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AC really a wave?

  1. Jul 13, 2003 #1
    I thought I had a pretty good understanding of AC; however I’m starting to question that after I read a chapter in an electronics book about impedance matching.

    I’ve never really considered AC to be a wave is the sense of ocean waves or sound waves. Reasons being that those more conventional waves don’t have particles that actually travel and AC does. Also AC doesn’t really propagate like other waves, its more just current goes this way, current goes that way.

    I was pretty confident with all this until I read that the reason for impedance matching is to avoid power losses in the form of it being reflected back to the source. This to me sounded like the phenomena of a real wave, and not something my understanding of AC could explain.

    So I guess my real question is what best describes AC, and also how could it ever be reflected?

    Please excuse the post. It’s quite late here and I’m very tired. Not really sure how coherent the whole thing is, but I just had to post it before I went to bed.
  2. jcsd
  3. Jul 13, 2003 #2

    Ivan Seeking

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    A wave in this context is a mathematical object. The voltage and current in a typical AC circuit are sinusoidal - meaning a mathematical wave - and if the circuit is purely resistive, they are in phase. If we are working with a wire based system such as a power system, as opposed to antennas which are another thing altogether, then we usually are talking about the current. You can actually see this on the screen of an oscilloscope. Have you ever seen the waveform on a screen like this?

    A reflected wave in this context is really a mathematical model that describes the flow of current as a function of time. If we see a reduction in the flow of current that deforms our mathematical wave, we interpret this as reflected mathematical wave since this accurately describes what we measure in the circuit. In other words, this wave model yields the correct results. To understand exactly how this happens is very complicated. It is probably best to imagine a reflected wave as a disruption in the normal or ideal flow of current in a circuit.

    Edit: Correction to comments about the phase of voltage and current.
    Last edited: Jul 13, 2003
  4. Jul 13, 2003 #3
    To really understand the flow

    of electricity in circuits you have to know about the Poynting Vector and the Electromgnetic Field. You can find a good description in the Feynman Lectures, but many other books and encyclopedias will have a description too. The energy from the power source, be it battery or alternator, actually moves through space and not down the wires as in the usual view of circuits. It's a result of Maxwell's Equations applying.
  5. Jul 13, 2003 #4
    Question for both replies...

    Ivan first.
    As I understand it you say its more just a convienient mathimatical reprosentation than whats actually physically happening. Yet in this book they also speak of impedance mismatching resulting in standing waves. It said how this could happen along transmission lines. I thought standing waves were more a function of position along the line than time. The book even says that when this occurs, there’s more or 'high spikes' of current in some parts of the line and troughs of current elsewhere (loops and nodes). Now this REALLY sounds like the result of a wave being physically reflected.
    So I’m definitely still a bit confused.

    Tyger; just a guess, but would this look like the model of flux lines connecting two point charges? Is this what’s happening between two battery terminals?
  6. Jul 13, 2003 #5
    It's a bit more complicated than that. The Poynting Vector is the cross product of the electric and magnetic field vectors, that is it is proportional to their products but at right angles to both of them, and it represents the rate of flow of energy, the energy per unit time per unit area, of the electromagnetic field. If you draw a pictoral of an electrical circuit and plot the field lines you can then plot the Poynting Vector. Alternately, you can solve the equations for the local fields and do the same. You can for instance see that the energy supplied by a battery will flow into a resistor from the space around it, but not flow into the perfect conductors that make up the resistor leads.

    It's a very differnt and correct view of how circuits work, instead of the naive picture of the wires carring the energy to the load.
  7. Jul 13, 2003 #6
    By AC I assume you me 60 cycle AC as used in the united states. It doesn't really matter what the frequency is however the result is the nearly the same. AC is a wave and as such has electromagnetic wave properties. The wavelength is so long that it is usually convient to think of it as simple current flow.
    However if ther is current flow there is an electromagnetic field in wave form created.
    If the terminal impedience isn't matched to the transmission line then part of that field will be reflected back down the transmission line and become a standing wave. An electromagnetic wave that will effect current flow in the line. This happens reguardless of the frequency.
    In order to increase efficency capacitors are added to inductive loads like motors to better match the impedience.
  8. Jul 13, 2003 #7


    Never heard of it like this before. I'm trying to visualize it in my head but its not working. Might you have a good link to a page that explains this?
  9. Jul 13, 2003 #8

    Ivan Seeking

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    Re: To really understand the flow

    I think he was asking about the usual view of circuits. Are you arguing that the representation that I gave was wrong, or do you mean to provide a deeper perspective on the question?
  10. Jul 13, 2003 #9
    Re: Re: To really understand the flow

    In this case it's a deeper view. If you look into waveguide theory and transmission line theory you will have the treatment of waves and Maxwell's Equations, and both of what you and I said will be true.
  11. Jul 13, 2003 #10

    Ivan Seeking

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    Re: Re: Re: To really understand the flow

    Ah, you scared me for a minute. I thought that you were disputing what I said, and that would mean one of us is wrong. Since I think you know what you're talking about, and since I think you are usually right, I would be wrong if I'm wrong and wrong if I'm right. That was a most disturbing reality to consider. Even when I'm wrong I'm right. :wink:
    Last edited: Jul 13, 2003
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