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10162007, 05:55 PM  #1 
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Wattage vs. wire size
Is there a formula for this? as based on 120v system.
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10162007, 06:01 PM  #2 
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P=I*E
Where P=power in watts I=amps E=volts Code states. 14 awg=15amps max 12 awg=20 amps max 10 awg=30 amps max Advertisement
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Yes I am a Pirate, 200 years too late. "Jimmy Buffett" Last edited by jbfan; 10162007 at 06:04 PM. Reason: add more info 
10162007, 06:12 PM  #3 
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First, thanks for the reply.
So, If I have this right, then the max. wattage for a 120v, with #12 is: Watts(X)=20*120 X= 2400 is this correct? 
10162007, 06:14 PM  #4 
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Yep! More letters to make the post long enough!!!
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10162007, 06:19 PM  #5 
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Ohms law
Just for fun, lets say I wanted to put 28 recessed cans in my big pimp kitchen and I was going to use 65 watt bulbs.
I=p/e or (28x65)/120 or i=15.17Amps In this case I would have to use 12 AWG CU wire. Or... lets say I had a 8000watt (or 8kw) heater that operated on 240Volts. I=p/e or 8000/240 or I=33.3Amps In this case I would have to use 8 AWG CU wire. Ohms law is extremely cool even for DIYers to wow your friends at cocktail parties with your knowledge of electrical theory! p=watts e=volts i=amps r=resistance If the electrician knows any two of these he/she can solve for the rest. And I was the kid in the eighth grade bi###ing about the fact that i'm never gonna need this algebra crap EVER! 
10162007, 06:21 PM  #6 
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Thankyou there, jbfan.

10162007, 06:22 PM  #7 
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If the electrician knows any two of these he/she can solve for the rest. And I was the kid in the eighth grade bi###ing about the fact that i'm never gonna need this algebra crap EVER!
And I thought that was me!!!! Andy, When you taking the test?(Sorta of topic)
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Yes I am a Pirate, 200 years too late. "Jimmy Buffett" 
10162007, 06:29 PM  #8 
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I've got 7 years resi., one year comm. and I'll sit for the GA test in fall of 08 before GA adopts the 08 code. Thanks for asking! I've become a huge code geek just in the last 18 mos.

10162007, 06:33 PM  #9 
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I took mine in 96 in at the Gwinnett civic center.
Good luck.
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10162007, 10:30 PM  #10  
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Quote:
P(kW) = E(v) x I x Cos phi (this is for single phase power). I = P divided by (E x Cos phi) = 8 000 divided by (240 x 0.8) = 8 000 divided by 192 = 41.7 Amps. When calculating power consumption of equipment, the Power Factor must be considered if the equipment is inductive or capacitive. This formula, P = E x I, is for finding single phase Apparent Power (VA). The formulae for finding True Power (kW) is; P = E x I x Cos phi (single phase). P = 1.732 x E x I x Cos phi (3 phase).
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10172007, 01:48 PM  #11 
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You are totally correct.

10172007, 03:32 PM  #12 
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Andy's heater had a pf of 1.0.......
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