IMPORTANT:
- It is far far better that a fuse sometimes blows when there was no need for it to, than for it to sometimes fail to blow when a fault condition exists.
Blowing a fuse. The term "blow" will be used here for the fusing of a fuse - the act of melting the fuse wire and breaking the electrical circuit. Terms such as "it blew a fuse" and "why did the fuse blow?" are common here. The term "blow" in this context may be less common in some countries. Using "fuse" which is correct, as in "the fuse fused", is liable to be too confusing :-).
Why do they blow?
Should they?
The purpose of a fuse is to protect equipment and wiring against the damaging effects of electrical faults which cause excess currents, and to disable equipment which is faulty. The fuse "blows" when the current carried exceeds the rated value for an excessive time. The higher the overload the shorter the period before the fuse blows. So, equipment which is meant to "draw" 10 amps but which has a short from phase to ground, so it draws, say, 100 amps, will blow its fuse in milliseconds. But, a piece of equipment which draws say double the fuse's rated value, may take many seconds to melt the fusewire and to blow the fuse. The ratio between trip times(time to blow) and "overload to rated current ratio" vary with fuse design and can to some extent be controlled by the manufacturer. This is a complete subject in its own right, but assume that a fuse will blow "after a while" at 2 x + overload and will blow almost immediately with say 10 x + overload.
A piece of wire can only be so smart ...
Because a somewhat complex task is being carried out by a deceptively simple piece of equipment (ie a piece of wire) and because the fuse is not always optimally dimensioned for the equipment used, the fuse sometimes "blows" when there is no significant or long term fault condition present.
To blow or not to blow ? - that is the question.
Dimensioning & surges.
Assume that a fuse will blow "after a while" at 2 x its rated value then we can expect it to run indefinitely at its rated value.
If we have a household circuit rate at 20 amps and a number of outlets rated at say 10A then it is possible to connect more load that the rated fuse value. If we connect say a 10A fan heater, a 5 amp one bar radiator (maybe in the next room), a 400 Watt plasma TV (about 2A), and some plug in mood lighting at say 1 A or less then all SHOULD be well. 10+5+2+1 = 18A. If somebody then turns on an electric jug rated at say 8A current rises to 26A. More than the 20A nominal value but less than the 2 x 20A = 40A we have said it will blow at. But if the plasma TV is off and is turned on suddenly the power supply input filters amy present a nearly pure capacitive load to the mains. The mains will be at random phase at TV turnon and usually a current spike will cause no problems. But on some random lucky (or unlucky) day the mains may be at the very peak of the mains cycle at turn on. The capacitor may have stored charge of opposite polarity from last turnoff leading to an even greater current spike. Add a possibly high mains voltage (as happens) and some heavy switching spikes from a nearby factory, or even domestic equipment (treadmill, welder, drill, sander, router, planer ...) Then load + capacitor spike + high mains + switching transient may lead to a very high short term load. And the fuse may decide enough is enough and melt. Or may not.
*Unlikely?*Is all the above likely to happen at once?
No. But as reported, the nuisance blowing happens only a few times a year. Ij the order of what is expected.
We could make the fuse rating higher (more amps)!
Yes. That is one solution. But the ability to react to moderate overloads is lost. Along with lack of protection may go loss of insurance, if the insurance loss assessors find a still intact 2 x 20A wire fuse in the smouldering ruins of your workshop.
What blows the fuse is really the power wasted at it. \$p(t)=u(t)\cdot i(t)\$.
But the voltage here is not the supply voltage of the circuit, but the voltage drop across the fuse, which is determined by the current passed through it. \$p(t)=R\cdot i^2(t)\$. R is constant, so it really depends only on the current.
In typical setting, the resistance of the fuse will be negligible compared to resistance of the load, so the current running through it will be determined solely by the load resistance (or impedance in general).
In case of short circuit, there will be no load impedance and current will be limited only by the fuse's filament. In such case, supply with higher voltage will make it blow faster. (\$p(t)=\frac{u^2(t)}{R}\$)
Best Answer
Yes, yes, and almost certainly.
Unless there is a specifically noted "peak current" spec that is higher than the "rated current" spec, you have to assume that the peak current cannot the maximum rated current. 5 seconds is long enough that it can generally be considered "continuous" anyway.
It seems pretty clear that pushing 600A (let alone 1200A) through a connector rated for 300A is way out of spec. Of course, being out of spec doesn't guarantee that the connector will fail... but I wouldn't count on it working.
Also, take a look at the opening time curves on your fuse. Your 600A fuse will take 10 seconds to blow at about 2400 (!!!!) amps. This is typical - fuses never have nice, sharp cutoffs - there is a large range between their rated (continuous) current and the current that opens them quickly.
I don't know how much current your supply is capable of, but if a short circuit produces anywhere between 350A and 2400A, I'd bet on the connector blowing long before the fuse. If your short circuit produces upwards of 4000A, your fuse will blow in around 10mS - but then, your connector will blow much faster as well, so the connector could very easily still blow first.