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I'm studying active and reactive power, and got to this question:
A loudspeaker, let's say the one from a guitar amplifier, receives an AC signal in order to sound, but usually it is said that DC is the only part of a signal that generates work. We know that a DC signal doesn't make any sound in a loudspeaker, so my doubt is:

Does the AC signal generate active power?

I will really appreciate if you help me solving this question. I think I have a wrong definition somewhere.

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  • \$\begingroup\$ Looks a "terminology" problem: I usually don't expect signals being used to transfer power to get work done. \$\endgroup\$ Commented Sep 30 at 5:07
  • \$\begingroup\$ @greybeard Curious; such is the definition of radio frequency design! Where "work" might be as gross as heating a target (induction, dielectric or other effect), or as precise as interplanetary communications. \$\endgroup\$ Commented Sep 30 at 7:56
  • \$\begingroup\$ @greybeard ERP is a different matter, but the sum is real power regardless; "work" in the physics sense. EM waves carry energy. \$\endgroup\$ Commented Sep 30 at 20:30
  • \$\begingroup\$ (@greybeard Ultimately, information itself is power :) if extraordinarily little of it, at the scales and rates we (intentionally, as opposed to incidentally) use.) \$\endgroup\$ Commented Oct 1 at 0:05

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AC and DC both do work in a loudspeaker. The voice coil is a transducer that converts current into displacement.

Applying DC will statically pull or push the cone - you will hear a 'pop' and see the speaker move (that's a handy way to test polarity by the way.) After that there will be no sound. So applying DC to the speaker produces mechanical work (it accelerates some air in a pulse), but just once.

Meanwhile the coil stores magnetic flux and gives off heat. These are also work, just not as useful as sound. (Speaking of which, as @Hearth points out, don't apply a large DC current for a long time as you can damage the driver, especially if it's a plastic cone.)

Applying AC will move the cone back-and-forth. AC power is indeed being converted to mechanical work: moving the coil makes pressure waves in the air. A measure of this electrical to mechanical conversion is 'speaker efficiency', and it is appallingly low - about 0.2 to 2%.

More here: https://engineeryoursound.com/what-does-speaker-efficiency-mean-explained/

Now, if you have an AC signal 'riding on top of' DC, the DC component won't produce sound (except initially), but it will influence the position of the cone. Maybe that's what you mean by DC 'not doing work'.

Now you mentioned 'active' and reactive power. I think you may be slightly confused here. There is real power, which is consumed in the device (as heat, mechanical work etc.) and reactive power, which is the response of a reactive component (L or C) to AC. Reactive power circulates back and forth from source to sink but doesn't do any useful work.

Reactive power only occurs when you have an AC component and reactive elements in your circuit. And that's certainly the case with a speaker: your amplifier can be cranking out lots of watts (volt-amperes) to the speaker but a large portion of that is just circulating back and forth and not being converted to sound.

This is a huge simplification of speaker behavior of course, but this gives the general idea.

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  • \$\begingroup\$ Note that some speakers, especially ones with a plastic cone, will not survive having DC applied to them for more than a few seconds. I've had speakers melt from that before. \$\endgroup\$ Commented Sep 30 at 4:00
  • \$\begingroup\$ "The voice coil is a transducer that converts current into displacement" For discussing "Active and reactive power", that shall do. For reproducing sound with high fidelity, for understanding phase relationships in multiway speakers a more elaborate model is needed (current and static field→force/mass→acceleration→linear displacement*area→volume displacement; linear displacement leading to an opposing force from the spider…). \$\endgroup\$ Commented Sep 30 at 4:49
  • \$\begingroup\$ @hacktastical -- A speaker's impedance is largely resistance except at its low frequency resonant point, so there is no significant reactive current except around that frequency, and the impedance goes up so the current is reduced there. \$\endgroup\$ Commented Sep 30 at 15:47
  • \$\begingroup\$ This might add to the OP confusion, but would it be worth mentioning, that in a pure mathematical/physics sense, when the speaker moves back to its starting point, no work has been done? \$\endgroup\$ Commented Sep 30 at 15:47
  • \$\begingroup\$ @Aaron-- No. In a physical sense, work has been done in moving the air to produce the sound. \$\endgroup\$ Commented Sep 30 at 15:50
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No, the part "usually it is said that DC is the only part of a signal that generates work" is wrong. We have AC motors and loads that do work with AC.

You can freely take the liberty of modeling a speaker as an inductance which has some resistance.

The AC part of the voltage applied over the inductance simply means that it takes time for the coil current to follow, i.e. a sine wave voltage will result into sine wave current that lags in phase.

It still does do some work, but voltage and current are not in phase. In addition to driving the load, the reactive part is just useless current in the wire which must be tranasferred in and out, the actual power part is the instantaneous voltage times the instantaneous current.

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usually it is said that DC is the only part of a signal that generates work

Until today I've never seen that said before; the reason is because it's wrong. AC can easily transfer power. I mean how do you think AC can power all your home appliances?

Does the AC signal generate active power?

Of course it does.

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