What happens when current splits to different consumers?
It's really quite interesting. I was experimenting with it for a while, to make automatic timers for reactors without extra mods.
You can find how it works yourself, btw - it could be really fun. Was for me. Here is what I know at the moment.
When there is more than one possible target for EU packet at the moment, it splits. Not alternating between target, not choosing at random, just physically splits.
It splits evenly between all possible targets. Or so i thought. and it is true - as long as you use only optic fiber with length up to four meters. I was using fiber for test stands, so has not noticed this detail until I started to replacing it with copper and gold to lower costs of production model.
When I replaced some parts of fiber with copper, currents was suddenly completely jumbled. So I started simple experiments - splitting current from one batbox to two others through different wires. It was quite apparent that split ratio is very close to total resistance of cable, i.e. if we have 1m copper and 2m copper (or 1m gold) then ratio should be 2. But it was is 2.1 - close, but not exactly. Fiber (anywhere between 0 and 4 meters) and 1m copper is 4.16666. Fiber and 1m gold is 9.33333
Suspicious numbers. I was very excited when I solved this thing - all these ratios (21/10, 25/6, 28/3) adds up to 31. So, packet is divided to integer sized packets, round down. Not sure what happens to remainder, but it seems to not be lost completely.
So, complete (?) rule of splitting is this: when there are many possible takers for a packet, then it is split to smaller integer packets. Ratio is inversely proportional to resistances (i.e. supposed eu/t loss) of pathes to targets, with bottom cap of 0.1 EU/t. So, fiber and tin up to 4 meters, as well as connection without wires (side against side) are completely equal.
This can be used, beside other things, to downgrade current all the way to 1 eu packets. 1eu can be made without losing energy, by merely a three-ways split, two of which are 1-4m fiber and last (1eu one) is 4m copper.
Luminators can be charged from BatBox by copper tin, but only if there is always at least six (or exactly zero) luminators and machines (or storages or transformers) consuming energy from same source each tick. This can be easily done by adding five transformers to the wire, that transport energy back to the source BatBox.
Any number of 100 Eu/t reactors can be fed directly to two MFSU - unless MFSU fill up. Useful if you run reactors in microcycles.
And if there is a very long cable, it's better if there is only one storage taking energy from it most of time, because otherwise losses are doubled or more.