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Driving current.

Driving current.

Driving current.

(OP)
if i have a coli of 242kW on a 110V Dc system...how much current is require to energize the coil?...i think a current of 2.2A would energize the cirucit....

 

RE: Driving current.

If you're asking how much current you need for 242kW at 110V DC then the answer is 2200A. If you're asking something different then I'm not sure what you're asking.
  

----------------------------------
  
If we learn from our mistakes I'm getting a great education!
 

RE: Driving current.

I get the same answer as Scotty, 2200 Amps.

Bill
--------------------
"Why not the best?"
Jimmy Carter

RE: Driving current.

"k" ????? ?????? ?????????? ?????????? ????????????????????????????
????????????? ???? ??????????????????????????????????? ??????????
?????????????????????? ????????????????????????????????????????
??????????? ???????? ?????? ?????????????????? ???????????????????
?????????

Keith Cress
kcress - http://www.flaminsystems.com

RE: Driving current.

Maybe he is getting confused with AC single phase/polyphase and DC systems.

When it is DC then there is no Root 2 or root 3 factors, no instantaneous, or avg power to worry about.

Assuming it is a continuous load at steady state, then the Wattage is nice and easy at
W = V*I.
 

RE: Driving current.

If kW should be W, then his result is correct. We're just being jerks!

Bill
--------------------
"Why not the best?"
Jimmy Carter

RE: Driving current.

Or if 'A' should be 'kA'?
  

----------------------------------
  
If we learn from our mistakes I'm getting a great education!
 

RE: Driving current.

I concede.
My assessment of the technical possibilities of this application was incomplete.

Bill
--------------------
"Why not the best?"
Jimmy Carter

RE: Driving current.

I have done a rather thorough research on the subject and I did come up with an alternate answer that I believe can be of some interest. If we assume that m means "milli", M means "mega" and A means "ampere", then the following expression also holds:

I [mA] = P [kW] / U [MV]

Please note that this is a preliminary result. It may need to be checked in a model circuit.

Gunnar Englund
www.gke.org
--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

RE: Driving current.

We are all naughty. We must behave ourselves in public!
  

----------------------------------
  
If we learn from our mistakes I'm getting a great education!
 

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