# Fuses

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Group:

I'm going to embarrass myself now.

I've been thinking about a problem too much, and now can't see the wood for the trees.

Can you just check this for me - I've been thinking about this too much, and got myself confused.

Let's ignore, for the moment, that many fuses will allow more than the rated current to flow through them.

I think that a fuse limits the amount of *current* through it. It does this buy popping when the power (I^2.R) exceeds a limiting factor - since the resistance of the wire in the fuse is constant, the factor is the square of the current. This means that the fuse will pop at a given current, regardless of the voltage. Despite this, fuses are rated in terms of voltage and current. I assume that the voltage rating is a safety thing - correct?. So, a fuse rated at 250V/2A will pop whenever >2A crosses it, regardless of the voltage - correct?

Where I got myself confused (in two different ways!) was here:

1. A fuse (say) 3A rated at 230V will allow 690W of power to be drawn before it pops. But 690W @ 115V is 6A. Therefore a 3A / 230V fuse is the same as a 6A / 115V fuse; and
2. A fuse (say) 3A rated at 230V, implies the fuse has a resistance of
76.66Ohm. But 76.66Ohm @ 115V is 1.5A. Therefore a 3A / 230V fuse is the same as a 1.5A / 115V fuse.

I sort of know (err hope) that 1 and 2 are wrong but am having difficulty explaining to myself why! Can someone (gently) put me straight.

Cheers

Den

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Den. My own personal opinion is that a fuse will rupture when the stated current is exceeded "regardless" of the voltage that is applied. The speed of rupture is based on how much current tries to pass. ie. if a short circuit occurs downstream then the maximum current would flow rupturing the fuse almost instantaneously. However should the current be just of an overload on the fuses rating then this could cause the fuse to rupture after a period of time dependant on the magnitude of the overload.

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Fuses will operate at the stated current, within manufacturing tolerance, no matter what the voltage is. The voltage rating is just there to describe how big an arc it will extinguish without destroying the holder and to adequately protect the load side.

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Nope, not equal, the 3A will blow at 3A using either voltage. you are calculating power and using V=IR, which dosent apply to fuses, as they are non linear with heating and braking. Cannot model it as a resistor.

3 * 76 = 228 volts, the FUSE drops ALL the voltage?? No way. Only drop about 10% at most.

You can use a 3A/230 in a 115 slot and also use a 3A/115 in the 115 slot and it will work at 3 amps forever. But if you put 6 amps through it, it will pop in 100 ms or so. If you put a 3A/115 into a 3A/230 slot it may ark over. ZAP!!

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1 and 2 are wrong.

The fuse does not have any significant resistance. The 230 volts or 115 volts is almost all dropped across the load device when the current is within the limits. If the current rises high enough to blow the fuse then NO current flows any more and the full line voltage appears across the fuse holder. It is only at this time that the voltage rating is important. (Apart from the insulation that is used to mount the fuse.)

If a 12 volt rated fuse holder was used in a HIGH voltage circuit there is a chance that the insulation will break down most particularly if the fuse is blown. Or an arc could devlop across the fuseable part and not stop the flow.

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Folks:

Thank you for your cogent responses. Of course, it is all blindingly obvious when it's explained!

Cheers

Den

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This one gets my vote.

(The voltage determines the length of the fuse, too, I think - so when it fails the voltage doesn't jump across the holders...)

HR.

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That is generally true but there are other ways to crank up the voltage rating. The standard little 1/4" diameter glass fuse is usually a low voltage device or very small currents. They also make some 250v 25a fuses in the same size but they will have a ceramic barrel and be filled with a white powder like you see in MI cable to put out the fire when they blow.

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Colin

Thanks for the input. Although I accept that a 3A fuse is a 3A fuse is a 3A fuse, and that the voltage rating is safety thing, my problem was that I couldn't articulate (to myself or others) what was wrong with my reasoning in the two erroneous calculations that I made i.e. I know that the following are wrong, but can't see why ... can you explain.

Incorrect #1. A fuse (say) 3A rated at 230V will allow 690W of power to be drawn before it pops. But 690W @ 115V is 6A. Therefore a 3A / 230V fuse is the same as a 6A / 115V fuse; and

Incorrect #2. A fuse (say) 3A rated at 230V, implies the fuse has a resistance of 76.66Ohm. But 76.66Ohm @ 115V is 1.5A. Therefore a 3A / 230V fuse is the same as a 1.5A / 115V fuse.

Cheers

Den

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Den, your confusion is about the voltage rating of fuses. The voltage rating of a fuse is *not* the voltage drop through a fuse. It is the maximum voltage that the fuse can have impressed acrossed it when blown and not allow continued arcing.

Imagine for a moment if you put a 3A/12V fuse in a 4160V circuit. As long as the current stays below 3A, things would seem normal. But 3A at 4160V is

12 480 watts. Yet the fuse doesn't blow at 3A. The problem with such an arrangement will come when the current exceeds 3A. The fusible link will melt and break the connection. But the ends of the link are so close together, the high voltage will just arc across the gap and current will continue to flow. The arc will probably generate more heat than the fusible link did, and the whole thing will start to melt/explode. But a high voltage like that will just keep arcing across the fuse-holder's mounting clips. Doesn't stop the current flow so whatever you're trying to protect with the fuse isn't protected and you've got a charred mess where the fuse was. A fuse rated at 3A and the proper voltage will open at almost the same current and the ends of the link will *not* have a continuous arc across.

A 3A/230V fuse may have a voltage drop across it (when passing 3A through it) of only 1/4 V. Under this operation, it will dissipate only 0.75 watts, while the rest of the circuit dissipates 3A* . If the circuit it is in has 200 V supply, then this fuse will allow 3A * 200V =>

600W of power. If the circuit is only 12V, the power it will allow is only 3A*12V = 36W. But at these voltages, if the link does melt from overcurrent, the voltage is not high enough to create a continuous arc across the gap between the bits of melted link. If a 3A fuse has 1/4V voltage drop across it, it has an internal resistance of 0.25 / 3 = 0.08333 ohms. Regardless of the voltage of the circuit it is in.

Generally, a fuse of a given voltage rating can always be used in lower voltage circuits if the physical size is the same. But not the other way around (never put a low voltage fuse in a high voltage circuit). Using a higher voltage rating sometimes leads to confusion later when someone else comes along and sees a 230V fuse in a 120V circuit and reads a label/schematic that says it is supposed to be a 120V fuse.

daestrom P.S. And in power applications, fuses have a *maximum* current interrupting rating, but lets not go there.

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No. Fuses are not power limiting devices. They limit current. As long as the volatge applied is within the capability of the fuse to operate, the fuse will open above the specifed rating. Does not matter at all if the supply voltage is 120V, 220V, or 25 volts (within the voltage limits of the fuse).

The fuse has close to zero resistance. The load is where the resistance is and what determines how much current flows through the fuse.

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Wow, thank you so much for taking the time to explain so cogently and thoroughly. Now I can see exactly where I was going wrong in my reasoning.

Many thanks

Den

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Remember of course that there are many different types of fuses, anti-surge, slow blow, quick-blow, and motor control fuses which area different thing altogether, fuses are rated at a maximum current over a set period of time.

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