Split the underground cable into 2 circuits

Then you should use 10A for the 10m circuit and 6A for the 50m circuit, but that is already calculated generously and for the 50m almost the absolute limit!!!

The calculation is unfortunately also wrong. With the given values, the voltage drop is twice as high. So 2.66% because the return path is not considered.

You need a 10sqmm underground cable if you want 16A

Just for the sake of clarity and because it continues: 10mm² is quite a statement! And lots of copper! It is mm²

This is intended for later garden lighting/power supply.

Lighting with LEDs hardly draws any power, that will work even with 100m. But what is connected to the "power supply"? The electric lawn mower with just over 1000W is already critical.

The calculation is unfortunately also wrong. With the given values, the voltage drop is twice as high. So 2.66% because the return conductor is not considered.

Finally, I have been waiting for a forum troll like you. If you knew it that precisely, you could have helped instead of whining here. And just for your information, the result is correct because the formula contains 2xl (l=length). Maybe you should have asked what is in the formula before whining. Oh man, people... P.S. Your statement that 100m is correct for three-phase current is unfortunately not correct... because in the formula it is not 2*l but square root(3) or also 1.73*l

Finally, I have just been waiting for a forum troll like you. If you know it so exactly, you could have helped him instead of whining here.
And just for your information, the result is correct because the formula contains 2xl (l=length). Maybe you should have asked what is in the formula before complaining.
Oh man, people, geez...

Now calm down. Sorry if I offended you.

Why am I a troll if I point out a wrong result? Unfortunately, electrical installations often do not forgive mistakes.

[ATTACH alt="Voltage drop.png" type="full"]44238[/ATTACH]Look, you only considered the simple length

P.S. Your statement that 100m is correct for three-phase current is unfortunately not correct.. because in the formula it is not 2*l but square root of 3 or also 1.73*l

My statement referred to the cable length. It is obviously clear that the line-to-line factor must be considered here as soon as it is about THREE-phase current.

If you know it so exactly, you could have helped him instead of whining here.

Yes, it wouldn’t be bad if the OP summarizes exactly what he wants to do again. Then we could help him once more.

Standard LED lighting on 50m 1.5mm² cable is no problem (of course not an unlimited number...) Well pump on 50m 1.5mm² cable better double-check and recalculate (there are very different models from the "pond whirl" to the "C-pipe supplier"). Stronger consumers close to the house to use short cables

Fuses must be correctly sized! You can plan a small sub-distribution adapted to the usage at the point where the cables split.