When a wire melts

A cable carrying far more current than it's rated for heats up, eventually hot enough that the insulation softens and then breaks down. How fast that happens depends on three things only: cross-section, current, and insulation material.

Where the heat comes from

Every current-carrying conductor has electrical resistance, and power dissipates as heat across it:

P = I^2 \cdot R

Double the current and the power dissipated quadruples, which is why a short circuit heats up so much faster than an ordinary overload. Resistance itself isn't constant either, it rises with temperature:

R(\theta) = \frac{\rho_{20} \cdot (1 + \alpha \cdot (\theta - 20))}{S}

$ho_{20}$ is copper's resistivity at 20 °C, $\alpha \approx 0.00393\,/ ext{K}$ its temperature coefficient, $S$ the cross-section. At the same current, a hot cable heats up slightly faster than a cold one: a small, self-reinforcing effect.

The same formula, rearranged

Cable short-circuit withstand follows the adiabatic approximation used in DIN VDE 0100-430 / IEC 60364-4-43. Adiabatic here means: over that short a time, essentially no heat escapes to the surroundings, it all stays in the conductor.

t = \left( \frac{k \cdot S}{I} \right)^2

The material constant $k$ is itself just a compact way of writing the temperature rise from a starting temperature $heta_i$ to a limit temperature $heta_f$ :

k(\theta_f) = 226 \cdot \sqrt{\ln \frac{\theta_f + 234}{\theta_i + 234}}

234 is a material constant for copper, the reciprocal of its temperature coefficient. 226 bundles up copper's density, specific heat and resistivity at 20 °C. Plug in $heta_i = 70\,° ext{C}$ (its rated continuous temperature) and $heta_f = 160\,° ext{C}$ (the short-circuit limit) for PVC insulation, and out comes $k \approx 115$ , the same figure already published as a table value in the short-circuit simulator on the blog. It's the same formula, just solved for $k$ instead of $t$ .

Solve $t = (k(\theta_f) \cdot S / I)^2$ for $heta_f$ instead, and instead of a single withstand time you get the whole temperature as a function of time:

\theta(t) = (\theta_i + 234) \cdot \exp\!\left( \left( \frac{I \sqrt{t}}{226 \cdot S} \right)^{\!2} \right) - 234

That's the curve the calculator below plots.

The calculator

Cross-sectionInsulationCurrent: 300 A

Power dissipation

–

Starts to soften

–

Fully melted away

–

Melting duration

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Adiabatic (no heat escapes to the surroundings), valid for a genuine overload/short-circuit event, not for continuous operation near the rated load, see the text below. θonset = 160°C / k≈115 for PVC is the short-circuit limit temperature from DIN VDE 0100-430 (the same figure used in the short-circuit simulator above, in the blog). θfull is not a standard value, it is a rough assumption of this tool’s own for "nothing recognizable as insulation is left", real PVC has no sharp melting point like ice: it softens, bubbles and decomposes over a range.

Pick a cross-section and insulation, then drag the slider: the curve shows the conductor temperature climbing over time, with two markers for where the insulation starts to soften and where this model treats it as fully decomposed.

At low overload factors, the temperature stays under the softening threshold within the displayed window, and the calculator deliberately shows no result. Something is still happening there, just slower than the underlying assumption covers: no heat escaping to the surroundings stops holding once an overload runs for a minute or more. In reality, the cable sheds some of that heat through the insulation, the air and its surroundings, and settles at a lower, stable temperature instead of heating up forever.

What about cable length?

A longer cable should handle a short circuit better, the thinking goes: more copper to soak up the heat, more surface to shed it. It doesn't. $t = (k(\theta_f) \cdot S / I)^2$ has no length term at all, on purpose.

Every metre of cable carries the same current, has the same cross-section, and starts at the same temperature, so every metre heats up identically and in parallel. What happens to the metre 10 cm from the fault has no bearing on what happens 10 m away, because heat generated inside the conductor escapes radially into the insulation and surrounding air, not sideways along the copper to some cooler stretch of cable. A 1 m run and a 100 m run of the same cross-section reach the exact same temperature under the exact same overload.

Even if heat could travel along the cable, a short circuit doesn't last long enough for it to matter: copper's thermal diffusivity puts the distance heat can conduct axially in a second or two at roughly 1–2 cm, nowhere near enough to reach a cooler part of the run before the fault clears (or the insulation is already gone). That's the same reason the adiabatic assumption behind this whole calculation holds in the first place.

The same logic applies to the ampacity table further down: it's a per-metre rating too, which is why it doesn't have a length column either. Cable length matters for other things on a real installation, voltage drop and total power loss chief among them, just not for the temperature this calculator works out.

What "fully melted away" isn't

The 160 °C onset-of-softening figure is straight from the standard. The 300 °C for "fully melted away" isn't, it's a rough call of this tool's own for the point where nothing recognizable as insulation is left. PVC has no sharp melting point like an ice cube. It softens, deforms, and starts breaking down chemically, releasing hydrogen chloride among other things, long before anything melts in the sense of turning liquid. The second figure in the calculator is best read as an order of magnitude, not an exact measurement.

Why this doesn't happen in practice

This calculation describes what would happen to the cable on its own. It's exactly why circuit breakers exist: a properly sized B, C or D breaker trips on a real short circuit long before the cable behind it feels anything from this curve. The short-circuit simulator in the other post pits breaker and cable against each other directly. Here it's just the cable by itself, without the protection that actually covers it in real operation.

How this compares to the standard table

The calculation above answers a different question than the table most electricians actually plan from: a cable's continuous current rating (ampacity, $I_z$ ) in normal operation, with real cooling through the wall, conduit or air, depending on how it's installed.

Cross-section	A2 · Sheathed cable in a thermally insulated wall	B2 · Sheathed cable in conduit or trunking	C · Sheathed cable on or in the wall	E · Sheathed cable in free air, cable tray
1.5 mm²	13 A / 13 A	15 A / 13 A	17.5 A / 16 A	18.5 A / 16 A
2.5 mm²	17.5 A / 16 A	20 A / 20 A	24 A / 20 A	25 A / 25 A
4 mm²	23 A / 20 A	27 A / 25 A	32 A / 32 A	34 A / 32 A
6 mm²	29 A / 25 A	34 A / 32 A	41 A / 40 A	43 A / 40 A
10 mm²	39 A / 32 A	46 A / 40 A	57 A / 50 A	60 A / 50 A
16 mm²	52 A / 50 A	62 A / 50 A	76 A / 63 A	80 A / 63 A

Each cell: ampacity I_z / largest circuit breaker with I_n ≤ I_z.

Copper conductor with PVC insulation (70 °C), three loaded cores as in a three-phase circuit, 30 °C ambient, no grouping. Installation methods A2, B2, C and E as in DIN VDE 0298-4. A single-phase circuit loads two cores; the standard gives slightly higher values for that, e.g. 18.5 instead of 17.5 A for 2.5 mm² in A2.

Source: DIN VDE 0298-4 as reprinted by manufacturers (2003-08 edition). A comparison with the 2013-06 edition is still outstanding.

That's the same principle behind every circuit breaker: the protective device's rated current $I_n$ sits at or below the cable's ampacity $I_z$ , so the breaker trips before the cable overheats for good, not after.

The calculator above deliberately models the case where that no longer holds: a fault current far above these table values, briefly, with no heat escaping to the surroundings. Take a 2.5 mm² PVC cable in conduit (method B2): the table gives 20 A continuous, protected by a 20 A breaker. Push 500 A through it instead, say during a solid short circuit, and the calculator above shows softening well under a second later.

For aluminium conductors, go at least one cross-section size up from the copper equivalent, and above 70 mm² closer to two sizes because of the larger voltage drop. All figures without guarantee, real installation conditions (bundling, ambient temperature, insulation) can differ and need checking case by case.

Simplified physical model for illustration, not a basis for real cable sizing. The adiabatic approximation only holds for short, severe overloads; the decomposition temperature is a rough assumption of this tool's own, not a standard value. Selection, protection and sizing of a real installation belong in the hands of a qualified electrician.