Let's say I have 1m of 2,5mm² copper wire, insulated with H07V-K. The wire carries 40A and environment temperature is 25°C. The wire is suspended in free air. How hot will the wire get? And how can I calculate it for other values?

The use case is a 12V-230V AC converter I want to hook up to my car and determine the minimum wire thickness required. Because it's only used from time to time and not buried in a wall I suspect I can use thinner wires. This is a temporary thing and doesn't need to be up to code.

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    \$\begingroup\$ Not easily. You need simulations for that. It's a butt load of math, even for a single wire. Just use a table. \$\endgroup\$
    – DKNguyen
    Aug 29, 2019 at 14:47
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    \$\begingroup\$ Temperature is one consideration. The other is the voltage drop on that wire along that distance at that current. \$\endgroup\$
    – JRE
    Aug 29, 2019 at 14:48
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    \$\begingroup\$ @JRE: yes, but that's easy :) \$\endgroup\$
    – Christian
    Aug 29, 2019 at 14:52
  • \$\begingroup\$ @Christian Voltage drop is proportional to resistance so it increases with the temperature, rather significantly if you're using something like 200°C as your wire limit (eg. PTFE insulation). \$\endgroup\$ Aug 29, 2019 at 15:31
  • \$\begingroup\$ @SpehroPefhany H07V-K is usually PVC insulated, so the use case can be narrowed down to 70°C, can't it? \$\endgroup\$
    – Ariser
    Aug 29, 2019 at 16:19

3 Answers 3


Is the 40A on the 12V side (480W) or the 230V side (9200W)?

2.5 mm2 wire has a resistance of 6.7 mΩ per meter, which means a voltage drop of 0.27 volts per meter. You probably can't tolerate more than about a 10% drop, which is only a total of 1.2 V on the low-voltage side — therefore, you can only use a total of about 4.4 m of wire (two runs of 2.2m).

Anyway, the power dissipation is (40 A)2 × 6.7 mΩ = 10 W per meter, which is a lot. The actual temperature rise is difficult to calculate; among other things, we'd need to know the thermal conductivity and thickness of the wire's insulation.

I personally wouldn't use less than AWG8 (roughly 8 mm2, 2 mΩ/m) in an application like this, no matter how "temporary" it might be. And don't forget that each connection adds a little resistance, too.

  • \$\begingroup\$ 0.1 watt per centimeter of 1mm or 2mm wide wire. It will get hot, unless you blow air to cool it. \$\endgroup\$ Aug 29, 2019 at 18:13

Amount of thermal energy transferred to copper or "Calculate Temperature rise over time in copper due to current flow"

Must find mass (m) per distance of run of copper wire (in grams) \$\text{Power (W)} = I_{\text{rmsamps}}^2 \times R\text{ }(\Omega)\$

\$\text{Energy (joules)} = P\text{ (W)} \times \Delta T\text{ (seconds)}\$

specific heat of copper (SH) = 0.385 Joules/gram*degree C

\$\Delta T =\frac{\text{Energy}}{\text{SH}} \times\text{mass}\$

  • \$\begingroup\$ unfortunately the carriage returns did not transfer (?). - CR- Must find mass (m) per distance of run of copper wire (in Grams) -CR- Power (W) = I (rms amps) ^2 * R (ohms) -CR- Energy (joules) = P (W) * delta T (seconds) -CR- specific heat of copper (SH) = 0.385 Joules/Gramdegree C -CR- -------formula---------------------- -CR- delta T =Energy/SHmass -CR- ------------------------- -CR- \$\endgroup\$ Mar 1, 2021 at 17:06
  • \$\begingroup\$ You can use math jax. \$\endgroup\$
    – DKNguyen
    Mar 1, 2021 at 17:21
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    \$\begingroup\$ The critical missing thing in your math, however, which is the trickiest part, is the heat lost to the air. Without this to achieve equilibrium, the temperature rises to infinite over time. \$\endgroup\$
    – DKNguyen
    Mar 1, 2021 at 17:29
  • \$\begingroup\$ Adiabatic heat rise is useful in some circumstances (massive one-off pulses, for example), in this case, not so much. \$\endgroup\$ Mar 1, 2021 at 17:51

Well, it really depends on many aspects. But for these small wire sizes, the insulation actually improves cooling as the gains in larger surface area initially offset the temperature difference across the insulation walls. There is an interactive app in analyticalviews.com that shows this. Also the amount of connections does not play a role, as the cable is surely long enough (in diameter units) so that its middle section will not be affected by the temperature of the heat sinks (at the end of the cable)


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