V=I*R

July 3, 2006 4:18 PM

Physics filter: The original fuse for an Ikea lighting system was rated for 6.3A and 250V. After a power surge, I noticed that the fuse had been toasted. Upon visiting 3 hardware stores, the best I could find was a similar fuse rated for 3.15A and 125V. Since V=I*R, it follows that the resistance is identical when the original fuse is compared with the replacement. Does this mean that these fuses are functionally equal?
posted by |n$eCur3 to science & nature (14 answers total)

I made the same substitution in a dimmable Ikea floor lamp. One gotcha is that if the original lighting was rated for 110V or 220V service, making the fuse substitution you are considering limits you to 110V service.
posted by RichardP at 4:57 PM on July 3, 2006

No. Fuses are rated for amperage, not voltage. You can get one that's rated for 6.3A and 125v, however. You can go higher than 125v, but you don't really need too. It's critical that you find the correct amperage for a fuse, otherwise it won't work when it needs to.

Does Ikea have a catalog that you could order one from? Does the fuse have a manufacturer or part number anywhere on it? If so, you could probably work some google-fu and find one.
posted by fvox13 at 4:58 PM on July 3, 2006

Also, note that the original fuse was rated for approximately 1600W, while the new fuse is rated for about 400W.
posted by RichardP at 5:00 PM on July 3, 2006

No, they wouldn't be. The new fuse will blow more easily than the old one. But you can try it, since it won't destroy your equipment.

Here is an oversimplification of what is going on:

When a fuse gets too hot, it burns up and opens the circuit.

The amount of power absorbed by heat in the fuse can be approximated by (i^2)*r. (i is the current through the fuse, and r is its internal resistance. i^2 = i*i)

Let's assume that there is no step-up transformer in use, and you are in the US, which means you are plugging into a 120V system. At 120V, the fuse will burn up with more than 6.3A goes through. The amount power consumed by the fuse (which is turned into heat) is (6.3A^2)*(r ohms). In this case, the voltage rating of the fuse does not come into play, and it would be ok if you found a fuse that was rated for 6.3A and 125V.
posted by rajbot at 5:04 PM on July 3, 2006

Oops, I checked my lamp - I substituted a 3.15A/125V for a 3.15A/250V fuse.
posted by RichardP at 5:04 PM on July 3, 2006

NO!
The current and voltage ratings do not indicate the resistance of the fuse. The amperage indicates the nominal current that the fuse will carry without blowing. The voltage rating indicates what voltage it is rated for *after* it blows.

You can safely substitute a lower amp fuse, at the risk of having it blow prematurely. Substituting a higher amperage fuse, means that the new fuse will not blow when it's supposed to, possibly risking shock or fire.
posted by kc8nod at 5:07 PM on July 3, 2006

Fuses are most definately rated by voltage, and it can be important.

The voltage rating comes in to play after the fuse blows. The 3.15A/125V fuse mentioned above will blow if it carries a current >3.15A. After it blows, the fuse is rated to insulate up to a voltage of 125V. At a voltage greater than that, there's a possibility that the blown fuse will still pass some current (possibly by arcing). This is bad because now the fuse is not protecting you after you dropped your nice Ikea lamp in the bathtub.

So if this is a US appliance, using 120V, you're probably fine. But if this is a 240V appliance (like if it's in Europe) then you should not use a 125V fuse.
posted by kc8nod at 5:20 PM on July 3, 2006

I've made the same substitution in my TAGGEN. The standard bulbs in that lamp are 75W+50W, which means they draw around 1A at 120V, and the fuse has been fine.
posted by mendel at 5:21 PM on July 3, 2006

Absolutely not. Fuses are rated for numerous things, but going above either the voltage or current rating may cause it to blow and be immediately useless, and worse, if you're above the voltage rating, it could cause the fuse to transmit the current regardless.

On preview, I want to emphasize what kc8nod said. Don't ever use a fuse in an environment higher than its rating. Using a fuse below its rating is OK, though.
posted by chimaera at 6:00 PM on July 3, 2006

Hopefully to clarify: you have the right relationship between volts, amps and ohms: V is indeed equal to I * R. But V is not what you're assuming it is.

When the fuse is working, the voltage across the fuse plus the voltage across the device protected by the fuse adds up to the applied mains voltage. In fact the voltage across an intact fuse is very very low (less than 1 volt, generally) because fuses are designed to have a very low resistance; therefore, most of the applies mains voltage appears across the protected device, allowing it to function.

After a fuse blows, its resistance is ideally infinite, and the current through it is therefore ideally zero, regardless of the applied voltage. Because the device protected by the fuse now has (ideally) zero current through it, there is now no voltage across that device.

But the voltage across the fuse plus the voltage across the protected device must still add up to the applied mains voltage; and since there is now no voltage across the device, the entire mains voltage appears across the blown fuse.

If, as other people have noted above, the fuse is not rated to deal with that voltage, Bad Things happen.
posted by flabdablet at 6:15 PM on July 3, 2006

kc8nod: Thank you so much for explaining the voltage rating of a fuse. I'd never considered that fuses would arc after being broken, but it makes perfect sense.
posted by odinsdream at 7:47 PM on July 3, 2006

No. Fuses are rated for amperage, not voltage.

well. this isn't really true, though in most cases you see fuses marked for 15A or whatever. what determines when a fuse blows is the amount of power dissipated in it.

this power is converted into heat, which melts or vaporizes the fusible element. usually this power is expressed in terms of current, given by the equation for power which is P=R*I². to make this substitution you need to know the resistance of the fuse.

the voltage rating is an entirely different animal, and it means "this fuse will safely break a circuit operating at or BELOW the voltage rating." it has nothing to do with how much current draw or power is required to blow the fuse in the first case.

so, what do you do? if your circuit is operating at 125V, then it's fine to use a 125V fuse. if it's operating at 250V, then it's not. in either case, you want to use the same current rating. using a lower one will result in the fuse blowing all of the time (so-called nuisance blows (huh huh)) and using a higher one will put you in danger of fire etc.

in this case, a 3.15A fuse might be okay, but be prepared for it to blow when you turn the lights on. that's not a big deal, but using a fuse rated for below your circuit voltage is dangerous.

more info on fuses here and here.
posted by sergeant sandwich at 8:03 PM on July 3, 2006 [1 favorite]

I really don't think I should be taking this any further, because it has been answered, but..

I was just talking fuses over here, and checking out a data sheet for a related property, I learned that fuses are often specified with a typical melting I^2t. Since the fuse wire has a heat capacity, it isn't so much the power that makes it blow, but the amount of energy you put in (the amount of energy that gets out is negligible, sort of) and I^2*time is proportional to energy.
posted by Chuckles at 10:22 PM on July 3, 2006

Wow, I was pretty sure that my limited knowledge of physics was not cutting it, but I didn't know that I could be so far off. Thanks everyone.
posted by |n$eCur3 at 10:32 PM on July 3, 2006