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-   -   Wattage vs. wire size (http://www.diychatroom.com/f18/wattage-vs-wire-size-12447/)

 SeanR 10-16-2007 05:55 PM

Wattage vs. wire size

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

 jbfan 10-16-2007 06:01 PM

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

 SeanR 10-16-2007 06:12 PM

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?

 jbfan 10-16-2007 06:14 PM

Yep! More letters to make the post long enough!!!

 Andy in ATL 10-16-2007 06:19 PM

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.:thumbup:

Ohms law is extremely cool even for DIYers to wow your friends at cocktail parties with your knowledge of electrical theory!:whistling2:

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!:laughing:

 SeanR 10-16-2007 06:21 PM

Thank-you there, jbfan.

 jbfan 10-16-2007 06:22 PM

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)

 Andy in ATL 10-16-2007 06:29 PM

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!:wink: I've become a huge code geek just in the last 18 mos.

 jbfan 10-16-2007 06:33 PM

I took mine in 96 in at the Gwinnett civic center.
Good luck.

 elkangorito 10-16-2007 10:30 PM

Quote:
 Originally Posted by Andy in ATL (Post 68448) 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).

 Andy in ATL 10-17-2007 01:48 PM

You are totally correct.:thumbsup:

 Stubbie 10-17-2007 03:32 PM

Andy's heater had a pf of 1.0.......:tt2:

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