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I've been looking at a device/power-supply pair I was sent recently for a project.

The device specs say it needs 24V and will draw 10A max.

The power supply provides 24V @ 9A (14A peak).

The power supply has 3 posts for V+ and 3 posts for GND (call them V0, V1, V2, G0, G1, and G2 respectively).

The cable that was supplied bundles 14 24 AWG wires, connected into a molex receptacle such that 2 wires go to V0, 2 to V1, 3 to V2, 2 to G0, 2 to G1, and 3 to G2.

I looked at a load-limit table for 24 AWG wire and it says that for chassis wiring, the wire is rated for 3.5A max and for power transmission 0.577A max.

Since the device can draw up to 10A, clearly a single piece of 24 AWG would be too small to be safe in this configuration.

The only conclusion I can draw from this is that using 7 wires across 3 terminals spreads the current load out such that 24 AWG is safe.

My questions, if you've managed to get this far are:

1) What is the calculation to determine the correct minimum number of smaller gauge wires to use in such a setup (i.e. where you're replacing a single larger wire)? Is it simply max current draw / number of wires?

2) How is the voltage drop due to the wire length calculated when using multiple wires (I know how to do it with one wire)?

Thank you!

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    \$\begingroup\$ I might be wrong but your supply seems less than adequate. If your load can pull 10A continuosly you might have problems, moreover using a piece of equipment on the edge of its maximum ratings isn't generally a good idea. \$\endgroup\$ – Vladimir Cravero Aug 3 '14 at 8:26
  • \$\begingroup\$ The device draws 4A average 10A peak and the PS supplies 9A with 14A peak. They're sold together from the manufacturer. \$\endgroup\$ – par Aug 3 '14 at 17:01
  • \$\begingroup\$ Well that's fair enough then. \$\endgroup\$ – Vladimir Cravero Aug 3 '14 at 17:35
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  1. Determine the cross-sectional area of the wire being replaced, and then divide that by the cross-sectional area of the smaller wire in order to determine how many smaller wires you'll need to equal the cross-sectional area of the larger wire. If the quotient isn't an integer, ALWAYS round up to the next largest integer.

  2. Determine the resistance of one wire as you normally would, and then the resistance of the cable will be that of the resistances of the separate conductors all connected in parallel. The voltage drop across the cable will then be the product of the current through the cable and its resistance.

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  • \$\begingroup\$ I was hoping for some insight on current equalization. I mean that the R of each cable will be slightly different, the % variation would probably be quite high though since R is low, so you might end up with one cable carrying too much current. Isn't this an issue? \$\endgroup\$ – Vladimir Cravero Aug 3 '14 at 8:27
  • \$\begingroup\$ @VladimirCravero: If both wires are the same length and same cross section, installed in the same conditions, I would expect no more than 10% imbalance between the pair. Note, in power engineering we frequently run multiple cables in parallel and don't worry about imbalance between the cables. We are more worried about mutual heating effects. \$\endgroup\$ – Li-aung Yip Aug 3 '14 at 9:07
  • \$\begingroup\$ @Li-aungYip so no "safety coefficient" is applied on the number of smaller cables? That was my main concern. \$\endgroup\$ – Vladimir Cravero Aug 3 '14 at 9:12
  • \$\begingroup\$ @VladimirCravero: There is a de-rating factor applied to cables run in close proximity to each other, but as mentioned previously, this is solely to account for thermal heating effects. In my country, AS3008 has tables that give the derating factors for cables bunched together. \$\endgroup\$ – Li-aung Yip Aug 3 '14 at 9:16
  • \$\begingroup\$ Well thanks for your clarification then. As long as you ran your wires far enough 100 1A cables can supply safely 100A then. \$\endgroup\$ – Vladimir Cravero Aug 3 '14 at 9:18

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