I think that it shouldn't be much but I wanted to get a ballpark numerical figure.

We've installed a 2kW solar power unit at our home, which is installed on our rooftop. To convert the DC current into AC current, there's an inverter at the Ground floor. We can take the wires to the ground around in two different ways.

The second way is about 6m longer than the first. My question is: will this 6m additional wiring cause a significant loss of final output? Is there a formula that I can use to arrive at a numerical figure?

What is the loss (in terms of final output) per metre of DC wiring, as opposed to AC wiring? I have no idea about what the resistance ratings and other properties of the wire is (if that is important to determine an answer to this question). Is there anyway we could arrive at an approximate figure?

  • \$\begingroup\$ You need to know the wire resistance to perform any voltage drop calculations. This can be looked up once you know the wire gauge. If you don't know this then measure the conductor wire diameter using a calipers. \$\endgroup\$ – Transistor Jul 7 '18 at 13:06
  • 3
    \$\begingroup\$ Um, 2kV or 2kW ? \$\endgroup\$ – Long Pham Jul 7 '18 at 13:08
  • \$\begingroup\$ How many amps, how many volts, what size cable? \$\endgroup\$ – SomeoneSomewhereSupportsMonica Jul 7 '18 at 13:15
  • \$\begingroup\$ @LongPham 2kW, sorry. @ Transistor, Someone -- I'm not sure. But I'll soon get an idea of it. But until then, is there anyway to approximate this? Or at least know if the difference over the 6 additional meters is going to significant? \$\endgroup\$ – WorldGov Jul 7 '18 at 13:20
  • \$\begingroup\$ I would be surprised if it was over 0.5%, almost certainly not more than 2%. That's at full sunlight; percentage losses will be lower in partial illumination. \$\endgroup\$ – SomeoneSomewhereSupportsMonica Jul 7 '18 at 13:21

My question is: will this 6m additional wiring cause a significant loss of final output? Is there a formula that I can use to arrive at a numerical figure?

There is indeed a formula, but not in quite the sense that you seem to think. Wire has a per-meter resistance which is inversely proportional to the cross-sectional area of the wire, so asking what the voltage drop in a wire is requires that you know both what the wire size (gauge) is and what the expected current is.

And since Googling "wire resistance" will tell you everything you need to know, including links to calculators so you don't even need know the exact formulas being used, it seems that you haven't spent much effort on the question.

Simply saying that you have a 2kW system is pretty useless, since you have not specified the system's voltage (and therefor current). Furthermore, saying that one routing path will add an additional 6 meters doesn't help, since you haven't specified the length of the shorter path. For example, if the short path is 6 meters, the voltage drop on the longer path will be twice the shorter. If the short path is 24 meters, the difference will only be 25%.

Finally, you haven't specified what "significant" means. What is the voltage drop over the shorter path? How much of a greater drop will the inverter tolerate?

TL;DR - You've left out almost everything you need to include if you want an accurate answer.


If using AWG16 which 26 mΩ/m per wire pair and 10A service then 6m additional length causes I^2R loss or 2.6 Watts so that is OK for AC 120/240Vac

But if using AWG10 for low Vdc which is 10.5 mΩ/m wire pair on say 12V*16A =2kW then I^2R = 2.7W which is also OK.

It is essential to know AWG size and V Ir Imax.


You could provide us with the mode or datasheet of your solar panel. 2kW can be hypothetically 2 Amps times 1000 Volts or any other multiplying that results in 2kW.

The overall formula for resistance of cable is rho * L / I. Where rho is resistance per meter of cable, L is the length of the cable and I is current that is flowing through a cable.

Remember the thicker the cable is the more amps you can send through it without heating it, therfore lower losses. Other parameters that you should include in calculations is temperature, if it's open air cable. The resitivity parameter you can find in cable manufacturers' catalogue.

  • \$\begingroup\$ What is the "grith" of a cable? \$\endgroup\$ – Transistor Jul 7 '18 at 16:45
  • \$\begingroup\$ Sorry, I've edited the answer. I meant the thickness of the cable. The bigger the diameter of operating cross-section, the more amps can be sent through the cable. \$\endgroup\$ – QST Jul 7 '18 at 17:32
  • \$\begingroup\$ Yes, cross-sectional area is the correct term for the OP as it, hopefully, points to the linear relationship with area rather than diameter. \$\endgroup\$ – Transistor Jul 7 '18 at 17:38

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