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I presume that by lamp cord you mean the cord which goes from the cieling to the lamp.

Or from a socket to a lamp or any other small appliance (TV, radio, tape recorder)
Couldn't find a good word.

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Are you so sure that a few seconds of 100A on a 0.75mm2 will make that there is no trace left of the lamp cord?

No, it was taking it too far. Evaporating the copper will take a huge amount of energy. (Copper melts at 1100°C, I think)

It is however easy to destroy the PVC insulation.

The maxium allowed operating temperatur for PVC is 70°C.

The maximum allowed short-circuit temperature of PVC is 160°C.

Thus there is 160-70 = 90°C of cable heat up available to handle a short-circuit.

How hot will a 0.75mm2 wire be if it is exposed to 100A for one second?

For copper, the specific heat is 0.39 kJ/(kg*K)

The density of copper is 8.9 x 10^3 kg/m3

The resistivity of copper is 1.7*10^-8 ohm*m2/m

I assume there is no change of state in the PVC and therefore neglect the heat absorbed by the insulation. (This can be justifed by doing the full calculations but I'll jump this part.)

I also assume the heat up is adiabatic, i.e. without any heat exchange to the surroundings.

energy = power x time
power = voltage x current
voltage = resistance x current
resistance = resistivity x length of wire/cross sectional area

Thus:
energy = resistivity x (length of wire/cross sectional area) x current^2 x time { * }

energy = increase in temperature x mass x specific heat

mass = density x length x cross sectional area

Thus:

increase in temperature = energy/(density x length x cross sectional area x specific heat) = {inserting * } = (resistivity x (length of wire/cross sectional area) x current^2 x time)/(density x length x cross sectional area x specific heat) = (resistivity x time x current^2)/(density x cross sectional area^2 x specific heat)

With numbers:

Temperature increase = 1.7x10^-8 x 1 x 100^2/(8.9x10^3 x (0.75x10^-6)^2 x 0.39x10^3) = 87 °C

We get an increase in temperature of 87°C, which is just within the 90°C limit. Hence, the cord survived. But had the breaker taken two seconds to trip it wouldn't.

>Remember that the lamp cord is
>relatively short.

I know you know the answer to this yourself. Why is a short cord better in a short-circuit situation? Is it:

1.) Because the cord will be cooler despite the same current going through it

or

2.) Because it has a lower resistance, increasing the current and thereby triggering a faster response from the breaker?

>If the lamp cord does melt, then it will be
>at the connecting points (b/c of the higher
>resistance at the connections).

Ah! I've never thought of this. It sounds reasonable.

>What curve do you use in Sweden?

Well - since 10A circuits are the norm (or at least used to be) type C breakers are common to prevent nuisance trips when you start motors (like the vaccum cleaner). D-I-Y:ers always use type C for everything. They would use type D if they could buy them... [Linked Image] Electricans use what is deemed suitable for the circuit, hence often type B.

Most homes have cartridge fuses, and these are now slow-blow. You could buy both fast and slow fuses a few years ago, but since nobody bought the fast ones, they have now disappeared. (Ranger - what is the situation concerning Diazed fuses in Austria?)

Many fuses also feature the "infinte" trip curve. (Copper wire or similar wrapped around the fuse. Or a nail through the fuse) These are so very convenient - never any blown fuses. *sarcasm*

What kind of fuses are used in Belgium?

Sorry Paul - I'm ruining your thread!