Assuming the recipe requires exactly 32 EU/t, you could use a battery buffer with two batteries in it and run 2x tin cable (or 4 batteries and 4x cable). I'm sure that would work.
I'll try that soon, thanks. I also overlooked that I need molten glas for the recipe, so no wonder that it wouldn't work
However, I always have trouble understanding the energy system. The recipe needs 32 EU
voltage and 1 Amp
current netting an
electric power (P=V*I=32 EU * 1 1/t) of 32 EU/t. Losses are applied to the voltage (package size) like in real physics as I undertand. A package loses x EU per block of voltage. In case of a 2x Tin-cable 1 EU/block. If I connect my battery buffer with one 2x Tin-cable to the assembler, why would that work? The voltage should drop to 31 EU at the machine or not? But the recipe would still work? Therefore the voltage of the recipe must not be met to carry it out, only the power requirement which can be achieved with higher current?
Second question:
If I put 16 batteries in a LV battery buffer, that buffer would be able to emit 16 amps. If I connect a machine which needs only one amp, would the other 15 amps be wasted during operation or would the buffer only emit what is needed by the consumers like in the real world?