We're just starting the whole RF scene having dealt with DC and low frequency AC for all our previous courses.

I understand that at high frequency AC, fundamental circuit laws don't apply anymore and the classic passive component models need to be changed. The justification for this was that at high frequency AC transmission, the wavelength becomes much smaller and can sometimes be smaller than the wiring on PCBs etc.

I understand that this is an issue when transmitting through free space with electromagnetic waves but why is this an issue with actual physical wires and PCBs being driven by an AC source? I mean it's a direct connection, we aren't using electromagnetic waves to propogate through free space and so wavelength and stuff shouldn't matter right?

  • 10
    \$\begingroup\$ At DC, an ideal inductor is a short and an ideal capacitor is an open. In the "from DC to daylight" limit, an ideal inductor is an open and an ideal capacitor is a short. If you open up a Tektronix oscilloscope designed for the upper limits of GHz performance, you will be able to see conductive paths formed by a series of capacitive striping and conductive blocks formed by what looks much like a simple trace. \$\endgroup\$
    – jonk
    Commented Sep 16, 2018 at 20:31
  • 9
    \$\begingroup\$ The wave takes time to get to the other end of the wire, don't you think? If you have a light-year-long wire, and you connect a battery to one end, it'll have to be at least a year before the battery realises there's nothing connected to the other end. And in that time your battery will be discharging into a seemingly open circuit. \$\endgroup\$ Commented Sep 16, 2018 at 22:39
  • 3
    \$\begingroup\$ @EricDuminil They also behave like the way you build them. \$\endgroup\$ Commented Sep 17, 2018 at 4:23
  • 4
    \$\begingroup\$ @immibis: This is how I usually measure the impedance of my infinitely long coax cables. \$\endgroup\$
    – PlasmaHH
    Commented Sep 17, 2018 at 8:34
  • 2
    \$\begingroup\$ "we aren't using electromagnetic waves to propogate through free space" is technically wrong - even if you don't intend to use them that way, if you have physical wires and high frequency AC, then that propagation through free space is happening whether you want it or not. \$\endgroup\$
    – Peteris
    Commented Sep 17, 2018 at 10:24

13 Answers 13


Actually, it is all about the waves. Even when dealing with DC, it is all managed by the electrical and magnetic fields and waves.

The "fundamental laws" aren't breaking down. The rules you have learned are simplifications that deliver accurate answers under certain conditions - you haven't yet learned the fundamental laws. You are about to learn the fundamental laws after having used simplifcations.

Part of the assumed conditions for the simplified rules is that the circuit is much smaller than the wave length of signal(s) involved. In those conditions, you can assume that a signal is in the same state across the circuit. That leads to a lot of simplifications in the equations describing the circuit.

As the frequencies get higher (or the circuits larger) so that the circuit is an appreciable fraction of the wavelength, that assumption is no longer valid.

The effects of wavelength on the operation of electrical circuits first became obvious at low frequencies but with very large circuits - telegraph lines.

When you start working with RF, you reach wavelengths such that the size of a circuit that sits on your desk is an appreciable fraction of the wavelength of the signals used.

So, you start having to pay attention to things you could conveniently ignore before.

The rules and equations you are now learning also apply to simpler, lower frequency circuits. You can use the new things to solve the simpler circuits- you just have to have more information and solve more complicated equations.

  • \$\begingroup\$ Parasitic effects of imperfect materials, negligible at LF, will bite the HF engineer. \$\endgroup\$
    – amI
    Commented Sep 17, 2018 at 7:35
  • 56
    \$\begingroup\$ Grade-school science also bites us: the wrong ideas that electricity is a separate kind of energy, that electrons=energy, or that electrons travel at the speed of light like Mrs. Frizzle and Bill Nye said. Actually all circuits are waveguides, the energy travels outside as EM fields, circuit-energy is ELF radio waves, and the electrons only wiggle slightly as the energy-waves propagate across our circuitry. Xmit antennas don't change electricity into EM fields, it was already EM fields; "electricity" was photons all along: even DC circuits deal in wave-energy of EM-fields. \$\endgroup\$
    – wbeaty
    Commented Sep 17, 2018 at 11:06
  • \$\begingroup\$ So basically, we've been taught the wrong way around the whole time. \$\endgroup\$ Commented Sep 19, 2018 at 11:11
  • 5
    \$\begingroup\$ @AlfroJango80: Not backwards at all. You learned a simplification that works for a lot of things. It is simple enough that you can work with it right away and accurate enough to be useful. \$\endgroup\$
    – JRE
    Commented Sep 19, 2018 at 11:19
  • \$\begingroup\$ @wbeaty In a DC current, the electrons do travel, albeit certainly << c . But you are correct that it's still a wave, since there was always a startup non-DC voltage, so the FourierTransform over all time has frequency components. \$\endgroup\$ Commented Sep 19, 2018 at 13:58

Because the assumptions required by the lumped element model are violated. The lumped element model is what allows you to analyze devices like resistors connected by nodes, without considering the physical layout of devices and the circuit.

The lumped element model assumes:

  1. The change of the magnetic flux in time outside a conductor is zero.

$${\frac {\partial \phi _{B}}{\partial t}}=0$$

  1. The change of the charge in time inside conducting elements is zero.

$${\frac {\partial q}{\partial t}}=0$$

  1. The characteristic length (the ‘size’ of the nodes and devices) is much less than the wavelength of the signal of interest.

$$L_c << \lambda$$

  • \$\begingroup\$ i dunno why this answer is not the one at the top of the heap. it directly and correctly answers the root question. \$\endgroup\$ Commented Sep 17, 2018 at 6:23
  • 11
    \$\begingroup\$ I agree - but rather than just enumerating these equations without explanation, I would have loved to see how Kirchoff's equations pop out of Maxwell's equations. Chapter 2.3 of Tom Lee's "Planar Microwave Engineering" does a fairly well job on this. \$\endgroup\$
    – divB
    Commented Sep 17, 2018 at 17:49
  • \$\begingroup\$ This is an excellent to-the-point answer, though it does not define the complex models of the LEM when the rules are violated, but other answers cover this issue. \$\endgroup\$
    – user105652
    Commented Sep 18, 2018 at 21:15
  • \$\begingroup\$ When the traditional lumped element circuit model doesn't work at high frequencies, I add more lumps to simulate the continuous transmission lines ala finte element modeling. \$\endgroup\$ Commented Sep 23, 2018 at 5:14

The fundamental laws of EM are Maxwell's Equations: $$\nabla \cdot \mathbf{E} = 4\pi\rho$$ $$\nabla \cdot \mathbf{B} = 0$$ $$\nabla \times \mathbf{E} = -\frac{1}{c} \frac{\partial \mathbf{B}} {\partial t}$$ $$\nabla \times \mathbf{B} = \frac{1}{c}\left( 4\pi\mathbf{J} + \frac{\partial \mathbf{E}}{\partial t}\right)$$

They have always been the fundamental laws of EM, but at lower frequencies, we find solving those multidimensional differential equations to be rather hard, and not all that beneficial to support our understanding of the circuit. You don't want to have to invoke symmetry to properly solve an an equation for propagation along a wire if the net difference between a short 18ga wire and a long 0000 wire is 0.0000001% with respect to the behaviors you are interested in.

Accordingly, people have already integrated these equations for simple cases, like wires at low frequencies, and found the equations you were given in earlier classes. Well, more precicely, we found these equations first, then found Maxwell's equations as we pushed deeper into EM, and then eventually showed that the original equations were consistent with Maxwell's.

Personally, I find it best to explore this by example. I'd like to take an example from the famous tome: The Art of High Speed Digital Design (subtitle: A Handbook of Black Magic). In their introduction, they point out how important capacitor type choices are. They make the extraordinary claim that at high speeds, a capacitor can look like an inductor because its leads are two parallel wires. Parallel wires have an inductance.

If we use the concept of impedance, we can calculate the effects of parasitic inductance on our capacitor. The impedance of a capacitor is \$\frac{-1}{\omega C}\$, and the impedance of an inductor is \$\omega L\$. We'll ignore parasitic resistance for now, though it's an important detail too in many cases. Put them in series and you see the impedence of the circuit \$\frac{-1}{\omega C} + \omega L = \frac{\omega^2 CL - 1}{\omega C}\$. As you can see, at high frequencies, that CL term starts to dominate, making the whole circuit look more like an inductor. At lower frequencies, where \$\omega^2CL \ll 1\$, you can ignore this. At high frequencies, you can't.

Likewise, at high frequencies, it gets harder to ignore the fact that wires emit EM radiation. At low frequencies, this effect is trivial, but at high frequencies, a large amount of power can be dissipated in the wire itself.

  • \$\begingroup\$ Cort, when @τεκ's answer gets voted up more, i will vote this up. \$\endgroup\$ Commented Sep 17, 2018 at 6:26

There is lot of complicated (and right) answers here. I will add one simple analogy - think of shooting gun:

  • at 10 cm distance, the time of bullet travel is just distance/velocity and hitpoint is on line identical to axe of the barrel
  • at 10 m distance you see, that the bullet hit the target lower, as gravitation pulled it little down and you have to adjust your aim for it
  • at 20 m you need adjust more, as the gravitation affect it more
  • at 100 m you see, that even with gravitation counted in, it does not fit. Why? Yes, there is and air and the bullet is slowed too. Also we see, that the bullet is doing everything else, than just flying straight, as it rotation combined with it vertical velocity compress air on one side and the bullets is dancing there. Also we can see, that it is probably not totally homogenous, which adds to its moving another factor
  • at 1000 m we can see, that there is something else yet - yes, the Earth is rotating and it counts too
  • so go higher, where it would not end its fly in th ground so fast, say on orbit and shoot there - again there is more to count - we forgot about moon gravity too
  • and on even longer distance we see, that there is not only Sun gravity, but also the light going from Sun, which push it a little too and all that electically active particles which makes little currents in it and magetic fields ...
  • and in extremely long (like interstelar) traces also gravity of other galaxies (not suprisingly), but our bulled have time to change its internal structure, as even the lead is extremly slowly breaking into other chemical elements by radioactivity decay

Well it is super complicated now, so lets return to the 10 cm distance on start - does that mean, that the formula time=distance/velocity does not work? Or does not work our final supercomplicated formula?

Well, both works, as all those elements we slowly added to our calculations are there still present, only on such short distance the difference is so small, that we cannot even measure it. And so we can use our "simple" formula - which is not totally exact, but in some reasonable conditions give reasonable exact results (say to 5 decimal places) and we are able to learn it fast, apply it fast and get results, which are correct (to 5 decimal places) at the scale which is interesting for us.

The same goes to DC, slow AC, Radio frequencies, ultra high frequencies ... each following is more exact version of the previous, each previous is special version of the following in situation, where the small differecies are such small, that we can discart them and get "good enought" result.

  • 4
    \$\begingroup\$ @ gilhad This answer should be required reading, and study, for all EE students. \$\endgroup\$ Commented Sep 17, 2018 at 4:02

I mean it's a direct connection, we aren't using electromagnetic waves to propogate through free space and so wavelength and stuff shouldn't matter right?

That's a very wrong assumption. The signals are still EM waves and remain EM waves, if they propagate through free space or a conductor. The laws remain the same.

At connections (wires) in the order of the length of the wavelength you can no longer used the "lumped element" approach. The "lumped element" approach means that connections are considered "ideal". For high frequency signals at distances in the order of the wavelength and larger, this approach is invalid.

So remember: the EM laws do not change as an EM wave travels through space or a conductor, they apply in both cases. EM waves remain EM waves in free space or in a conductor.

  • \$\begingroup\$ Okay. I understand that the EM waves still exist when transmitting AC voltages through a wire - but they don't contribute to the actual current flow right (apart from reducing it a bit with the opposing emf). So then why should we abandon all our low-frequency and DC models when essentially the AC current is still flowing fine through that wire. I just don't see how the wavelength being too small comes into play when we have a direct wire from the AC source and load. \$\endgroup\$ Commented Sep 16, 2018 at 17:52
  • \$\begingroup\$ One should add, even for the most-high-speed signals one could expect on a "normal" PCB, the lumped model is still applicable if capacitance and inductance of a whole track are taken into account. The distances are small, after all. \$\endgroup\$
    – Janka
    Commented Sep 16, 2018 at 17:53
  • 4
    \$\begingroup\$ @AlfroJang80, a dipole antenna is just a pair of direct wires from the feed to their open ends. And yet it can transmit and receive wireless RF signals. Somewhere between a very short wire that doesn't transmit or receive any energy, and a quarter wave dipole that transmits and receives very efficiently, there must be a middle ground where radiation effects are significant but not dominant. \$\endgroup\$
    – The Photon
    Commented Sep 16, 2018 at 18:13
  • 3
    \$\begingroup\$ @AlfroJang80 Think about a simple situation where "current" is just "the movement of electrons." If something makes the first electron in the wire start moving, what makes the next one, and the following one -and the ones 1km away if it is a long wire - move? Answer, the electromagnetic field around each electron. Don't forget that a simple circuit with just a battery, a switch, and a resistor is not a "DC circuit" at the instant when you open or close the switch, because the current changes - but in your first course in DC circuit analysis, you ignore that fact. \$\endgroup\$
    – alephzero
    Commented Sep 17, 2018 at 9:13
  • 2
    \$\begingroup\$ @AlfroJang80 current is only half, and voltage is the other half. That's the key. Current is the magnetism part of the EM wave, voltage is the e-fields part. "VI" is "EM." All wires are waveguides! But we can ignore this, if we say that the EM wave is actually a separate "E," the voltage, and "M" the current. Then concentrate on DC volts/amps only, ignore circuit's EM waves. But even DC is a wave at 0Hz (or at 0.0001Hz.) In circuit-physics, DC doesn't exist, and everything is actually EM waves guided by long rows of electrons, with all "electricity" energy only traveling outside the wires. \$\endgroup\$
    – wbeaty
    Commented Sep 17, 2018 at 11:15

Although this has been answered a couple of times I would like to add the reasoning that I personally find most eye opening and is taken from Tom Lee's book "Planar Microwave Engineering" (chapter 2.3).

As indicated in the other responses, most people forget that Kirchoffs laws are just approximations that hold under certain conditions (the lumped regime) when quasi-static behavior is assumed. How does it come to these approximations?

Let's start with Maxwell's quations in free space:

$$ \nabla \mu_0 H=0 \qquad(1) \\ \nabla \epsilon_0 E = \rho \qquad(2) \\ \nabla\times H=J+\epsilon_0 \frac{\partial E}{\partial t} \qquad(3) \\ \nabla\times E=-\mu_0 \frac{\partial H}{\partial t} \qquad(4) \\ $$

Equation 1 states that there is no divergence in the magnetic field and hence no magnetic monopoles (mind my username! ;-) )

Equation 2 is Gauss' law and states that there are electric charges (monopoles). These are the sources of the divergence of the electric field.

Equation 3 is Ampere's law with Maxwells modification: It states that ordinary current as well as a time-varying electric field creates a magnetic field (and the latter one corresponds to the famous displacement current in a capacitor).

Equation 4 is Faradays law and states a changing magnetic field causes a change (a curl) in the electric field.

Equation 1-2 is not important for this discussion but Equation 3-4 answer where the wave behavior comes from (and since Maxwell's equations are most generic, they apply to all circuits incl DC): A change in E causes a chance in H which causes a change in E and so forth. Is is the coupling terms that produce wave behavior!

Now assume for a moment mu0 is zero. Then the electrical field is curl free and can be expressed as the gradient of a potential which also implies that the line integral around any closed path is zero:

$$ V = \oint E dl = 0 $$

Voila, this is just the field-theoretical expression of Kirchhoff's Voltage Law.

Similarly, setting epsilon0 to zero results in

$$ \nabla J = \nabla (\nabla \times H) = 0 $$

This means the divergense of J is zero which means no (net) current can build up at any node. This is nothing more than Kirchhoffs Current Law.

In reality epsilon0 and mu0 are of course not zero. However, they appear in the definition of the speed of light:

$$ c = \sqrt{\frac{1}{\mu_0 \epsilon_0}} $$

With infinite speed of light, the coupling terms would vanish and there would be no wave behavior at all. However, when the physical dimensions of the system are small compared to the wavelengths then the finiteness of the speed of light is not noticeable (similarly as time dilation always exist but won't be noticable for low speeds and hence Newtons equations are an approximation of Einsteins relavivity theory).

  • \$\begingroup\$ why so few upvotes? I like this answer. \$\endgroup\$
    – Neil_UK
    Commented Sep 19, 2018 at 15:10

They don't break down, but when the rise time approaches 10% or is less than the propagation delay to a load impedance matching is important due to that wavelength. Load impedance is inverted to a source at 1/4 wavelength whether it is conducted or radiated.

If the load is not a matched impedance to the "transmission line and source" reflections will occur according to some coefficient called return loss and the reflection coefficient.

Here is an experiment you can do to demonstrate conducted EM waves.

If you try probing a 1 MHz square wave on a 10:1 scope probe with the 10 cm ground clip you might see 20 MHz lumped coax resonance. Yes, the probe is not matched to the 50 ohm generator so reflections will occur according to the 10 nH/cm ground lead and 50 pF/m special probe coax. It is still a lumped element (LC) response.

Reducing the 10:1 probe to less than 1 cm to just the pin tip and ring without long ground clip, raises the resonant frequency perhaps to the limitation of the probe and scope at 200 MHz.

Now try a 1:1 1 m coax which is 20 ns/m so a 20~50 MHz square wave on a 1 m coax with a 1:1 probe will see a reflection at one fraction of a wavelength and horrible square wave response unless terminated at the scope with 50 ohms. This is a conducted EM wave reflection.

But consider a fast logic signal with a 1 ns rise time may have a 25 ohm source impedance and it has a >300 MHz bandwidth so overshoot can be a measurement error or actual impedance mismatch with track length reflections.

Now compute 5% of the wavelength of 300 MHz at 3e8 m/s for air and 2e8 m/s for coax and see what the propagation delay times are that cause echoes from a mismatched load, e.g. CMOS high Z and say 100-ohm tracks. This is why controlled impedances are needed usually above 20~50 MHz and this as an effect on ringing or overshoot or impedance mismatch. But without, this is why logic has a such a large grey zone between "0 & 1" to allow for some ringing.

If any words are unknown, look them up.

  • \$\begingroup\$ @PeterMortensen ty \$\endgroup\$ Commented Sep 16, 2018 at 22:40

Electrical signals take time to propagate through wires (and PCB traces). Slower than EM waves through a vacuum or air, always.

For example a twisted pair in a CAT5e cable has a velocity factor of 64%, so the signal travels at 0.64c, and it will go about 8" in a nanosecond. A nanosecond is a long time in some electronic contexts.It's 4 clock cycles in a modern CPU, for example.

Any configuration of conductors of finite size has inductance and capacitance and (usually) resistance so it can be approximated using lumped components at a finer level of granularity. You might replace the wire with 20 series inductors and resistors with 20 capacitors to the ground plane. If the wavelength is very short compared to the length, you might need 200 or 2000 or .. whatever to closely approximate the wire and other methods might start to look attractive, such as transmission line theory (typically a one semester undergrad course for EEs).

"Laws" like KVL, KCL are mathematical models that approximate reality very accurately under appropriate conditions. More general laws such as Maxwell's equations apply more generally. There might be some situations (relativistic perhaps) where Maxwell's equations are no longer very accurate.

  • 2
    \$\begingroup\$ Maxwell's equations can be modified (Lorentz–FitzGerald) to be made invariant under relativistic transformations. If you read German (as I do) then this is probably the best short overview of the transformed equations that I can quickly find. I also like this. \$\endgroup\$
    – jonk
    Commented Sep 16, 2018 at 19:08

It is a wave. The same thing that is going on here is the same thing that is talked about when it's mentioned how that "electricity moves at the speed of light" even though that electrons "move" much slower. Actually it is about 2/3 (IIRC) the speed of light in most conducting materials - so about 200 000 km/s. In particular, when you throw a switch, for example, you send an electromagnetic wave down the circuit, which causes the electrons to be incited to motion. It's a "step" wave in that case - behind it the field is steady high, ahead of it, it's zero, but once it passes the electrons are now moving. Waves move in a medium at slower speeds than in free space, but they still go through media - that's why, after all, that light can pass through glass.

In this case, the voltage source is constantly "pumping" back and forth, and thus is setting up oscillating waves that just the same way, move with the same speed. At low frequencies, like 60 Hz, the length of these waves is far longer than the scale of a single device at human scale, namely for that particular frequency about 3000 km (200 000 km/s * (1/60 s)), versus maybe 0.1 m (100 mm) for a typical hand-held PCB, meaning about a 30 000 000:1 scale factor, and thus you can treat it as a uniform current that is changing periodically.

On the other hand, go up to say 6 GHz - so microwave RF applications as in telecom transmission technology - and now the wavelength is 100 million times shorter, or 30 mm. That is way smaller than the scale of the circuit, the wave is important, and you now need more complex electrodynamic equations to understand what is going on and good ole' Kirchhoff just won't cut the mustard no more :)


A simpler answer: because parasitic components that are not drawn in your circuit diagram begin to play a role:

  • the series resistance (ESR) and series inductance of capacitors,
  • increasing resistance of wires due to skin effect,
  • the parallel damping (eddy currents) and parallel capacitance of inductors,
  • the parasitic capacitance between voltage nodes (e.g. between PCB traces including "ground"),
  • the parasitic inductance of current loops,
  • the coupled inductance between current loops,
  • the coupling of the magnetic fields between unshielded inductors, which may depend on the random polarity of component placement,
  • ...

This is also the topic of EMC, very important if you want to build circuits that actually work in the field.

Also, don't be surprised if you cannot even measure what is going on. Above a MHz or so it becomes an art to properly connect an oscilloscope probe.


You've got lots of excellent answer to your question, so I won't reiterate what has already been said.

I'll try to address your comments to various answers, instead. From the comments you posted you seem to have a basic misunderstanding of the physical laws governing circuits.

You seem to think that "moving electrons in a wire" are something quite unrelated from EM waves. And that EM waves come into play only in certain situations or scenarios. This is basically wrong.

As others have said, Maxwell's equations (MEs from now on) are the key for truly understanding the issue. Those equations are capable of explaining every EM phenomenon known to mankind, except for quantum phenomena. So they have a very broad range of application. But that is not the main point I want to make.

What You should understand is that electrical charges (electrons, for example), generate an electric field around them just by their very existence. And if they move (i.e. if they are part of an electric current) they also generate a magnetic field.

Traveling EM waves (what common people usually understand as EM "waves") are just the propagation of the variations of electric and magnetic fields across space ("vacuum") or any other physical medium.

Basically that is what MEs say.

Moreover, MEs also tell you that whenever a field varies (be it electric or magnetic) then "automatically" the other field comes into existence (and it is varying too). That's why EM waves are called Electro-Magnetic: a (time-)varying electric field implies the existence of a (time-)varying magnetic field and vice versa. There can be no varying E-field without a varying M-field and, symmetrically, there can be no varying M-field without an accompanying varying E-field.

This means that if you have a current in a circuit, and this current is not DC (otherwise it only generates a static magnetic field), you WILL HAVE an EM wave in all the space surrounding the path of the current. When I say "in all the space" I mean "all physical space", regardless of which bodies occupy that space.

Of course the presence of bodies alters the "shape" (i.e. the characteristics) of the EM field generated by a current: in fact, components are "bodies" designed to alter that field in a controlled manner.

The confusion in your reasoning may come from the fact that lumped components are designed to work well only under the assumption that the fields are varying slowly. This is called technically the quasi-static fields assumption: the fields are assumed to vary so slowly to be very similar to those present in a true DC situation.

This assumption leads to drastic simplifications: allowing us to use Kirchhoff's laws to analyze a circuit without appreciable errors. This doesn't mean that around and inside components and PCB tracks there are no EM waves. Indeed there are! The good news is that their behavior can be usefully reduced to currents and voltages for the purpose of designing and analyzing a circuit.


You are really asking two questions: 1) "Why the fundamental circuit laws breakdown" at high frequencies AC. 2) Why should they also breakdown when using "actual physical wires..."

The first question has been covered in the previous answers, but the second question leads me to believe that your mind has not transitioned from "electrons moving" to E-M waves moving, which I will address.

Regardless of how E-M waves are generated, they are the same (other than amplitude and frequency). They propagate at the speed of light and in a "straight" line.
In the specific case when they are generated by charges flowing in a wire, the wave will follow the direction of the wire!
At all times, when dealing with moving charges, you are dealing with E-M waves. However, when the ratio of wavelength to size of circuit is high enough, 2nd and higher order effects are small enough that, for practical purposes, they can be ignored.

I hope it is now clear that the wires only serve to direct the E-M waves, rather than change their nature.

  • \$\begingroup\$ Fantastic! That was exactly my concern. \$\endgroup\$ Commented Sep 21, 2018 at 7:30
  • \$\begingroup\$ One last thing. So at low-frequency AC, electrons are moving back and forth and this generates emag waves that propogate. However due to the low-frequency, the amount of energy contained in these waves is negligible and so doesn't matter if we take them into account or not. At high-frequency AC, these emag waves now contain much more energy and we must take them into account as well as remembering that the voltage and current waveforms will also be delayed at different points on the circuit. Is that correct? \$\endgroup\$ Commented Sep 22, 2018 at 1:21

You need to change the way you think about electricity. Think of the concept as an electron oscillating in empty space. In DC the oscillations push and displace electrons in the same general directional vector. At high frequencies the displacements take place in many directions at higher rates and more randomly, and everytime you displace electrons something happens, and using the equations listed here and in text books helps model what will happen. When you are engineering you are trying to make a model and identify patterns of what is happening and use that to solve problems.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.