Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Capacitor makes the circuit open?

  1. Mar 30, 2012 #1
    We know charge is accumulated on the conductor plates of capacitor. Here is a circuit (image) with voltage source, resistor and capacitor. Now due to the capacitor the circuit is actually open so flow of charge aka current is zero. Then how can charge be accumulated on the plates of capacitor?
     

    Attached Files:

    • ckt.JPG
      ckt.JPG
      File size:
      2.5 KB
      Views:
      1,027
  2. jcsd
  3. Mar 30, 2012 #2
    It is only open at steady state, meaning after the circuit is in this configuration for a long time so that only the DC component of your voltage source is present.

    Before the circuit is in the state of your schematic, there is no charge accumulated on the plates and so there is no voltage across the capacitor, this is known as an initial condition. Once the voltage is applied, charge flows through the resistor and begins accumulating on the plate. Initially, since there is no charge on the plate and so no voltage, all of the voltage is dropped across the resistor. As the plates begin filling up with charge, their voltage increases and so less voltage is across the resistor and so less current is flowing. Once enough charge has been accumulated on the plates so that its voltage is equal to the voltage of the source, the voltage across the resistor will be 0, and so no more current will flow. This is when it is considered an open, and in stead state -- the charge is already accumulated.

    So, you should know that the capacitor is only an open to DC voltage/current, and not to AC.
     
    Last edited: Mar 30, 2012
  4. Mar 30, 2012 #3
    Thanks for your reply.
    Though voltage is applied the circuit is in open condition so the current flowing through resistor should be zero isnt it?
     
  5. Mar 30, 2012 #4
    The circuit is only in the open condition once enough charge has accumulated on a capacitor so that its voltage is equal to the DC voltage applied.

    Remember the voltage on a capacitor is V = Q/C, so as more charge is added, its voltage increases.

    To your question specifically, if the capacitor is already in the charged condition, then the current flow through the resistor will be 0.
     
  6. Mar 30, 2012 #5

    sophiecentaur

    User Avatar
    Science Advisor
    Gold Member

    Eventually, yes, with DC applied.
     
  7. Mar 30, 2012 #6
    But before that? Isnt it open in all conditions due to the dielectric between the plates?
     
  8. Mar 30, 2012 #7
    No. If there was initially no charge on the capacitor, and then you put it in a circuit with current, the current coming into it would get "captured" by the capacitor, and the charge of the current will start to accumulate on the capacitor. This actually means that the capacitor is acting more like a short circuit rather than an open circuit in the very beginning. Once the capacitor has captured enough charge, its voltage increases til it cannot capture any more charge, and this happens over a long time. When it is finally filled with charge that it can't take anymore, it acts like an open circuit.

    The dielectric simply determines how much charge the capacitor can hold at a given voltage. It determines what the capacitance value of the capacitor is, and so you can see from what I said before V = Q/C, if you rearrange it to how much charge it can hold per a given voltage C = Q/V. Its basically telling you how much energy the capacitor can store inside its dielectric.
     
  9. Mar 30, 2012 #8

    sophiecentaur

    User Avatar
    Science Advisor
    Gold Member

    I think you are using the word 'open' wrongly which is leading you to a false conclusion. Any 'open' circuit - even the gap between switch contacts or bare ends of two wires - has a finite Capacitance, which can allow signal leakage so the 'open-ness' is all relative to the frequency involved and the particular situation involved.
    An Open circuit is a practical description of a situation in which any current which may be flowing is negligible in the context of the other, more significant, currents when the open circuit has been closed.

    It is very easy to try to categorise things too rigidly and to become 'obsessed' with getting the names right. This is the beauty of using Maths to describe things in Science because it involves using actual Numbers and Quantities, rather than Absolute terms to 'classify' things.
     
  10. Mar 30, 2012 #9
    This might be a useful analogy. Imagine you are pumping a really small tire with a hand pump that is connected with a hose.

    Charge is air, and the flow of air is current.

    The diameter and length of the hose is the resistor, it determines how much air you can pass through it at a given pressure on the air.

    The tire is the capacitor. When it is flat, it has no air in it, and when it is full, it has the most air you can fill it with. The elasticity of the tire rubber is like the dielectric, how much it can stretch determines how much air it can hold at a given pressure. If you fill it til the tire can't hold anymore and pops, its like breaking through the dielectric.

    The pump applies pressure that is like the voltage. If you start pushing down the pump handle, it will create a pressure that pushes air through the hose and into the tire.

    As you begin pushing down the handle, the tire is empty (capacitor is not charged) and so it takes all of the air into it with no resistance. In this sense, it is acting like a short circuit, because there is no pressure in it so it looks like the hose isn't even blocked by anything and freely flows air to the outside pressure. The only resistance you see is from the hose, since it can only pass a certain amount of air at a given pressure.

    As the handle moves down, the air keeps filling up the tire, and so the pressure in the tire builds up. This means it is becoming harder to fill the tire with more air, and so the air flow slows down, and more pressure is in the tire and less in the hose.

    Once the tire fills up completely, you won't be able to push down the handle anymore because you aren't strong enough. You don't have enough pressure to push down the handle, you can sit on the handle and it won't move. The air still won't move since the pressure in the tire is just as strong. In this way, it looks like an open circuit, because no more air will flow.

    It is the same way in a capacitor. Your voltage source begins pushing current through, and the capacitor is empty, so it looks like a short circuit. Once you have filled the capacitor with charge, its voltage is equal to the voltage you use to push the charge, and so you can no longer push anymore through, and it looks like an open.
     
  11. Mar 30, 2012 #10

    sophiecentaur

    User Avatar
    Science Advisor
    Gold Member

    Umm.
    That is a very long winded analogy and I wonder just how useful it is for explaining things simply in terms of the actual electrical Capacitance across a gap in a wire vs the capacity of a high value Capacitor.
    Sometimes less is more. But then, you really needed to included the case then the hose is blocked off before the tyre, in order to show the quantitative difference between a Capacitor and an open circuit.
     
  12. Mar 30, 2012 #11
    The point is that a capacitor does act exactly as an open circuit or a short circuit in specific conditions, and not in all conditions (t = infinity/~5 time constants and t= 0).

    And I think the analogy serves to give the concept of "capacitance" pretty well since the OP seems to be confused with the relationship of charge accumulation and capacitance, and how this leads to the short/open conditions. It shows that you only reach an open after charge has been pumped, not in all conditions, and it also shows that in fact it is acting like a short when you initially begin pumping.
     
  13. Mar 30, 2012 #12

    sophiecentaur

    User Avatar
    Science Advisor
    Gold Member

    You actually made my point for me because that first sentence of yours expresses the whole idea perfectly in just over one line of writing.

    PS I am very cranky about analogies because they so often let you down in the end! I guess I should get over it!
     
  14. Mar 30, 2012 #13
    Yes, but explaining it directly did not seem to help a lot and I'm still not sure that it would until a better idea of what is going on is there. That's why I decided to make an analogy.
     
  15. Mar 30, 2012 #14
    I know analogies are bad, but they help make pictures/animations in your head that give you the intuitive feelings of things.

    Anyway, if an analogy is mathematically equivalent to what it is being an analogy of, is it still an analogy or at least a bad analogy? Differential equations for mechanical and electrical systems can have identical relationships.
     
  16. Mar 30, 2012 #15

    sophiecentaur

    User Avatar
    Science Advisor
    Gold Member

    I take your point but every analogy needs it own Caveat if you want to prevent problems. I so often read responses that have grabbed the wrong bit of the analogy and ignored the important message. Having said that, I just lurve the mass on spring analogy for almost every oscillator description.
     
  17. Mar 30, 2012 #16

    sophiecentaur

    User Avatar
    Science Advisor
    Gold Member

    AH yes - once you introduce the maths model and the correspondence can be spelled out, you get my vote.
     
  18. Mar 30, 2012 #17

    Averagesupernova

    User Avatar
    Science Advisor
    Gold Member

    I think that it is generally understood that ALL analogies only go so far. An analogy is a good place to start from. The fault lies within the person who expects it to solve all of the problems, not within the analogy. Analogies have a tendency to put a persons mind in a certain mode that better prepares them the real thing rather than jumping right into the real thing. I have seen the question asked about how computers know how to do stuff. My reply to that is: How do mousetraps know how to catch mice? It is such a brief reply that it can hardly be considered an analogy but it activates a certain mode of thinking which is important.
     
  19. Mar 30, 2012 #18

    psparky

    User Avatar
    Gold Member

    I liked the analogy with the tire and air pump....good stuff.

    Don't mind Sophie....he always gets grumpy with those...haha!

    Let's face it......an interesting analogy AND a mathematical model work best for learning.

    If you have only one of the two....you are typically missing something!

    My firehose and water analogy leading into P=IV was a perfect example of this!!!!!!
     
  20. Mar 30, 2012 #19

    psparky

    User Avatar
    Gold Member

    Because I'm pickin on Sophie today.....let me sum up you capacitor question:

    i(t)=C*dv/d(t)

    A perfect mathematical model.

    Any questions?

    Alright....I'll throw in a small analogy as well.....what the heck.

    That statements says....the change in voltage in respect to time multiplied by the capacitance.....equals the current.

    So when you first put your voltage source across your uncharged zero volt capacitor..........the change in voltage is large compared to time....so you get a fairly large current.....as it charges........the voltage across the cap starts to climb and the change is much less....so the current gets to be less. Then when the cap reaches its full charge......there is no change in voltage.....there fore there is no current!

    Any questions?
     
    Last edited: Mar 30, 2012
  21. Mar 30, 2012 #20

    psparky

    User Avatar
    Gold Member

    Might as well explain inductors as well....the work exactly the opposite. They resist current at start up and short in steady state DC.

    v(t)=L*di/dt

    Before they get energized...they have zero current thru them. When you turn the switch on....the change in current is fairly large.......then a fairly large voltage is across them....to have a voltage....I must have a resistance. As time passes...the change in current is much less.....so the voltage across is much less.....at steady state there is no change in current......therefore no voltage....and the inductor acts like a short.
     
    Last edited: Mar 30, 2012
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook