. You can get any answer you want if you then introduce just one non-deal element. Fact is that there will be power dissipated within the dielectric. There is always some hysteresis in the Q/V curve and that represents energy dissipation. Capacitors do 'get hot' with AC 'going through them'.
     
  14. May 17, 2015 #13
    I agree. There are no ideal circuits.

    Yet understanding the ideal circuits allows us to model real circuits with closer and closer approximations. We might start with an ideal capacitor. We might add some inductance and resistance for the leads. We might add some ESR for dielectric losses. We might model edge capacitance as it interacts with other components. On and on, there are lower and lower order effects.

    But to do any of that we need to understand the ideal elements we're modeling with. And we need to model what we need to model. The resistance of a capacitor matters more in a power factor correcter than in a ceramic filter capacitor.
     
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