I have a few straightforward questions about AC but surprisingly can't find any of explicit confirmations.

I understand that in reality all this is much more complicated but assume example below is greatly simplified case, only steady state with no any transient phenomena, TV acts like perfectly matched load without any reflections etc. For now I'm interested if my basic understanding is correct or not.

  1. Here is static illustration of alternating voltage / current. And this is image from Wikipedia article about transmission line, which looks pretty similar.

Is it animated version of AC which demonstrate propagation of alternating voltage (and current) waves from AC generator (assume it connected on left side) toward load?

  1. Now, I plug, say, TV set power cord into electrical wall outlet. There would be exactly same picture: constantly changing electric field due charges, magnetic field (due currents) and the waves would appear between/around "live" and "neutral" conductors in the outlet as well as wires inside the cord. Is this correct?

  2. Since electric field constantly changing as I noted above, potential difference associated with the field constantly changing as well. Hence, actually, "live" and "neutral" wires (and wires inside power cord) keep permanently recharging to corresponding PD as waves propagates between them, with peaks +311V and -311V but for convenience there is famous ~220-240V RMS. Am I right?

  3. At same time "neutral" wire can be grounded for safety reason, so PD between "neutral" and the Earth is ~0V?

  4. At last I'm wonder what if we don't plug anything into outlet, will it act as open-ended transmission line?

Sorry for my english and thanks for the help.

  • 1
    \$\begingroup\$ Usually for 50Hz or 60Hz AC power within a home, the transmission line concept is not needed. AC can just be viewed as a a time-varying voltage between two wires. One of them may be designated as "neutral" but not always. There is usually a conductor for ground also. If you don't plug anything into the outlet, the voltage is still present, but the current is basically zero, because it is an open circuit, and not really long enough to be a transmission line at 50 or 60 Hz. \$\endgroup\$
    – user57037
    Jun 21, 2018 at 6:20

2 Answers 2


When we are analysing circuits, it's usually best to either think in the simple 'quasi-static' circuit theory approach, or use transmission lines and take that treatment to its logical conclusion. If you start thinking transmission lines, but then assume the load has to be matched, you've both complicated the situation and restricted your thought to a very marrow case.


AC is such low frequency that all points on all wires are at the same voltage. It is a 'node', as far as circuit theory is concerned. There are no waves. Loads draw as much current as they need to from the voltage applied and their reactance.

Transmission lines

The voltage and the current in a wave on the transmission line are always in the ratio of the line's impedance. However, there are two independent waves on the line, one in each direction. The voltages and currents we read on a meter are the sums and differences of those waves.

For instance, if we short circuit the end of the line, the waves are equal in amplitude, and phased so they add up to zero voltage and a finite current. If we open circuit the end of the line, then equal amplitude waves are phased so they add up to finite voltage with zero current. Only in the case of a load matched to line is there a forward wave with no reverse reflected wave. An arbitrary load will tend to reflect a smaller amplitude of reflected wave than the incident wave. This is the 'reflection coefficient' or 'return loss' wave.

When power is first connected to the line, there is a transient state that lasts several propagation delays of the line. A forward wave from the power source travels to the end, and finds the load. If the load accepts the ratio of voltage and current that's in the wave, that is if the load is matched, that's the end of the transient stage. If however, the load is not matched, a reverse wave is reflected so that the sums of the waves create a load voltage and current that is correct for the load. This reverse wave may again be reflected by the source. Each reflection will be smaller than the last, and after enough round trips, we can say the situation has settled down into the steady state.

This is complicated, which is why we approximate to the 'short line' circuit theory case, whenever we can.

It's this transient behaviour that can cause problems in digital systems with long line. Unless the lines are well matched, the reflected waves can cause multiple edges to be seen by receivers on the line.

Going back to your TV lead. We would treat this lead as short for the mains supply, and just use circuit theory. However, for sources of EMI within the set, leakage from oscillators and edges from logic and power supplies, we would treat the lead as a transmission line. We would also treat the pair of wires in the lead, taken together, with respect to ground, as a transmission line, when assessing the ways that RF leakage was getting out of the set.


For a transmission line to behave as a transmission line, the wavelength of the (AC) signal needs to be smaller than the length of the transmission line. This is because the "wave" should "fit" inside the transmission line.

wavelength = speed_of_light / frequency

So take for example a 2.5 GHz WiFi signal, then the wavelength is \$3*10^8/2.5 * 10^9 = 0.12\$ = 12 cm

This means that a short (coax) cable designed for 2.5 GHz signals will behave as a transmission line.

But for the 50 (or 60) Hz mains frequency the wavelength is 6000 (or 5000) km ! That means that the transmission line way of thinking does not apply here, the cables are simply too short (compared to the wavelength). This means that the mains cables can be treated as a "direct connection" perhaps with some series resistance. But there's no "wave propagation" going on like in a transmission line, at least not in the sense that we need to consider and account for that wave propagation.

  • \$\begingroup\$ +1 for "wave" should "fit" inside the transmission line. \$\endgroup\$ Aug 6, 2018 at 22:21

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