# Fuses

Discussion in 'Electrical Engineering' started by Den, Nov 3, 2004.

1. ### DenGuest

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

2. ### GKNGuest

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.

Regards. GKN.

3. ### Colin PillingerGuest

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.
No! No! Bad!

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!!

4. ### John GGuest

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.

5. ### DenGuest

Folks:

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

Cheers

Den

6. ### RowbotthGuest

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.

7. ### DenGuest

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

8. ### daestromGuest

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* <circuit voltage>. 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.

9. ### Bob PetersonGuest

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.

<snip>

10. ### DenGuest

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

11. ### Garry BellamyGuest

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.

gb

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