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10-16-2007, 05:55 PM   #1
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## Wattage vs. wire size

Is there a formula for this? as based on 120v system.

10-16-2007, 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

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 10-16-2007, 06:12 PM #3 Member   Join Date: May 2007 Posts: 56 Rewards Points: 75 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?

 10-16-2007, 06:14 PM #4 Electrical Contractor     Join Date: Jun 2004 Location: Newnan GA Posts: 7,027 Rewards Points: 650 Yep! More letters to make the post long enough!!! __________________ "The problem isn't that Hillary Clinton lies. We all know she lies. The problem is that her supporters don't seem to care"
10-16-2007, 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!

 10-16-2007, 06:21 PM #6 Member   Join Date: May 2007 Posts: 56 Rewards Points: 75 Thank-you there, jbfan.
 10-16-2007, 06:22 PM #7 Electrical Contractor     Join Date: Jun 2004 Location: Newnan GA Posts: 7,027 Rewards Points: 650 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) __________________ "The problem isn't that Hillary Clinton lies. We all know she lies. The problem is that her supporters don't seem to care"
 10-16-2007, 06:29 PM #8 Electrician philosopher   Join Date: Aug 2007 Location: Lilburn, GA Posts: 838 Rewards Points: 500 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.
 10-16-2007, 06:33 PM #9 Electrical Contractor     Join Date: Jun 2004 Location: Newnan GA Posts: 7,027 Rewards Points: 650 I took mine in 96 in at the Gwinnett civic center. Good luck. __________________ "The problem isn't that Hillary Clinton lies. We all know she lies. The problem is that her supporters don't seem to care"
10-16-2007, 10:30 PM   #10
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Quote:
 Originally Posted by Andy in ATL Or... lets say I had a 8000watt (or 8kw) heater that operated on 240Volts. I=p/e or 8000/240 or I=33.3Amps
Sorry to ruin the party but what if your 8kW heater is inductive (uses wire coils) & has a Power Factor of 0.8?

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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 10-17-2007, 01:48 PM #11 Electrician philosopher   Join Date: Aug 2007 Location: Lilburn, GA Posts: 838 Rewards Points: 500 You are totally correct.
10-17-2007, 03:32 PM   #12

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Andy's heater had a pf of 1.0.......

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