Measuring Voltage Drop - With or without Load?

[dpressm]

I am sending power to some 5v led lights about 50ft from the transformer. I was going to use some CAT6 direct Burial cable that I kept after a contractor accidently cut it (I replaced the wire as I did not want to do an underground splice).

So I plan to use 1.5 pairs (three conductors) for Pos and neg and one pair for the data line. I measured 5.17 v directly out of my transformer and added the 50 ft of wire and got 5.16v at the end of that. I did this with no load (other than the multimeter) on the circuit. Telling me that there would be no significant voltage drop with my planned setup.

But... I remember something about testing with vs without load.

So, my question is: Do I have to repeat this with the LED lights attached???


Second question: I might use a 12v power supply and a buck converter to bring it down to 5v - again do I need to test with a load when dialing in my buck converter?

Thank You in advance

 

[James]

volt drop is proportional to the load on the cable so always with no load, you will get no volt drop.

 

[Avo Mk8]

It might be an idea to use a cable drop calculator on the interweb to tell you the expected resistance of the cable.
I think in cat 6 swa buried cable the copper conductors are around 23AWG, and working in metric that's about 0.3mm csa.
So the round trip length of cable will be 100 feet, and 100ft of 23AWG has a resistance of over an ohm (more like 1.5 ohm if my sum was right - which it probably isn't!)
If your LED lights are going to take much current, you may get a lot of voltage drop. Eg if they consume 1Amp, you get about 3.3 volts at the far end.
The buck converter will help a lot, but I think it's still worth checking the sums if your LED's are going to consume much current.
And with the caveat that my sums might be rather out, especially if your cable has thicker conductors!