Uncharged capacitor is a short circuit?

Only until the first juice hits it, and only when talking about DC. (AC basically sees a cap as a dead short - lots of "special rules" about that for various frequencies of AC versus values of capacitor, but that's the basic idea, anyway) At first, there is practically no resistance. Like a short. Charge flows freely into it - For a moment. Once it takes on its first "droplet" of charge, it becomes more like a gradually increasing-value resistor until it's fully charged, where it behaves more or less like an open to DC. Trade a load for the charge source at this point, and it behaves like a current source until the plates equalize (full discharge) then it's back to being like a short until the next dose of charge hits it.

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The relationship between capacitor voltage and capacitor current is I=C*(dv/dt), where I is in amperes, C is in farads and dv/dt is in volts per second.

When you first connect a capacitor to a stiff voltage source (one that puts out as much current is necessary to maintain the voltage) the source tries to alter the voltage of the capacitor very quickly. This is a big dv/dt. Si it takes a big current to do that. Short circuits also draw big currents. The difference is that as the cap voltage approaches the source voltage, the dv/dt goes down and approaches zero, and so does the current. A short circuit just keeps sucking current.

because even though there is an open circuit between the plates of a cap there is a force across them. This force is felt on the other side when an electron moves onto one place. In essence it pushes off an electron on the other plate when it arrives... the same kinda stuff happens in a wire where one electron feels the repulsive force of another... in this case though there is a definite force but there is a "gap" that prevents the electron from continuing.

Thing of blowing up a water rocket with water. It requires a certain force to push the air in and initally is pretty easy to do... the more air(electrons) you put in the harder it gets until eventually you cannot push any more in(or it explodes) and the forces are counter balanced. The same thing happens in a cap except in this case we can think that we have a reservoir of air connected to the pump(so there is a limited amount of air) and when we pull the air from the reservoir and push it into the rocket it is likened to a current flowing(the air being analogous to the electrons)... if you put the reservoir and the water rocket close to each other you can envision that there is a "hypothetical" current flowing between them where it seems that the air is flowing continously through(ofcourse we know it isn't but it sorta looks like that from a distance).

i.e.

(need to view this with a fixed width font)

Pump | | |-----| | |-------| | Air ==|== Water | |-----| |-------|

|------||-----| === Water || Air ==== // |------||-----| \\\\ || || || || \\\\ // ==========|========== | | pump

So as we pump it first is very easy but gets harder and harder. Also we feel as if the air leaves the Air chamber and enters the water and we might even suspect that some air leaves the water chamber and goes across to the Air chamber(maybe through a little valve betwen them). This is not what happens but if we didn't not know they were not connected and we started to pump we would see a air current flowing from one around the "circuit".... but we would notice it would get harder and harder to pump. We could hold the air that we pumped into the water chamber but as soon as we released the pump all the air would rush back into the air chamber(until it equalized).

A capacitor is very similar except that it uses conductive plates(reservoirs) that hold electrons. When the electrons arrive on one side of the plate they "push" off electrons on the other side and it "simulates" a current(even though there isn't any "physical" current. The more electrons that arrive makes it harder and harder to push more on(because the electrons start to repel each other(ones on the same plate) too).

What you have to understand is that just because there is a "gap"(an open circuit) in a capacitor that it doesn't stop electrons from flowing... Electrons can easily flow through a gap(say a vaccumn) if they have enough energy. Since capacitors are made of conducting material electrons move freely on on or off its plates... if, say, one of the plates were an insulator then it would not conduct electricity.

You can also think that a battery pulls an electron off one side one plate(making it +) and puts it on the other plate(making it -).... this causes a potential difference across the cap and the more electrons that the batter does this two the stronger the ptotential difference is... if you have N electrons that we could hypothesize that the potential would be proportial to N(which it theoretically is)... but the battery can only push so "hard" and eventually it's voltage would be "countered" by the voltage across the cap.

Hope that helps some, AD

It can ? How so ?

Actually it's more like a capacitor. Capacitors are 'reactive' not resistive btw.

Graham

NO, a typical capacitor has allmost NO resistance, this resistance is CONSTANT irrespective of the state of charge. When a current flows through a capacitor it developes in internal electric field (( current x time ) / capacitance) which is a voltage which opposes the external voltage. So eventually the internal voltage will balance out the external voltage and so the current will be zero BUT notice the circuit resistance is still the same.

Ron J wrote: : Hey all,

: I'm curious why an uncharged capacitor can be viewed as a short circuit : by, say, an amplifier? I thought an uncharged cap would be more like an : open circuit.

: Thanks!

Ron,

An alternate way to look at this is that the capacitor has an impedance (similar to resistance) equal to 1/(2*j*pi*f*C) where j = sqrt(-1), f = The frequency of the source, and C = the Capacitance. The magnitude of the impedance is called the reactance and is simply equal to

1/(2*pi*f*C). Even if you don't understand everything that I have just said, what you should take away from it is that the impedance of the capacitor is inversely proportional to the frequency -- i.e. the higher the frequency, the lower the impedance (think resistance.) Note that if you plug f = 0 (i.e. DC) into that, the capacitor has infinite impedance and acts like an open circuit.

Under some circumstances, the capacitor can look like an open circuit -- namely when the frequency is very high. Steps and impulses contain high frequency components, and thus, at the onset of the application of a step or an impulse to a circuit containing a capacitor, one might think of the capacitor as an open circuit (to calculate max. current draw or something like that.)

If you are talking about a capacitive load on an amplifier that has a continuous time input signal, then the capacitor neither behaves as a short or as an open circuit, but something in between. It has an impedance given by the equation above, and, for a single frequency, could be modeled as a resistive load with a value of the capacitive reactance. Certain amplifier-containing circuits cause their input/output to transition in steps. Switched-capacitor circuits are an example. Therefore, once again, it is sometimes useful to consider the capacitor as a short circuit, because at the beginning of one of these steps, the current required to charge/discharge the capacitor is a maximum, and this is almost certainly important for the design of the amplifier.

That's probably more information that you really wanted, but hopefully it's helpful.

Joe

P.S. Those who said that capacitors don't have resistance, etc, are, of course correct, but if your are only interested in the magnitude of the impedance, then you can simply treat a capacitor as a resistor with a resistance given by the formula I posted waaaaaaay above. If you care about phase, then things are not as simple.