Charging current in open AC circuit

Question

I watched the lecture and at this point there is mention about AC generator and wall outlet, that there is only voltage and, obviously, no current until we plug in some load.

Since that, one thing start to disturb me so I have a question.

If we connect a wires to a DC voltage source terminals and leave them open ended there will be tiny currents during transient state as the wires charging to the terminals potentials. And no currents after that as should expect from open circuit. Then if some load are plug in there will be current in the circuit which depends on the load resistance.

What if there is same circuit but AC source instead of DC. Connected wires keeps constantly recharging since source voltage now permanently changing as well.

Does that means there actually always will be small currents in the wires solely because of these charging processes despite the fact that the circuit is open ended ?

At same time it is considered that there is no currents in a wires of open AC circuit (as example the case of wall outlet in the video above). This confuse me a bit.

Only way I can think of it that the small "charging" currents are neglected.

And when some load is plugged, a currents in the circuit will begin to increase in order to be adjusted to the load impedance. And from that moment it customary to talk about "traditional" current which appears in a circuit, dependance I=V/Z, power consumption etc. Besides that, I assume some portion of that "traditional" current would continue to "used" for the wires charging. Is this correct ?

I'm not read about transmission lines in details yet so probably better to assume that frequencies and wires length in examples above are small enough to avoid phenomena specific for TL.

Sorry for my english and thanks for the help.

Answer 1

I think you have a sound understanding of what is going on.

Yes, the current associated with charging and discharging of the wires is always present, but it is not significant and could be neglected for most practical purposes. You can view the wires as two plates of a capacitor, estimate its capacitance and the current.

I would add that this current is distributed through the length of the wires, like in a transmission line, i.e., it does not flow through one wire all the way, say, from a transformer to an outlet, jumping through a tiny capacitance of the outlet and flowing all the way back to the transformer through the second wire.

Answer 2

The wires carrying the current, whether DC or AC, is a transmission line that can be modeled using a lumped-element approach to include the resistance, inductance, capacitance (between conductors), and conductance(between conductors). Although most important for AC analysis, the lumped model can also be used for DC since there is resistance and capacitance to consider but if studying the transient changes, DC would include inductance as well.

Typically the lumped-element model will designate series resistance and inductance for each conductor as well as between conductors capacitance and conductance (aka inverse resistance) in parallel. These lumped-element models are a per-unit length measure of R (resistance), L (inductance), and C (capacitance), and G (conductance).

There are a lot of resources available to review and study these models. This is one of them.

Therefore, even though there may be nothing connected to the terminals of an AC outlet or a battery, there are lots of current paths available in the conductors for charging capacitance, inductance (building a magnetic field, tearing it down again), and so forth.