I have a switch that has "3A 250VAC" and "5A 125VAC" printed on it, which I imagine are ratings of what the switch can handle, but my understanding that 3A at 250V would be 750W and that 5A at 125V would be 625W so I'm confused about how the ratings correlate. If current or power alone are not sufficient to describe the capability then how would one determine whether any given values are within the acceptable range? In the ratings given there's an inverse correlation between the current and power so it seems conceivable that it can handle more current at a lower power, but that doesn't seem specific enough provide the power handling at a given amperage or to communicate the absolute limits for voltage/current/power.

To give a concrete example, I have a guitar speaker cabinet that contains 2 8-ohm speakers wired in series and I want to add a switch to toggle the speakers between series and parallel. The most powerful amplifier I would use to power the speakers says it makes 120W, but it's never turned up to more than half power so I'm assuming 120W is a safe estimate of the max power the switch would see. Into the 16-ohm load when the speakers are in series 120W would be "2.74A ~44VAC" which is less than the smallest of each value in the provided ratings so I assume it's okay. Into the 4-ohm load however 120W would be "5.5A ~22VAC" which is a higher current than the highest rating on the switch. Assuming the higher-current-at-lower-power thing is true I'm tempted to think 5.5 is a little more than 5 and 22 is a lot less than 125, but hardly seems like a reliable formula.

I've tried googling things like "understanding electrical ratings", "understanding wattage and current ratings", "how do I know if a switch can handle the power/current in a circuit" but I just finding explanations of what wattage/amperage/voltage are, their relationships to each other, analogies about water pressure, etc. I also tried searching for switches with various ratings and I found ones that were rated by wattage instead of A/VAC but nothing sufficient to answer my question.


  • 3
    \$\begingroup\$ Since audio is not constant power and 5.5A is a worst case peak, I think a 5A rating at so much lower voltage would be adequate. \$\endgroup\$
    – user16324
    Oct 17, 2020 at 20:53
  • 3
    \$\begingroup\$ Further to @Brian's comment, you'll generally be switching while the amplifier is quiet so you're unlikely to cause arcing and the voltage rating won't be important. The current rating is the one to watch. \$\endgroup\$
    – Transistor
    Oct 17, 2020 at 21:02

4 Answers 4


Switches are complex. They can get extremely complex when related to audio signal (they can even introduce noise and distortion - everyone old enough to remember relay-switched networks of wired phones can confirm).

The rating of a switch is more or less experimentally determined.

In general, the lower the voltage, the higher the current rating. The same switch is probably 6A 24VAC if someone bothers to rate it at 24V.

The first thing you have to know is that the rating is related to some number of switch actions and some timespan. The rating may be way more if you need to switch it few times and expect it to be used for no more than a week. It goes also in reverse - if you need an extended service, you may get a higher-rated switch.

... to an extend. A less known fact (and pretty much related to audio) is that switches need some current ("wetting current") in order to behave. If you use them at much lower than the rated current, the contacts don't self-clean and a variety of unpleasant things happen. Like, increased and unstable "on" resistance, up to and including a complete loss of contact because of the oxide layer, overheating or the aforementioned audio noise and distortion. The oxide layer can as well behave like a bad diode.

In regard to wetting current, non-telephone (non-dc-biased) audio is quite bad because of its wide dynamic range.

There are relays and switches that have contacts plated with some expensive metals (silver, gold, platinum) that are less prone to the wetting current effect. Their price is accordingly higher.

In short, you can pretty much try the switch you have in mind. 5A is not a lot less than 5.5A and the audio signal is rarely full-time full-power.

The result may or may not be acceptable for you. Either way, you won't lose much. At worst, you will have bad sound or no sound at all.

  • \$\begingroup\$ Thanks for the reply. The switch would never be actuated with a signal present, and likely toggled a no more than a few times per year. My concern about trying without being more confident in the suitability is that having the switch fail with a signal present may damage the amplifier. This is based on very common advice to not run an amplifier without speakers connected, but admittedly I don't know whether the circuit described by "without speakers connected" is electrically equivalent to an of the states of a failed switch; but of course I don't know that it isn't either. \$\endgroup\$
    – slagroom
    Oct 20, 2020 at 1:27
  • \$\begingroup\$ Is your amplifier tube-based or transistor-based? Transistor-based amps are completely safe, tubes are better kept loaded, but they don't fail right away. The failed switch may finally be equivalent to "without speakers", but you will probably hear the sound distorting long beforehand. It is also not likely that you will operate the amp for long without any sound coming out. \$\endgroup\$
    – fraxinus
    Oct 20, 2020 at 7:12
  • \$\begingroup\$ @fraxinux Tube amps. That's a good point about hearing the degradation/sound loss; I've never known how/why exactly the amp would be damaged so (in my mind) it wasn't out of the realm of possibility that it would just melt or explode within a fraction of second (and I didn't want to learn by experimentation lol). Thanks for all the info. \$\endgroup\$
    – slagroom
    Oct 21, 2020 at 12:40

Current and voltage ratings do not neccessarily apply at the same time and in a single operating point. A switch is a pretty good example for this:

  • it must not conduct a current higher than the rated current. This is relevant when the switch is closed.

  • a voltage higher than the rating must not occur between the switches contacts. This is relevant when the switch is open.

The current/voltage ratings may by no means be understood as a power rating. Actually there is either voltage accross the switch (when it is open) or current through the switch (when it is closed). It's not possible to have both conditions at the same time and of course, the switch would never be able to survive a power dissipation of hundreds of watts.

Like when you buy a car and the seller says: "This car can be driven up to 120mph and have a maximum weight of 2000kg".

That means, at no time must the car be faster than 120mph (regardless of its weight) and it must not be heavier than 2000kg at any time (regardless of its speed). Given mass and velocity, you could calculate the cars momentum and cinetic energy, but those are meaningless with respect to the seller's "maximum rating".

  • 3
    \$\begingroup\$ It's not possible to have both conditions at the same time - unless you get arcing across the contacts if the switch isn't capable of interrupting that much voltage, especially if there's an inductive load that makes the voltage spike when current is interrupted. This is a good answer for the conceptual basics, but you might want to mention that it's not always quite this simple, as other answers point out. You get some arcing as the contacts separate because they're temporarily very close to each other, moving away at some finite speed, unless the AC current / voltage happens to be 0. \$\endgroup\$ Oct 18, 2020 at 18:39
  • 1
    \$\begingroup\$ @PeterCordes No, even then. As soon as the arc ignites, the field across the terminals collapses to a couple of V/mm. You need the high voltage to ignite the arc - you need current to sustain it. \$\endgroup\$
    – J...
    Oct 19, 2020 at 15:17
  • \$\begingroup\$ Thanks for the reply. It sounds like I might be misunderstanding the application of P=VI; does power dissipation correlate with the resistance/voltage drops? So in this example ~1/2 of the power is dissipated by each speaker and a negligible amount by the wires/switch? If so then I'm even more puzzled by the ratings on the switch body. If the voltage is relevant when the current is not then "xA yVAC" seems like an incoherent format for the ratings and having multiple ratings also seems misleading. I assume it's more misunderstanding on my part, but obviously I don't know what I don't know. \$\endgroup\$
    – slagroom
    Oct 20, 2020 at 1:10
  • \$\begingroup\$ @slagroom The difference between the AC ratings comes from two things. First, ~110V and ~220V are very common mains voltages - selecting these voltages for a rating makes sense because they are the most common voltages people will be switching. The difference between the 240V and 120V ratings is that higher voltage increases the probability and the size of arcs generated when switching a live load. Contact arcing is a main failure mode for relays. Higher voltages create larger, more frequent arcs and do more damage. With lower current loads, the arcs become smaller and briefer. \$\endgroup\$
    – J...
    Oct 20, 2020 at 11:49

During switching there is more arcing at higher voltage than at lower voltage. Switches have a voltage limit, but they also have different current limits at different voltages. There are also wattage limits sometimes, and the rated limits may be different for different loads, depending on if they are resistive, inductive or capacitive.


That switch is for AC mains wiring at either 50 Hz or 60 Hz depending on which is prevailing in your locale. It is not rated for audio at all. So you just have to try it out and see if it works.

In general, lower frequencies are harder on a switch. DC is the hardest. Higher currents are hard on a switch. Higher voltages don't necessarily put a lot more stress on a switch. Some switches have the same current rating at 120 as they do at 240. AC switches shouldn't be used on DC at all unless you are willing to do your own validation testing or the voltage and current are negligible.

The problem with DC is that you get arcing that lasts for a relatively long time compared to AC. In AC, arcs get extinguished when the current passes through zero.

The switch might work for you if you stay somewhere around 5A. I don't see any harm in trying it. If the switch fails after a while you can put in a different one.

  • 3
    \$\begingroup\$ The point that this rating is AC and not DC is very important here. It's not uncommon to see a DC voltage rating >5x lower than the AC voltage rating for the same current. \$\endgroup\$
    – BeB00
    Oct 17, 2020 at 22:12
  • 2
    \$\begingroup\$ Sometimes I include extra information for people who find these questions by search. \$\endgroup\$
    – user57037
    Oct 17, 2020 at 22:52
  • \$\begingroup\$ Thanks for the reply. My concern about trying is that it's very common advice that running a guitar amplified with "no speakers connected" is likely to damage the amplified, and I assume that the switch failing could create a circuit that would be effectively the same. \$\endgroup\$
    – slagroom
    Oct 20, 2020 at 1:14
  • \$\begingroup\$ Well if you want to be conservative, use a switch rated for DC and AC. Make sure the DC current rating is substantially higher than the actual current. Also, there must be pro gear available that allows you to select from multiple guitars. Or you could just use pro audio plugs. \$\endgroup\$
    – user57037
    Oct 20, 2020 at 3:22

Your Answer

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

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