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Should be a simple yes or no question.

If I attach 20ft of 16AWG cable to a power supply, in theory I should see some voltage drop. Do I need to place a load at the end of the cable to read the actual drop or can I just connect my multimeter to read the end and take that as the voltage available.

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  • \$\begingroup\$ The voltage drop is proportional to the current through the cable. Remember the cable goes there and back if you use Ohm's law to calculate the voltage drop. \$\endgroup\$ – Andrew Morton Nov 29 '16 at 16:32
  • \$\begingroup\$ Without load, you could use a multimeter in "ohm" mode to get the resistance of your cable and deduce the drop from that. Otherwise, if you can only measure voltage, yes you would need a load (with no load, there will be no voltage drop). Now, just to check: are you well aware, however, that the voltage drop will depend on the current through the cable (and therefore on the load)? \$\endgroup\$ – dim Nov 29 '16 at 16:32
  • \$\begingroup\$ y e s . . . . . \$\endgroup\$ – Sunnyskyguy EE75 Nov 29 '16 at 16:50
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    \$\begingroup\$ If you know the gauge and length you can actually calculate it and get pretty good results, so no need to plug different loads on it everytime you want to study a different scenario. \$\endgroup\$ – Wesley Lee Nov 29 '16 at 17:01
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To expand on the comments, a long wire obeys Ohms law, just like any other resistance, so V = IR.

With no load, no current flows. I = 0, so V = 0 and there's no voltage drop.

If you add a load, then a current flows, and the voltage drop becomes non-zero and proportional to the current. It's possible to calculate the drop if you know the ohms-per-metre of the cable and the expected current. Don't forget the cable goes both ways, so the drop is double that of one length of cable.

On AC, you might see a small voltage drop in a long enough cable because there will be some capacitive leakage between the cores, and hence a small current.

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    \$\begingroup\$ Just a comment on procedure: Measure the voltage at both the source, and at the load, with the same meter, with the load connected. Thiis will remove any variation of the source voltage with load, and any meter inaccuracy, from the measurement. \$\endgroup\$ – Peter Bennett Nov 29 '16 at 16:56
  • \$\begingroup\$ ah, it's never a yes or no question. I know the gauge of the cable 16AWG and the current load should in theory 13 amps at 12VDC. So now I have to find the resistance per meter of a copper 16AWG and then I would be set to use Ohms law. right? \$\endgroup\$ – Fernando Nov 29 '16 at 20:41
  • \$\begingroup\$ @Fernando Just try Googling something like "16 awg copper wire resistance" and the answer will pop up (either in feet or metres, take your pick). I reckon it's about 0.17V per metre one way at 13A. \$\endgroup\$ – Simon B Nov 29 '16 at 21:19

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