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Alternating current signals travelling on a wire obey wave-like behavior. In a rough sense, the higher the frequency the more wavy the signal. A good rule of thumb is that if the length of wires is comparable to or much longer than the wavelength of the signal, then you have to worry about wave-like phenomena such as reflection. The wavelength $\lambda$ of an electrical signal is roughly $$\lambda = c / f$$ where $c$ is the speed of light and $f$ is the frequency. Suppose we'd like to transmit the electricity from an electrical substation to a house and we want to keep the wavelength big enough to prevent reflection physics without having to deal with careful impedance matching. Let's put in a length of $1000 \, \text{m}$ to be conservative. Then we get $$f \leq c / 1000 \, \text{m} = 300 \, \text{KHz} \, .$$$$f \leq c / 1000 \, \text{m} = 300 \, \text{kHz} \, .$$

We're talking about the voltage inside the building here. Note that power is transmitted at much higher voltage and then stepped down near the end point. The 120 V V choice apparently comes from the fact that electricity was originally used for lighting, and the first lamps back in those early days were most efficient at around 110 V. The value 120 V V may have been chosen to offset voltage drop in the wires going to the lighting sources.

Alternating current signals travelling on a wire obey wave-like behavior. In a rough sense, the higher the frequency the more wavy the signal. A good rule of thumb is that if the length of wires is comparable to or much longer than the wavelength of the signal, then you have to worry about wave-like phenomena such as reflection. The wavelength $\lambda$ of an electrical signal is roughly $$\lambda = c / f$$ where $c$ is the speed of light and $f$ is the frequency. Suppose we'd like to transmit the electricity from an electrical substation to a house and we want to keep the wavelength big enough to prevent reflection physics without having to deal with careful impedance matching. Let's put in a length of $1000 \, \text{m}$ to be conservative. Then we get $$f \leq c / 1000 \, \text{m} = 300 \, \text{KHz} \, .$$

We're talking about the voltage inside the building here. Note that power is transmitted at much higher voltage and then stepped down near the end point. The 120 V choice apparently comes from the fact that electricity was originally used for lighting, and the first lamps back in those early days were most efficient at around 110 V. The value 120 V may have been chosen to offset voltage drop in the wires going to the lighting sources.

Alternating current signals travelling on a wire obey wave-like behavior. In a rough sense, the higher the frequency the more wavy the signal. A good rule of thumb is that if the length of wires is comparable to or much longer than the wavelength of the signal, then you have to worry about wave-like phenomena such as reflection. The wavelength $\lambda$ of an electrical signal is roughly $$\lambda = c / f$$ where $c$ is the speed of light and $f$ is the frequency. Suppose we'd like to transmit the electricity from an electrical substation to a house and we want to keep the wavelength big enough to prevent reflection physics without having to deal with careful impedance matching. Let's put in a length of $1000 \, \text{m}$ to be conservative. Then we get $$f \leq c / 1000 \, \text{m} = 300 \, \text{kHz} \, .$$

We're talking about the voltage inside the building here. Note that power is transmitted at much higher voltage and then stepped down near the end point. The 120 V choice apparently comes from the fact that electricity was originally used for lighting, and the first lamps back in those early days were most efficient at around 110 V. The value 120 V may have been chosen to offset voltage drop in the wires going to the lighting sources.

2 Add voltage info, spelling
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Why a standard?Frequency

Why a standard?

Light flicker

Light flicker

Reflections / impedance matching

Reflections / impedance matching

Alternating current signals travelling on a wire obey wave-like behavior. In a rough sense, the higher the frequency the more wavy the signal. A good rule of thumb is that if the length of wires is comparable to or much longer than the wavelength of the signal, then you have to worry about wave-like phenomena such as reflection. The wavelength $\lambda$ of an electrical signal is roughly $$\lambda = c / f$$ where $c$ is the speed of light and $f$ is the frequency. Suppose we'd like to transmit the electricity from an electrical substation to a house and we want to keep the wavelength big enough to prevent reflection physics without having to deal with careful impedance matching. Let's put in a length of $1000 \, \text{m}$ to be conservative. Then we get $$f \leq c / 1000 \, \text{m} = 300 \, \text{KHz} \, .$$

Voltage

We're talking about the voltage inside the building here. Note that power is transmitted at much higher voltage and then stepped down near the end point. The 120 V choice apparently comes from the fact that electricity was originally used for lighting, and the first lamps back in those early days were most efficient at around 110 V. The value 120 V may have been chosen to offset voltage drop in the wires going to the lighting sources.

Why a standard?

Light flicker

Reflections / impedance matching

Alternating current signals travelling on a wire obey wave-like behavior. In a rough sense, the higher the frequency the more wavy the signal. A good rule of thumb is that if the length of wires is comparable to or much longer than the wavelength of the signal, then you have to worry about wave-like phenomena such as reflection. The wavelength $\lambda$ of an electrical signal is roughly $$\lambda = c / f$$ where $c$ is the speed of light and $f$ is the frequency. Suppose we'd like to transmit the electricity from an electrical substation to a house and we want to keep the wavelength big enough to prevent reflection physics without having to deal with careful impedance matching. Let's put in a length of $1000 \, \text{m}$ to be conservative. Then we get $$f \leq c / 1000 \, \text{m} = 300 \, \text{KHz} \, .$$

Frequency

Why a standard?

Light flicker

Reflections / impedance matching

Alternating current signals travelling on a wire obey wave-like behavior. In a rough sense, the higher the frequency the more wavy the signal. A good rule of thumb is that if the length of wires is comparable to or much longer than the wavelength of the signal, then you have to worry about wave-like phenomena such as reflection. The wavelength $\lambda$ of an electrical signal is roughly $$\lambda = c / f$$ where $c$ is the speed of light and $f$ is the frequency. Suppose we'd like to transmit the electricity from an electrical substation to a house and we want to keep the wavelength big enough to prevent reflection physics without having to deal with careful impedance matching. Let's put in a length of $1000 \, \text{m}$ to be conservative. Then we get $$f \leq c / 1000 \, \text{m} = 300 \, \text{KHz} \, .$$

Voltage

We're talking about the voltage inside the building here. Note that power is transmitted at much higher voltage and then stepped down near the end point. The 120 V choice apparently comes from the fact that electricity was originally used for lighting, and the first lamps back in those early days were most efficient at around 110 V. The value 120 V may have been chosen to offset voltage drop in the wires going to the lighting sources.

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In the end, the choice of a single specific number comes from the necessity to standardize. However, we can make some physical observations to understand why that final choice had to fall in a certain range.

Why a standard?

First of all, why do we even need a standard? Can't individual appliances convert the incoming electricity to whatever frequency they want? Well, in principle it's possible, but it's rather difficult. Electromagnetism is fundamentally time invariant and linear; the differential equations we use to describe it Maxwells' equations are such that a system driven by a sinusoidal input at frequency $\omega$ responds only at that same frequency. In order to get out a frequency different from $\omega$ the electromagnetic fields have to interact with something else, notably charged matter. This can come in the form of a mechanical gear box or a nonlinear electrical elements such as transistors. Nonlinear elements such as the transistor can generate harmonics of the input, i.e. frequencies $2 \omega$, $3 \omega$, etc. However, in any case, frequency conversion introduces efficiency loss, cost, and bulkiness to the system.

In summary, because of the time invariance and linearity of electromagnetism, it is considerably more practical to choose a single frequency and stick to it

Light flicker

In a historical note by E. L. Owen (see references), it is noted that the final decision between 50 and 60 Hz was somewhat arbitrary, but based partially on the consideration of light flicker.

During the lecture, while Bibber recounted Steinmecz’s contributions to technical standards, he briefly repeated the story of the frequencies. By his account, “the choice was between 50- and 60-Hz, and both were equally suited to the needs. When all factors were considered, there was no compelling reason to select either frequency. Finally, the decision was made to standardize on 60-Hz as it was felt to be less likely to produce annoying light flicker.”

The consideration of light flicker comes up elsewhere in historical accounts and explains why very low frequencies could not be used. When we drive a pure resistance with an ac current $I(t) = I_0 \cos(\omega t)$, the instantaneous power dissipation is proportional to $I(t)^2$. This signal oscillates in time at a frequency $2\omega$ (remember your trig identities). Therefore, if $\omega$ is lower than around $40 \, \text{Hz}$$^{[a]}$, the power dissipated varies slowly enough that as a visual stimulus you could perceive it. This sets a rough lower limit on the frequency you can use for driving a light source. Note that the arc lamps in use when electrical standards were developed may not have had purely resistive electrical response (see Schwern's answer where cooling on each cycle is mentioned) but the source frequency is always present in the output even in nonlinear and filtered systems.

Reflections / impedance matching

Alternating current signals travelling on a wire obey wave-like behavior. In a rough sense, the higher the frequency the more wavy the signal. A good rule of thumb is that if the length of wires is comparable to or much longer than the wavelength of the signal, then you have to worry about wave-like phenomena such as reflection. The wavelength $\lambda$ of an electrical signal is roughly $$\lambda = c / f$$ where $c$ is the speed of light and $f$ is the frequency. Suppose we'd like to transmit the electricity from an electrical substation to a house and we want to keep the wavelength big enough to prevent reflection physics without having to deal with careful impedance matching. Let's put in a length of $1000 \, \text{m}$ to be conservative. Then we get $$f \leq c / 1000 \, \text{m} = 300 \, \text{KHz} \, .$$

Further reading

Detailed document by E. L. Owen with references

$[a]$: I'm not an expert in human flicker perception. This number is a rough guess based on personal experience and some literature.

P.S. I consider this answer a work in progress and will add more as I learn more.