Basic question about electricity

[1kar0s]

I started tekkit and I just want to be sure. Sowy if my english isn't perfect (:

Lets say I have 10 nuclear reactor

connected in one cable(HV cable) then I connect them in one MFSU.

I am losing Energy right? The best way is to put a MFSU for each of them?

Again lets says, Each of my nuclear reactor produce 2048EU IF two reactor are connected with one HV cable to one MFSU( I know the capacity of HV cable is 2048) does this mean my MFSU will receive 2048EU or 4096?

I guess its pretty obvious answer but I am a bit confused

Thanks:W

[Xylord]

You should use a HV transformer to turn the EV into HV right at the output of the reactors, and get the EU's into the MFSU's with glass fiber, that way there won't be any waste of energy.

It will receive 2048EU/t packets, but twice as fast as if there was only one reactor.

[SimpleGuy]

Looks like you are getting confused how individual packets are being produced. For a comprehensive basis on how packets works consult: http://forums.technicpack.net/threads/wip-electrical-engineering-101-power-electricity.156/

Anyway to answer your last question:

Each reactor produces 2048 EU (theoretically), that means that each reactor creates 1 packet containing 2048 EU. If you are using one HV cable (which can handle 2048 EU packets) then your cable will not blow up. However, there are 2 packets of 2048 EU moving in your HV cable (from your 2 reactors). How much does the cable care? Not one damn bit. So both packets will go to your MFSU (each packet losing 1 EU per block travelled. This is your only energy loss assuming no forks of your cable).

Thus if you are using n reactors each producing p power travelling along L length of HV cable with t loss per packet per block traveled into a single MFSU, your MFSU will store:

n*(p-t*L)

EU's of power

You should use a HV transformer to turn the EV into HV right at the output of the reactors, and get the EU's into the MFSU's with glass fiber, that way there won't be any waste of enrgy.

It will receive 2048EU/t packets, but twice as fast as if there was only one reactor.

In his latter question about reactors outputting 2048 EU/t packets, each packet travels one block per IC2 EnergyGrid tick. So changing the size of the packet won't affect its speed (they are independent). However, he would have 4 packets instead of one travelling down his HV cable and thus lose 4 times as much pure EU.

[Xylord]

In his latter question about reactors outputting 2048 EU/t packets, each packet travels one block per IC2 EnergyGrid tick. So changing the size of the packet won't affect its speed (they are independent). However, because he would have 4 packets instead of one travelling down his HV cable and thus lose 4 times as much pure EU.

By the "twice as fast", I meant he would gather energy faster, I didn't even know the EU packets were actually moving, I always thought the process was instantaneous. I learn everyday! ;)

[SimpleGuy]

By the "twice as fast", I meant he would gather energy faster, I didn't even know the EU packets were actually moving, I always thought the process was instantaneous. I learn everyday!

Don't actually quote me on the "moving one block per tick" as it's mostly a grandpa sayin' coming from an elder who saw tick rates slow down due to loopage in the older days of IC2. It may be instantaneous, I don't know the exact algorithm but there is the possibility of slowing down the tick rate of servers by spamming packets everywhere in an energy grid.

Anyway, back on topic, I don't think splitting packets or combining them affects their ultimate travel speed (unless you count the server-side Java bytecode that needs to be executed per packet) but it does affect loss per packet (fewer packets = less overall loss in general).

[1kar0s]

thanks guys,

simpleguy ur guide about electricity look good but i will get confused too much words for me :X