# How to get 64 EU/t? +lead acid battery

So i want to electrolyze some solid MgCl and need 64eu/t but dynamos do some weird numbers, i thought that mb some batbox or transformer can do that but no…

Also, i know that you can electrolyze molten magnesium cloride! If its the only recepie that need 64eu/t than i guess i dont need it that much!

Also: lead acid battery tooltips says that it does not require a canning Machines, but theres no other recepie in nei other than filling it in a canning machine

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A GT battery box can output 64EU/t.
But for early game when you don’t have circuits you can use two dynamos + some fitting wire/cable.
Naked constantan wire has a 4EU/block loss if I remember correctly, so you can get a perfect 12EU loss.
Overhead view would be something like this, maybe. (E)lectrolyser, D(ynamo), x suitably thick constantan wire.

`````` E
x    <- with wire losses you`ll get 2x32EU/t (=64EU/t) after here
xxx
D D   <- Each LV dynamo outputs 44EU/t if fed max RU/t
``````

As for the battery, you can put a stainless steel faucet (or other acid-proof faucet) on a tank containing acid and right click the faucet with the battery part to fill it up manually.

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I actually have circuits! So what are you saying is if i connect batboxes or other sources of power in “parallel” their voltage is combined? If its so than yahooo!

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their voltage is combined

Think of it as several packets of energy…
A single LV battery box with two batteries in it can output 2 packets (amps) * 32EU, which effectively becomes 64EU/t. So if you put a batbox right up against the electrolyser with no cable/wire inbetween (so there’s no cable loss) and use two batteries in it, it can easily power the 64EU recipes.

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Wow! Its sounds pretty easy… Thank you alot!

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