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As the only functionality of a switch is to interrupt the flow of electricity, it acts like a small resistor right?

Lets say that we have a switch that its max ratings are 3A/250V AC. So if we connect it directly to source it will handle up to 3 A * 250 V = 750 W. If I connect it to 12 V DC will it be able to handle around 60 A? (12 * x = 750 => x=750/12 => x= 62.5) How are the voltage / current ratings determined? Will it be able to handle around 30 A instead of 60? Is there any way to calculate the maximum ratings?

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no a switch isn't like a small resistor. A mechanical switch does not consume power. Think about, a switch is an open circuit when the interrupter is open, and is a short circuit when is closed.

A switch labeled as 3A/250V AC means that it can keep 250V when is open. And can keep 3A when is closed. Remember a short circuit has ideally 0v, and open circuit has ideally 0 current.

The same for DC.

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    \$\begingroup\$ A switch does consume power, but that's not the reason for the ratings. \$\endgroup\$
    – Hearth
    Jun 21 at 15:08
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    \$\begingroup\$ It is bad that this has been tagged as the correct answer as it is worded wrongly. Yes, 250V means that it can handle that for AC but the value might be smaller for DC. 3A means that it can BREAK 3A under certain assumptions of longevity (the contacts erodes when breaking current). Some switches can handle only a few hundred breaking actions at the stated current, some allows many thousand. The ratings for breaking AC and DC often is different for switches, generally with a lower value for DC. \$\endgroup\$
    – ghellquist
    Jun 21 at 21:09
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    \$\begingroup\$ (-1) This answer is so simplified (with some wording that is, at face value, wrong) it doesn't even go near answering the question. A real switch (since we are talking about actual ratings of a physical object) is a complex electro-mechanical system that must cope with lots of real-world scenarios (is the load inductive? What's the max fault current that can occur? etc.), \$\endgroup\$ Jun 22 at 18:41
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    \$\begingroup\$ BTW, "The same for DC.": not at all: faults in DC circuits of the same power rating as an AC circuit are much nastier, since the ensuing electric arc is MUCH more difficult to quench. \$\endgroup\$ Jun 22 at 18:45
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Switch ratings are related to the designed breaking capacity: it's much more difficult to break a circuit while current is flowing than it is to simply not impede flowing current while the circuit's closed, and prevent airgap breakdown when the switch is open.

Both current and voltage work together to create and maintain the arc of ionized air (plasma) that you are trying to break with a switch.

  1. As soon as you open the switch and the contacts stop touching, your load voltage will present itself across the switch and, when the contacts are still very close to each other, go beyond the airgap's breakdown voltage (goes up with the contacts distance, initially quite small). In any case, there is always some inductance in your circuit that will increase its voltage until the voltage across the switch goes beyond the air breakdown voltage . Reminder: this is because current cannot change instantaneously in an inductor: V=L*di/dt.
  2. At this time, the air between the contacts will ionize and turn into a plasma channel (fun fact: which has negative resistance, i.e. it conducts more and more with current) that will be bigger and hotter with higher load currents.
  3. That plasma will self-sustain until the contacts are sufficiently far away from eachother that the arc becomes unstable and eventually breaks: the breakdown voltage is already higher than the voltage the inductance can develop in the time it takes the arc to go away, thanks to the now-large distance between contacts. Natural air movement around the arc is usually enough, but you can blow on the arc yourself or with a fan to help (power plants shoot inert gases on the arc).

As you can see, in a nutshell:

  • Higher currents tend to create a more damaging arc - possibly to a point the contacts melt instantaneously
  • Higher voltages tend to maintain the arc for longer - possibly to a point the arc does not break at all

Some engineers only look at the load power vs switch breaking power but based on my understanding summarized above I'd say that's a dangerous assumption.

Last bit:

  • AC inherently stops current midway through the wave frequency, so the arcs are rather easy to break
  • DC never stops, so the switch has to do all the work. Arcs are stronger, for longer, which tends to weld the contacts or at least eat through them at each cycle.

So definitely don't make assumptions about DC based on AC rating.

Addendum: the ratings tell you guidelines to obtain an arbitrarily chosen design lifespan (this is what we call qualification) - don't expect them to last forever even if the ratings are met, or to fail only at that time, because the sparks erode and oxidize the contacts in an intricate way. You should be able to expect about half of those cycles with 99.9% probability assuming a gaussian (normal) distribution of failures, though.

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    \$\begingroup\$ You can see a good demonstration of this here: youtube.com/watch?v=mQpzwR7wLeo Sometimes it helps to see just how big the difference is. \$\endgroup\$
    – Frodyne
    Jun 22 at 10:53
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    \$\begingroup\$ @Frodyne: If you liked this video you'll love this one: youtu.be/Zez2r1RPpWY?t=54. My jaw dropped when I first saw it. The two videos cover different currents so one can see the effect of that. \$\endgroup\$ Jun 22 at 17:36
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    \$\begingroup\$ @MisterMystère Wow! That was really mindblowing! I knew of the difference between the two situations, but I've never seen a more striking comparison! \$\endgroup\$ Jun 23 at 15:25
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When closed, up to 3 A can flow through the switch.
When open, up to 250 V can be blocked by the switch.

You cannot multiply these two numbers because they describe different cases.

When used with 12 V, the maximum current still is 3 A.

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    \$\begingroup\$ If a switch has a 3A rating at 250V AC, it usually won't have 3A rating at 12V DC. \$\endgroup\$
    – Justme
    Jun 21 at 14:04
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    \$\begingroup\$ @Justme is correct, but it's important to note that the reason is a difference between AC and DC, not about the voltages involved. \$\endgroup\$
    – Hearth
    Jun 21 at 15:08
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Switches as relay contacts can become complicated. They are rated according to the electrical and mechanical characteristics. This includes such things as contact material, contact spacing and of course contact size. It is important to note many switches have several ratings. Also if the load is inductive that also has an effect on the rating. Sometimes they lump these into a single rating. The separation of ratings becomes more prevalent on higher capacity switches. You will find many switches have a seperate horsepower rating. With your switch 0 - 3 Amps and 0 to 250 nominal circuit voltage is ok or anything in between these ratings. Both are maximum values not to be exceeded. That switch may also have a DC rating which you did not mention.

A single switch can have several ratings for example: 10A, 400V, AC or 16A, 28V, DC; This indicates the current rating is dependent on the applied voltage. If you plan on replacing this switch the replacement should be rated at least 250 volts and 3A. You can use a switch with either a higher amperage or voltage or both but never less then.

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