Why Does a Bulb Stay Lit with AC Even Though the Energy Direction Keeps Reversing?

Question

It's pretty fascinating when you think about how electrical energy is actually transferred. Most of us imagine electricity as electrons flowing through wires, but in reality, it's much more complex. Electrical energy isn't directly carried by those electrons—it’s transferred through electric and magnetic fields around the wire. This transfer can be measured using Poynting vector, which we can use to find the direction using right hand rule

more about POYNTING VECTOR

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MY DOUBT

In the case of DC, the current flows in one direction, and so does the electric field. Using the right-hand rule, we can easily visualize how energy is transmitted from the battery to the bulb. But when we switch to an AC source, the direction of electron flow constantly changes, and with it, the electric field also oscillates.

This raises an interesting question: if the direction of energy transmission keeps changing in AC, how is the energy actually transmitted to the device? And if the energy is constantly reversing direction, how does the bulb still manage to glow continuously? Without affecting the efficiency of the devices my question is more generalized. Instead of focusing on a specific example like a light bulb, I'd like to concentrate on the broader concept of energy transfer, particularly how alternating current (AC) functions.

NOTE: This is not the duplicate question, read the question properly...the whole reason for this question is to understand the direction of energy transformation (according to the Poynting vector), why does it not change when the electric and magnetic field direction changes

Answer 1

There are two answers, depending on the bulb type.

For an incandescent bulb, the light is produced due to the filament being hot enough to emit lots of light in the visible spectrum. Such a bulb arguably flickers at 120Hz (half a period of the 60Hz AC input), but from a practical perspective, the filament does not cool off fast enough for this to be visible with the eye. I've never measured this effect, but I think it would be something a decently advanced laboratory apparatus could detect. (although it would be a very noisy signal to try to detect)

For a fluorescent bulb, the answer is that they do indeed flicker. The UV emissions of the gas in the bulb do pulse, and many of the phosphors used to convert that light to visible light have a short enough time constant that you can observe them with a high-speed camera.

For LED bulbs, the answer is an unequivocal yes. It is common to use a half-wave rectifier for cheap LED lights (like fairy lights for a Christmas tree), which flicker at 60Hz. Many people, including myself, can see that flickering with their naked eye. For myself, it's especially visible for blue lights.

Answer 2

The light is produced via a transformation of electrical energy into other forms, and this transformation occurs in one direction, even if the electric field is being reversed. This energy is then release in a form of electromagnetic waves - these are generated in the lamp, not provided by the battery.

The most obvious case is the incandescent lamp, where the energy results from heating of the filament (which is essentially a result of "friction" experienced by electrons, regardless of the direction in which they move.) The instantaneous heat production in a filament is given by Joule's law: $$ P(t)=I^2(t)R $$ If we now take current to be $I(t)=I_0\cos(\omega t)$ and calculate the average heat production per a period of oscillations, $T=2\pi/\omega$, it turns out to be finite, since the cosine is squared, i.e., $P(t)$ take only positive values: $$ \overline{P}=\frac{1}{T}\int_{t_0}^{t_0+T}dt I_0^2\cos^2(\omega t)R=\frac{I_0^2R}{2}. $$ It is worth knowing that the voltage power indicated for the power sockets is not the real amplitude of the voltage in the socket ($V_0=I_0R$), but the effective voltage of a direct current that would produce the same power output, $\overline{V}$: $$ \overline{P}=\frac{\overline{V}^2}{R}=\frac{V_0^2}{2R}, $$ that is $V_0=\bar{V}\sqrt{2}$ (about 310V for a socket marked as 22V.)

See threads How is light emitted by an incandescent lamp? and Heating effect of electric current for more details.

In other types of lamp the situation is rather similar, although it may differ in details. E.g., in neon lamps the electrons lose their energy via collision with the gas filling the bulb (aka "friction"), and the gas later releases this energy via emitting light ("heat").

Answer 3

when we switch to an AC source, the direction of electron flow constantly changes, and with it, the electric field also oscillates.

This is correct, but it is incomplete. For a resistive load the current is proportional to the electric field, so the current also oscillates. The oscillating current produces an oscillating magnetic field. So both the electric and the magnetic fields oscillate.

if the direction of energy transmission keeps changing in AC, how is the energy actually transmitted to the device?

This, however, is not correct. The direction of energy transmission does not change. The Poynting vector is the cross product of the electric and magnetic fields. When you take the negative of the electric field and the negative of the magnetic field, a negative times a negative is a positive. The energy flows in the same direction in both cases.