Use one designed for low voltage DC such as Eaton KD1 or this automotive type. A better option is also a Raychem Polyswitch which self resets when the current is removed. The problem with the 230V ones is the high trip current, have you really got more than 16 or 25 amps available from your 12V supply?

Edit:
I didn't realise this related to an earlier question where you are talking about an automotive battery. In that case, yes, many amps are available (as opposed to a wall wart), and Andrew's comment applies - a breaker designed for AC should not be used on DC because it is a harder job to seperate the contacts on a DC current than on AC where the magnetic field falls to zero repeatedly due to the waveform.

MCBs (and fuses) have a characteristic curve showing time to trip against overcurrent. At the rated current, they should last indefinately, but as that is exceeded they trip faster. For a fast trip, you need the source to be capable of many times the rated trip current.

In your edit, what's missing is that the rate of cooling will depend on the temperature. In general, the cooling rate will increase as the temperature increases. When the temperature rises enough that the cooling rate matches the heating rate, the temperature will stabilize.

But the actual cooling rate is very difficult to calculate. It depends on what other materials the copper is in contact with (conductive cooling), the airflow around the conductor, etc.

As an added complication, the heating rate will also depend on temperature, because the resistance of the copper will increase at higher temperatures.

So without much more detailed information about your conductor and its environment, its not really possible to give a precise answer to your initial question, how hot will it get?.

As for the second question, how fast will it heat up if there's no cooling, you can calculate that from the heat capacity of copper, which Wikipedia gives as 0.385 J / (g K), or 3.45 J / (cm^3 K).

## Best Answer

There are some topics you should pay attention concerning the cable thickness selection: the voltage dropout, cable heating, electro-magnetic interference, impedance matching.

The maximum allowed voltage dropout is a good parameter to calculate the cable thickness. In most situations this is the limitation factor and may define alone (without the need to calculate the other factors) the cable thickness.

\$ V_{drop} = R_{cable} \times I \$

\$ R_{cable} = \dfrac {p \times length}{CrossSectionArea}\$

Where p is the material electrical resistivity

For a round cable:

\$ R_{cable} = \dfrac {p \times length}{\pi \times radius²}\$

Joining the equations we have that the thickness should be at least:

\$ radius >= \sqrt{ \dfrac {p \times length \times I}{V_{drop}\times \pi}}\$

Assuming that a voltage drop of 0.5(v) is allowed and you'll be using a 2 meters copper cable (p=1.68×10−8) for a maximum 16(A) current:

\$ radius >= \sqrt{ \dfrac {1.68×10^{−8}\times 2 \times 16}{0.5 \times \pi}}\$

\$ radius = 0.59 mm \$

\$ diameter = 2 \times radius = 1.18 (minimum) \$

From this AWG table we have that a cable of AWG-thickness of 11 has a diameter of 2.3mm, that is enough for your application.

The table also states that this AWG 11 can carry over 12(A) for what they call "Maximum amps for power transmission". This is a

lowernumber than the one we calculated, but we should pay attention that this is a general advice for any cable length, not for a specific 2 meters cable (which I supposed in this calculation).The cable heating is a little harder to calculate. You'll need information on heat dissipation characteristics of the environment and the cable: temperature, air flow, sunlight incidence, cable thinness, cable material, insulator layer and even the cable color (black cables heat up faster but also cool faster).

In general the cable insulation has a good thermal protection (around 80 Celsius degrees) which is enough for most cases. You should pay attention only if your cable will go through some heat source. In this case a thicker cable with a better insulator can stand up better.

Unfortunately there is differences among AC and DC current for cable selection. If you're going to transport AC current, a thinner cable may limit the maximum frequency you can carry, above which the signal may start degrading due Skin Effect as @TokenMacGuy noted . Lucky this is not the case with DC.

The last topic I said that could interfere with the cable thickness selection is the impedance matching. For more precise circuits, the resistance from the cable can make a huge impact in the power transmission. This problem arises mainly when there is a huge current, more than 100 (A), or the load resistance is really low, as speakers with 8 or 4 ohms. In this case a thicker cable with a lower resistance is desirable.