# capacitor

Discussion in 'Homework Help' started by chaitu123, Dec 21, 2009.

1. ### chaitu123 Thread Starter New Member

Aug 31, 2009
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why the capacitor acts as a open and short circuits at initial conditions?

Apr 20, 2004
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3. ### m.farahanchy New Member

Jun 5, 2009
3
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if you dont have much time ,can read this

in ac current, whene the frequency is high , charges and dischargs happens rapidly. in the steady state it seeems as if the capacitor is short cutted
whene the frequency is zero, capacitor is charged once, and no charge can cross the C .

4. ### ping New Member

Sep 16, 2009
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Lets say you have a capacitor a nine volt battery. Lets also assume that this capacitor is completely discharged. If you put a battery across the capacitor it will charge up to 9 volts rather quickly in order to equalise to the battery voltage. Once it has reached 9 volts the cap is fully charged and will hold its charge when you remove it from the battery.

The question that gets many people is how can a current flow across an insulator or for that matter how can an insulator appear as a short circuit.

Yes the capacitor is two plates separated by an insulator, but the fact that the plates are so close to one another means that when you apply an external voltage and an electric field is set up in the insulator (dielectric) which will be conducive to an accumulation of opposing charges on the plates.

So charges therefore migrate into the capacitor. (electrons leave the positive plate and go into the battery, and electrons enter the negative plate) what you have is actually displacement current that allows for the charges to accumulate. When the cap is completely discharged the displacement current is at its maximum.

Since there are no charges in the capacitor initially there is no voltage across it when you begin. This results in the displacement current flowing at the maximum possible rate. This rate is limited only by the resistance between the battery and the capacitor. Since you hooked the battery up directly the only resistance is that of the leads of the battery which is neglible. And as such the displacement current is initially that of a short circuit.

As the capacitor charges up voltages build up across the plate and the accumulation of electrons on the negative plate repels further electrons and net postive charge on the postive plate makes it harder for electrons to leave so the displacement current decreases as the voltage builds up. Until the capacitor is fully charged at which point the displacement current is zero and the capacitor behaves like an open circuit.

So initially its appears to be a short circuit because the displacement current is flowing at maximum and it has zero volts across it, but once its charged up it appears as an open circuit and has the full volt drop accross it with no current flowing through it like an open switch does. Hence its open circuit.

When you put your mutimeter on the resistance setting and measure a discharged cap the same thing happens. The multimeter forces current into the load and then measures this current to determine the resistance. When you measure an empty capacitor the displacement current flows and it initially is the maximum possible current which multimeter thinks is a short circuit, then as the cap charges and charges build up the current gets less and less until its zero which the multimeter interprets as open circuit. So the multimeter sees displays a gradual increase in resistance until an open circuit is reached.

A bigger cap will have to pull in more charge and so the current will flow for longer gradually decreasing which you will see as a slowing increasing resistance measured. A smaller cap will charge up much faster and might appear to be going from short to open near instantaneously. Using an a multimeter like this is a nice quick way to test caps, but not an accurate or proper test for that matter.

Capacitors are one of those things that once you know how they work you shouldn't have to worry about the physics behind them, which is why most electronics books don't go into too much detail, because that is more in the real of physics.

Electronics is about using the devices to control power or information in a meaningful way, which can be more fun than theoritical exercises like this. Just remember a cap is initially short and then becomes an open circuit. This can have consequences when desiging circuits for example when you power up a circuit with a lot of caps in it you will have an initial surge as the caps power up. Inductors are the complete opposite of caps so these can be used to counter the effects but that is covered else where, or simple resistors can be used or things like MOVs etc etc. Have fun.

Last edited: Dec 22, 2009

Apr 10, 2009
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Actually, in a capacitor ...in initial condition it behaves as short circuit whereas in final conditon it behaves as open circuit.

It is so because .. at initial condition when the capacitor carries no charge(and hence no voltage) the source voltage i.e., EMF is applied to a circuit then charge transfer takes place from one plate to other due to which the capacitor behaves as short circuit.
And when the total transfer of charge takes place. it hence becomes stable and acquires a voltage drop across it and so acts as open circuit.....in the final state of transient.

6. ### ping New Member

Sep 16, 2009
13
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As far as I know there is no flow of current across the from one plate to the other plate directly. Because that would mean charges are flowing through the dielectric. Instead there is a "migration" of between each plate and the source of EMF terminal it is connected to. ie negative charges migrate into the postive terminal of the EMF from the one plate, and negative charges simulatenously migrate out of the negative terminal of the EMF and onto the negative plate, thereby setting up an electric field stored across the dielectric.

This migration or displacement of electrons in order to allow charge to accumulate in the cap is a displacement displacement current. So charge transfer takes place not from plate to plate but between each plate the source EMF terminal its connected to.

What I am saying is that the apparent short circuit is not reallly a short circuit but is actually an "apparent short circuit" because the displacement current is experienceing very little resistance initially. If it really was a true short circuit then it would remain a short circuit continously.

Yes leakage currents can flow through the dielectric, but thats not the charging mechanism. But I suppose I am splitting hairs because for most common uses of capacitors in electronics you don't have to go into such detail, even though it can be interesting to know.

7. ### Ratch New Member

Mar 20, 2007
1,068
3
Ping,

Assuming an ideal capactior. You are basically correct. Some of your terminology could be corrected or improved. You are dead right about no charge passing through an ideal dielectric. Instead of saying "current flows", or "flow of current", you should say "charge flows", or "flow of charge". A capacitor does not "charge", at least not with charge carriers. The accumulation of charge carriers on one plate is balanced by a depletion on the other plate, for a net charge carrier accumulation of zero, no matter what the voltage is. The capacitor does become "charged" with energy, however. That is why I prefer to say a capacitor is "energized". The capacitor does not cause a "resistance" to current. It does cause a back voltage which impedes the current with out dissipating any energy like a resistor would do.
And of course, it acts like a energy bank, so is able to store/supply current and voltage for a finite time by increasing/decreasing its electrostatic storage field.

Ratch