Drop across AGW cable

[joniengr081]

I was searching a table for AWG cable resistance and found the following.

https://www.powerstream.com/Wire_Size.htm

The cable AWG 12 is rated up to 9.3 A and having a resistance 5.20864 Ohm/km.

I wanted to calculate the voltage drop for 8 meter long cable which is carrying 6 A current. There is a voltage at source 3.3 V at one end of the cable and the voltage sink ICs on the other end of the cable.

V_drop = I x R
V_drop = (6 A) x (5.20864 Ohm/km x 8 /1000)
V_drop = 0.250 V

The calculation shows 250 mV drop across the cable. Is there anything I am missing ?

 

[RoGeorge]

If the resistance is for one wire, an 8 meter cable has 2 wires, so 16 meters in total.  250mV drop on one wire, and the same drop on the other wire, total 0.5V drop on the cable.  If the power supply gives 3.3V, at the other end of the 8m cable, it will be only 2.8V for a 6A load.  And that is if the wires are made of Copper.  Other materials will be worst.
https://www.rapidtables.com/calc/wire/wire-gauge-chart.html

Same website has a calculator for the voltage drop, too:
https://www.rapidtables.com/calc/wire/voltage-drop-calculator.html

[MarkT]

Is there anything I am missing ?

Well just that you don't need tables if you use cross-sectional area for these calculations.  Resistivity of copper = 1.78 * 10^-8, resistance = resistivity * length / area (all in ohms and metres).

Or a rough and ready method:   18 milliohms * (length in m) / (area in square mm)
Use 35mohm for there-and-back.

[JXstaystonight]

Your calculation is correct, but it seems like you've converted the cable's resistance per kilometer to ohms for the given length of 8 meters, which is why you divided by 1000. However, since the resistance value you have is already given in ohms per kilometer, you don't need to divide by 1000 again.

So, the correct calculation for the voltage drop would be:

V_drop = I x R

Where:

I=6 A (current)
R=5.20864 ohm/km (resistance per kilometer)

V_drop = 6×(5.20864×8)
V_drop = 6×41.66912
V_drop = 250.01472 mV

So, the corrected voltage drop across the cable would be approximately 250.01472 mV. Your initial calculation was correct; you just unnecessarily divided by 1000.