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I want to test a 20m wire and soldered alligator clips on each end to see if it can carry the needed current I want to use it for (eg 5A.)

Can I just hook it up to the terminals on my bench top (lab style) power supply that has constant current / constant voltage dials, essentially shorting the power supply, but use the constant current mode/control to stop anything bad happening to the wire or PSU by excess current?

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yes, absolutely, that will work well.

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For a power cable, there are easy equations and tables about which current it can carry over which distance.

For example \$U_v = \frac{2*I*l}{y*q}\$

with

 I - current in [A]
 l - length of cable in m
 y - specific conductivity [m/(Ohm * mm²)]
 q - conductor cross-section area [mm²]

to calculate the voltage drop caused by the cable in the DC case

\$U_v = U_{VDC} cos (\phi)\$for the AC case

and for 3 phase AC

\$Uv = U_{VDC} cos (\phi) \sqrt3\$

So you should calculate the theoretical value of the resistance \$R_{cable} = \frac{U_v}{I}\$

Then national & international standards also give you advice which cable (material, diameter, construction) is allowed for a certain application - here already safety factors are included.

Just my electric table book has 15 pages with tables and figures for the most common types of isolated cables and application, mounting styles etc.

To measure an existing power cable, the resistance is one of the key values. The next most important value is isolation resistance.

PS: The setup with the power supply in current regulation mode is just an overcomplicated measure of resistance, or a very complex one where you need to measure the temperature rise (best at different points of the cable) of the cable which then could be used to qualify the maximum current in respect to the isolation material maximum temperature and the ambient temperature where the cable is going to be used.

We still have not talked a word about the clamps and what their rating is.

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  • \$\begingroup\$ Its primarily the solder connections that i want to test, having already had the lead fail once because of this. \$\endgroup\$ – Hayden Thring Aug 30 at 22:31

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