Capacitors in DC Circuits, and How it Affects Reactance & Current

[The Head]

I read that capacitors that are part of DC circuits have a reactance that is essentially infinite. When I look at equations, this seems to make sense: If the reactance X=1/ωC, ω is zero in a dc circuit and so this means that X approaches a very large value. Thus, if you use V=IX, this means very little current will run, right? But current certainly runs as you charge a capacitor-- but I thought that you could charge a capacitor with direct current.

Also, and this may be related, would this mean displacement current can only run through capacitors with an AC source? If so, why is this? I consulted a textbook and found an equation relating the change in electric flux to displacement current, but why does there need to be a change? Is it induced? Because I thought displacement current "ran" through a capacitor, creating a magnetic field, but I wasn't aware that there had to be a changing EMF to create it.

Sorry if this is all over the place, I just can't pinpoint the exact concept that is confusing me. Any help on this topic will likely help clarify my perspective, so please, show me where I have misconceptions!

 

[sophiecentaur]

We're into the realms of 'definition' here. When you 'switch on' a DC supply, you don't have pure DC. There is, of course, a changing component as the C's are charging up.

'Displacement current ' will only be flowing whilst the Cs are charging and will stop once they are fully charged. Only during that time, there will be a resulting Magnetic Field. But you will get a magnetic field from the currents flowing through any DC paths (if there are any) between the supply connections.

 

[The Head]

Thanks for your response, sophiecentaur!

So why does current flow in the DC circuit if the reactance, 1/ωC is nearly infinite (as the frequency is zero)? Because won't a huge reactance value make current nearly zero?

 

[NascentOxygen]

I read that capacitors that are part of DC circuits have a reactance that is essentially infinite.

That's true, because the definition of "DC" that is being used in that statement is of voltages & currents that are constant and unchanging, i.e., where they have had plenty of time to stabilize at steady levels.

When I look at equations, this seems to make sense: If the reactance X=1/ωC, ω is zero in a dc circuit and so this means that X approaches a very large value.

That formula applies only for sinewaves, and only after the circuit has had sufficient time to stabilize to that sinewave. For DC or waveforms of any other shape, you must resort to integral calculus. The equation you need is: voltage across a capacitor is related to current by v(t)= (1/C)·∫i(t)·dt[/color]

Thus, if you use V=IX, this means very little current will run, right? But current certainly runs as you charge a capacitor

Current does flow into a capacitor when you charge it, and current flows out of it when you discharge the capacitor. These currents are not of a constant level, they change with time, and they have not been afforded a long enough time to settle at a steady-state level—so they fail the definition of DC as it is used in that earlier statement.

-- but I thought that you could charge a capacitor with direct current.

You can. That is using a different meaning. In this case their reference to DC is where the voltage is constant, as from a battery. That's a looser definition that the one I pointed out earlier.

Good luck with your study of electronics!

 

[sophiecentaur]

If we are dealing with ideal components (which is pretty much the universal and successful approach to this stuff) we use Maths.
You start with the charging of a capacitor through a resistor from a constant voltage source, with the charging current being equal to the volts across the resistor/R. This then leads to the voltage across the Capacitor reaching the supply volts exponentially. Applying this behaviour to an applied sine wave AC, you arrive at the notion of Reactance, (the X that you quote), which tells you the voltage to expect across the capacitor and also, of course, the current flowing through the Resistor.
My point is that, once you have accepted that Maths gives you a good model then the 'arm waving' descriptions ceases to be any help - there are too many things to include in a verbal sentence to allow the true pattern of what goes on to be appreciated. Your OP quotes some Maths so you are clearly happy with Maths. So why not just go along with it and accept that what it tells you is the best description? Words are not necessarily a 'better' way to describe the situation - after all, would you discuss the amount of your grocery bill in any other way than using arithmetic and a calculator? (Apart from moaning that it is, qualitatively 'too much' again.)