Calculating current load for a switch?

This SPST switch is labelled on the box as "Rated 10A at 125VAC or 6A at 250VAC".

Is there a calculation as to the amperage it will handle at 12V DC?

(update) As JYelton points out, a naive assumption would be that the switch capacity would be a simple function of current flow, implying that the switch would handle 100A or so. Why is this not the case?

• If Ohm's law states that P = I * E, then based on the existing ratings, it is rated for between 1250 and 1500 watts. Using that, one might assume it can handle between 104 and 125 amps at 12 volts! I think a good answer should explain why not. Jan 8, 2013 at 7:19
• @JYelton you're way off! Read Anindo Ghosh's answer as to why. If the switch had a DC rating for 12vDc, it would have been 10A. And I wouldn't feel safe above ~9A. Jan 8, 2013 at 7:49
• @Garret, JYelton is pointing out what the naive assumption would be, and pointing out that a good answer will clearly show why that naive assumption is incorrect. Jan 8, 2013 at 7:53
• @JYelton The wattage rating you computed is applicable across a load, not the wattage dissipated in the switch. The actual heat generated in the switch should be a function of the current flowing through the switch, and the effective resistance of the contacts plus any oxide or other aging-related resistive formation on the contacts. DC will age worse than AC due to a sort of "electroplating" effect that reverses itself each half-cycle. Jan 8, 2013 at 8:08
• @Garret What Mark said; I didn't myself know the calculations, but wanted to add to Mark's question that one can't just try to apply some sort of power rating to recalculate. Jan 8, 2013 at 16:14

To err on the side of safety, read the specifications thus:

Rated to allow up to 10A at up to 125VAC or up to 6A at up to 250VAC. Not actually rated for DC, so you're on your own.

In other words, if at all there is a choice, opt for a DC-rated switch, so you know you are within rated parameters. If, however, that is not an option, read on...

Some of the factors that affect a switch contact rating:

1. Voltage at which breaking the contact will not cause unacceptable arcing / pitting
2. Voltage at which the isolation provided by the switch housing starts getting unsafe
3. Current at which the contacts will not overheat enough to melt or damage the housing
4. AC or DC voltage: AC signals are easier to break, i.e. less contact pitting or arcing, since the potential difference falls to zero twice per cycle.

Thus, when used for DC, I prefer to assume 10% of the highest rated AC voltage, while keeping the current rating the same as the lowest rated current for the AC specifications.

For this particular switch, 6 Amperes at 12.5 Volts DC would not trigger a paranoia attack.

To address the update to the question:

The heat generated within the switch is a function of the current flowing through it, and the sum of its contact resistance and any other resistance (solder junctions, oxide build-up et cetera). Calculating the wattage by P = V x I for the voltage rating of the switch is invalid, since that voltage is not seen up across the contacts of the switch (except momentarily during making / breaking of contact).

A better computation basis would be P = I^2 x R.

As the power dissipated for a given current through a given resistance is equal for two currents of equal RMS value, and AC voltage is typically expressed as its RMS value, the heat generated within the switch would be equal for AC and DC cases at the same current.

However, contact resistance over the expected lifetime of a switch will increase, more so for DC than for AC: The contacts tend to show an effect somewhat like electroplating / metal sputtering, as electricity flows through them. With AC, this electroplating-like effect is reversed at each half-cycle, so the deterioration over time is less than for DC, where one of the contacts will build up a deposit.

Other factors increasing contact resistance, such as oxidation, humidity related effects and airborne contaminants, are nominally equal in AC and DC cases - Actually AC will marginally reduce such effects as well.

One last factor to keep in mind: Plasma formation during contact breakage may cause "spot welding" like effects to bond the contacts closed (shorted); this is more prevalent in DC, since AC has those two zero-crossings per cycle which break the arc.

For reference here is an example of the ratings for a switch... Arcolectric 1350 High Inrush Rocker Switch

http://www.arcolectric.com/pdfs/catalogue/pages/P028-031%7C1550+1350-High-Inrush-Switches.pdf

These can help choose the right switch for your application. (the "hp" refers to Horse Power for a motor switch)

Anindo Ghosh's answer of ~6A @ 12v is spot on! But I just wanted to add, don't trust radio shack quality stuff to be rated correctly!!! (I worked there when I was a teen, I'm very familiar with this stuff, etc. ) be on the safe side as much as possible!

If that switch was DC rated, it would most likely been given a 10Amp @12vDc. And it perhaps could safely handle 9Amps at the most, (DON'T RISK IT TRY TO STICK TO 6)

An example of horrible ratings, etc. I googled radioshack relay, this is from their approved datasheet for the first relay that came up!

Notice how the contact capacity states:

60 A 14 VDC Resistice.

and Max switching current states:

120 A 14 VDC.

The actual relay (I didn't look it up, but I'm sure it's an automotive relay meant for DC, usually rated @ 40~60aDc (and they do not last very many cycles if used for ~>20amps.)

There is a huge difference between AC and DC current ratings. The type of load makes a big difference.

If your switching a resistive load then it is simple. However if the load is capacitive or inductive it can be very agressive.

Also note for DC the number of operations on an inductive load is reduced (It's ok for once or twice a day for 8 years).

For reference here is a snip from the Panasonic relay datasheet. (The same data is available for switches):

@Anindo Ghosh, Feel free to add this to the answer if it contributes as a cited example, I don't feel I have the right to change yours as I'm still a noob.

• Welcome to EE@SE (Electrical Engineering at Stack Exchange,) Thank you for contributing! However your statement, "If your switching a resistive load then it is simple. However if the load is capacitive or inductive it can be very aggressive." needs a lot of clarification and details. And the datasheet and part you posted is not what the OP was asking about. This will most likely end up being down voted and removed (I would have except for the fact you are new.) A downvote is intimidating when you are new, but everyone gets them and it's how the site stays helpful to everyone. Jan 8, 2013 at 14:09
• @Spoon For valuable and relevant additions to answers, one can always propose an edit - so long as it does not change the nature of the original answer. Reviewers can then decide whether the change ought to be accepted or rejected. That's what makes this site what it is. Jan 8, 2013 at 16:09

To further confuse the issue, with AC you use RMS (root mean square) to calculate current and voltage with the emphasis on mean or average so that current will spike higher than the rating suggesting that a 10A switch at least momentarily will be handling more like 12-13 A at times, but has a cooling period to where it drops to 0 then reverses direction. based on that fact I would say you should try to stay at or under half the rating as DC has no cooling time as it has no cycles as AC. I actually came across this looking for this due to a momentary switch for a slide out on a camper has a 15a@125v rating and I wanted to be sure it would handle about 5A to a solenoid that would be replacing the relays it ran before (12 v relay coils pull maybe 10% of what a solenoid coil will.)