For #14 -
Voltage drop: 0.35
Voltage drop percent 2.92%
Voltage at the end: 11.65

For #12 -
Voltage drop: 0.22
Voltage drop percent 1.83%
Voltage at the end: 11.78

Neither cases are reason for concern.

:vs_bulb:

 

Hi, I have a 12 volt led circuit. I have a 10 watt led strip but plan to add many more. The power supply is going 40 feet from the actual led strips.

The led strips are each a meter a piece. I am not worried about voltage drop across them.

I am not sure which power supply to get or how a typical off the shelf power supply responds to longer wires. I dont need more than 50 watts at 12 volts. Would a 360 watt power supply be safer? Does it just increase the wattage output in order to make up for the extra length in wire?

Yours is a slightly confused set of questions but it seems to "boil down" to two parts.

1. What is the voltage drop over 40 feet of (pair) cable when supplying 5 metres of LEDs when each metre of LEDs is rated at using 10 Watts (or is it 10.5 Watts?) making a total of 50 Watts (or 52.5 Watts)

2. Will providing a power supply rated at much more than 50/52.5 Watts increase the current in the wire and reduce the voltage drop.


2. To answer the second question first, the answer is
NO ........................ BUT
it is NOT good practice to provide a power supply which will only just supply the required load. You should use a power supply which will supply at least 20% more than the minimum and using a power supply capable of supplying double the load is not unreasonable.
However, going beyond this is likely to be a waste of money, unless you use this increased capacity for other loads at a later date.
Hence, you should use a power supply rated at between 65 W and 100 Watts.

1. The Voltage drop could be calculated from first principles.
#14 AWG wire has a resistance of 2.525 milliohms/foot (0.002525 ohms/foot) - for each conductor.
At a distance of 40 foot (with two conductors) the resistance is
0.002525 * 40 * 2 = 0.202 Ohms

At 12 V, the current for 50 W = 50/12 = 4.1666 Amps.
Hence the voltage drop is = 4.1666 * 0.202 = 0.841666 V

However, if you use http://www.calculator.net/voltage-d...phase=ac&noofconductor=1&distance=40&distanceunit=feet&amperes=4.166666&x=0&y=0
and "plug in" the values
Wire Size - #14 AWG
Voltage - 12 V
Distance - 40 Feet (single set of conductors)
Load Current 4.1666
you will obtain the following result, which includes a little more information : -

Voltage drop: 0.84
Voltage drop percentage: 7.00%
Voltage at the end: 11.16

For #12 AWG the results are
Voltage drop: 0.53
Voltage drop percentage: 4.42%
Voltage at the end: 11.47

Either result will be OK for LEDs.

I used a voltmeter to measure the amp with a long wire and with a short wire. The short wire had amperage all over the place. It would start at .15 amps, which is about right from my calculation ( 1 meter strip of 10.5 watt 5 meter led line) for the strip it was hooked up to, then increase from there to about .50 amps. I hooked the long sting up to it and i barely got a reading at all but the led lights work the same either way.

I was trying to see if the power supply increased the current in the wire with the longer run.

My volt meter is pretty cheap. Like 15 dollars. Could that be the issue?

You cannot (directly) measure Amps with a Voltmeter. What you are using is (probably) a Multi-Meter.
Because readings are varying "all over the place", it is probably Digital.
If so, I strongly suggest that you obtain an equally cheap Analog Multi-Meter, which you will probably find easier to use.