# A few words on Wire Sizing and Voltage Drop Calculations

Wire sizing guides leave out a part of the calculation important for figuring out the right size wire to use, particularly when dealing with low-voltage circuits where voltage drop can be a significant fraction of the total.   Standard practice is to use Ohm’s Law to calculate the current needed by a device, and look up that current in an ampacity table.  But that’s only part of the picture.

Let’s take a look at a wire sizing calculation for a 1kW (output) inverter. We’ll assume a 92% efficiency, which means that the inverter actually needs 1087W supplied to its DC terminals. We know that efficiency will drop as the input voltage drops due to heating of various components, but we’ll ignore that for now.

The inverter needs that power input regardless of what voltage it occurs at–as voltage drops, current has to go up. The bigger the wire is feeding the inverter, the lower the voltage drop (and the voltage supplied to the inverter will be higher). Ohm’s Law also reminds us that as current goes up, so does the voltage drop in our wiring. For a fixed-power load, like our inverter, this means that after calculating voltage drop in our wiring, we need to re-calculate our expected current, and re-calculate the voltage drop in the wiring. It’s a necessarily iterative solution, but that’s not a big deal if we’re not using pencil and paper.

Here’s how to do it:

First, calculate the resistance in the wire.  That can be obtained from a wire sizing guide.  A 10AWG wire, 10 feet long (remember we’re concerned with the round-trip wire length) will have a resistance of 0.0108 ohms.

We can calculate the current that will flow to the load if the source voltage were applied to it (as in the case of perfect wiring) using the power formula:

$I=P_{load}/V_{battery}=(1087W)/(13V)=83.6A$

For a 1,000W AC load (remember we’d have 1087W going in), with a 13V source, we’d have 83.6A flowing.

Ohm’s Law can then be used to calculate the voltage drop in the wiring:

$V_{drop}=I R_{wire}=(83.6A)(0.0108 ohms)=0.851V$

So instead of a perfect 13V at the inverter, it would only see 12.15V.  But at 12.15V, we need more than 83.6A to operate our inverter.  Using 12.15V in the power formula, we now get 89.5A.  And at 89.5A, Ohm’s law shows a bigger voltage drop:

$V_{drop}=I R_{wire}=(89.5A)(0.0108 ohms)=0.966V$

Now we’re down to 12.03V at our inverter with 10AWG wire.  See where this is going?  Repeat the process enough times and the result will converge to where the change is insignificant.  If we look at the power lost between the first and last iteration, it would suggest that the losses would be more like 82W instead of 71W in the simpler calculation.  Both are still too high though–dumping nearly a tenth of our battery capacity into heating our wires isn’t a good thing.

The correction isn’t huge, but if the wire is of marginal size, it can become significant–with the calculated power lost in the wiring easily off by more than 10%.  A rule of thumb in the absence of doing the calculation might be to go one size larger than a simple wire sizing table would suggest.